Python polys.cancel函数代码示例

def no_cancel_equal(b, c, n, DE):
    """
    Poly Risch Differential Equation - No cancellation: deg(b) == deg(D) - 1

    Given a derivation D on k[t] with deg(D) >= 2, n either an integer
    or +oo, and b, c in k[t] with deg(b) == deg(D) - 1, either raise
    NonElementaryIntegralException, in which case the equation Dq + b*q == c has
    no solution of degree at most n in k[t], or a solution q in k[t] of
    this equation with deg(q) <= n, or the tuple (h, m, C) such that h
    in k[t], m in ZZ, and C in k[t], and for any solution q in k[t] of
    degree at most n of Dq + b*q == c, y == q - h is a solution in k[t]
    of degree at most m of Dy + b*y == C.
    """
    q = Poly(0, DE.t)
    lc = cancel(-b.as_poly(DE.t).LC()/DE.d.as_poly(DE.t).LC())
    if lc.is_Integer and lc.is_positive:
        M = lc
    else:
        M = -1

    while not c.is_zero:
        m = max(M, c.degree(DE.t) - DE.d.degree(DE.t) + 1)

        if not 0 <= m <= n:  # n < 0 or m < 0 or m > n
            raise NonElementaryIntegralException

        u = cancel(m*DE.d.as_poly(DE.t).LC() + b.as_poly(DE.t).LC())
        if u.is_zero:
            return (q, m, c)
        if m > 0:
            p = Poly(c.as_poly(DE.t).LC()/u*DE.t**m, DE.t, expand=False)
        else:
            if c.degree(DE.t) != DE.d.degree(DE.t) - 1:
                raise NonElementaryIntegralException
            else:
                p = c.as_poly(DE.t).LC()/b.as_poly(DE.t).LC()

        q = q + p
        n = m - 1
        c = c - derivation(p, DE) - b*p

    return q

def ratsimp(expr):
    """
    Put an expression over a common denominator, cancel and reduce.

    Examples
    ========

    >>> from sympy import ratsimp
    >>> from sympy.abc import x, y
    >>> ratsimp(1/x + 1/y)
    (x + y)/(x*y)
    """

    f, g = cancel(expr).as_numer_denom()
    try:
        Q, r = reduced(f, [g], field=True, expand=False)
    except ComputationFailed:
        return f/g

    return Add(*Q) + cancel(r/g)

def calculate_series(e, x, logx=None):
    """ Calculates at least one term of the series of "e" in "x".

    This is a place that fails most often, so it is in its own function.
    """
    from sympy.polys import cancel

    for t in e.lseries(x, logx=logx):
        t = cancel(t)

        if t.simplify():
            break

    return t

    def get_diff(self, f):
        cache = self.cache

        if f in cache:
            pass
        elif not hasattr(f, "func") or not _bessel_table.has(f.func):
            cache[f] = cancel(f.diff(self.x))
        else:
            n, z = f.args
            d0, d1 = _bessel_table.diffs(f.func, n, z)
            dz = self.get_diff(z)
            cache[f] = d0 * dz
            cache[f.func(n - 1, z)] = d1 * dz

        return cache[f]

 def find_fuzzy(l, x):
     if not l:
         return
     S1, T1 = compute_ST(x)
     for y in l:
         S2, T2 = inv[y]
         if T1 != T2 or (not S1.intersection(S2) and
                         (S1 != set() or S2 != set())):
             continue
         # XXX we want some simplification (e.g. cancel or
         # simplify) but no matter what it's slow.
         a = len(cancel(x/y).free_symbols)
         b = len(x.free_symbols)
         c = len(y.free_symbols)
         # TODO is there a better heuristic?
         if a == 0 and (b > 0 or c > 0):
             return y

def roots_binomial(f):
    """Returns a list of roots of a binomial polynomial."""
    n = f.degree()

    a, b = f.nth(n), f.nth(0)
    alpha = (-cancel(b/a))**Rational(1, n)

    if alpha.is_number:
        alpha = alpha.expand(complex=True)

    roots, I = [], S.ImaginaryUnit

    for k in xrange(n):
        zeta = exp(2*k*S.Pi*I/n).expand(complex=True)
        roots.append((alpha*zeta).expand(power_base=False))

    return roots

    def _splitter(p):
        for y in V:
            if not p.has(y):
                continue

            if _derivation(y) is not S.Zero:
                c, q = p.as_poly(y).primitive()

                q = q.as_expr()

                h = gcd(q, _derivation(q), y)
                s = quo(h, gcd(q, q.diff(y), y), y)

                c_split = _splitter(c)

                if s.as_poly(y).degree() == 0:
                    return (c_split[0], q * c_split[1])

                q_split = _splitter(cancel(q / s))

                return (c_split[0]*q_split[0]*s, c_split[1]*q_split[1])
        else:
            return (S.One, p)

def solve_undetermined_coeffs(equ, coeffs, sym, **flags):
    """Solve equation of a type p(x; a_1, ..., a_k) == q(x) where both
       p, q are univariate polynomials and f depends on k parameters.
       The result of this functions is a dictionary with symbolic
       values of those parameters with respect to coefficients in q.

       This functions accepts both Equations class instances and ordinary
       SymPy expressions. Specification of parameters and variable is
       obligatory for efficiency and simplicity reason.

       >>> from sympy import Eq
       >>> from sympy.abc import a, b, c, x
       >>> from sympy.solvers import solve_undetermined_coeffs

       >>> solve_undetermined_coeffs(Eq(2*a*x + a+b, x), [a, b], x)
       {a: 1/2, b: -1/2}

       >>> solve_undetermined_coeffs(Eq(a*c*x + a+b, x), [a, b], x)
       {a: 1/c, b: -1/c}

    """
    if isinstance(equ, Equality):
        # got equation, so move all the
        # terms to the left hand side
        equ = equ.lhs - equ.rhs

    equ = cancel(equ).as_numer_denom()[0]

    system = collect(equ.expand(), sym, evaluate=False).values()

    if not any([equ.has(sym) for equ in system]):
        # consecutive powers in the input expressions have
        # been successfully collected, so solve remaining
        # system using Gaussian elimination algorithm
        return solve(system, *coeffs, **flags)
    else:
        return None  # no solutions

#.........这里部分代码省略.........

       None is returned if checksol() could not conclude.

       flags:
           'numerical=True (default)'
               do a fast numerical check if f has only one symbol.
           'minimal=True (default is False)'
               a very fast, minimal testing.
           'warning=True (default is False)'
               print a warning if checksol() could not conclude.
           'simplified=True (default)'
               solution should be simplified before substituting into function
               and function should be simplified after making substitution.
           'force=True (default is False)'
               make positive all symbols without assumptions regarding sign.
    """

    if sol is not None:
        sol = {symbol: sol}
    elif isinstance(symbol, dict):
        sol = symbol
    else:
        msg = 'Expecting sym, val or {sym: val}, None but got %s, %s'
        raise ValueError(msg % (symbol, sol))

    if hasattr(f, '__iter__') and hasattr(f, '__len__'):
        if not f:
            raise ValueError('no functions to check')
        rv = set()
        for fi in f:
            check = checksol(fi, sol, **flags)
            if check is False:
                return False
            rv.add(check)
        if None in rv: # rv might contain True and/or None
            return None
        assert len(rv) == 1 # True
        return True

    if isinstance(f, Poly):
        f = f.as_expr()
    elif isinstance(f, Equality):
        f = f.lhs - f.rhs

    if not f:
        return True

    if not f.has(*sol.keys()):
        return False

    attempt = -1
    numerical = flags.get('numerical', True)
    while 1:
        attempt += 1
        if attempt == 0:
            val = f.subs(sol)
        elif attempt == 1:
            if not val.atoms(Symbol) and numerical:
                # val is a constant, so a fast numerical test may suffice
                if val not in [S.Infinity, S.NegativeInfinity]:
                    # issue 2088 shows that +/-oo chops to 0
                    val = val.evalf(36).n(30, chop=True)
        elif attempt == 2:
            if flags.get('minimal', False):
                return
            # the flag 'simplified=False' is used in solve to avoid
            # simplifying the solution. So if it is set to False there
            # the simplification will not be attempted here, either. But
            # if the simplification is done here then the flag should be
            # set to False so it isn't done again there.
            # FIXME: this can't work, since `flags` is not passed to
            # `checksol()` as a dict, but as keywords.
            # So, any modification to `flags` here will be lost when returning
            # from `checksol()`.
            if flags.get('simplified', True):
                for k in sol:
                    sol[k] = simplify(sympify(sol[k]))
                flags['simplified'] = False
                val = simplify(f.subs(sol))
            if flags.get('force', False):
                val = posify(val)[0]
        elif attempt == 3:
            val = powsimp(val)
        elif attempt == 4:
            val = cancel(val)
        elif attempt == 5:
            val = val.expand()
        elif attempt == 6:
            val = together(val)
        elif attempt == 7:
            val = powsimp(val)
        else:
            break
        if val.is_zero:
            return True
        elif attempt > 0 and numerical and val.is_nonzero:
            return False

    if flags.get('warning', False):
        print("\n\tWarning: could not verify solution %s." % sol)

def simplify(expr, ratio=1.7, measure=count_ops, fu=False):
    """
    Simplifies the given expression.

    Simplification is not a well defined term and the exact strategies
    this function tries can change in the future versions of SymPy. If
    your algorithm relies on "simplification" (whatever it is), try to
    determine what you need exactly  -  is it powsimp()?, radsimp()?,
    together()?, logcombine()?, or something else? And use this particular
    function directly, because those are well defined and thus your algorithm
    will be robust.

    Nonetheless, especially for interactive use, or when you don't know
    anything about the structure of the expression, simplify() tries to apply
    intelligent heuristics to make the input expression "simpler".  For
    example:

    >>> from sympy import simplify, cos, sin
    >>> from sympy.abc import x, y
    >>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
    >>> a
    (x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
    >>> simplify(a)
    x + 1

    Note that we could have obtained the same result by using specific
    simplification functions:

    >>> from sympy import trigsimp, cancel
    >>> trigsimp(a)
    (x**2 + x)/x
    >>> cancel(_)
    x + 1

    In some cases, applying :func:`simplify` may actually result in some more
    complicated expression. The default ``ratio=1.7`` prevents more extreme
    cases: if (result length)/(input length) > ratio, then input is returned
    unmodified.  The ``measure`` parameter lets you specify the function used
    to determine how complex an expression is.  The function should take a
    single argument as an expression and return a number such that if
    expression ``a`` is more complex than expression ``b``, then
    ``measure(a) > measure(b)``.  The default measure function is
    :func:`count_ops`, which returns the total number of operations in the
    expression.

    For example, if ``ratio=1``, ``simplify`` output can't be longer
    than input.

    ::

        >>> from sympy import sqrt, simplify, count_ops, oo
        >>> root = 1/(sqrt(2)+3)

    Since ``simplify(root)`` would result in a slightly longer expression,
    root is returned unchanged instead::

       >>> simplify(root, ratio=1) == root
       True

    If ``ratio=oo``, simplify will be applied anyway::

        >>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
        True

    Note that the shortest expression is not necessary the simplest, so
    setting ``ratio`` to 1 may not be a good idea.
    Heuristically, the default value ``ratio=1.7`` seems like a reasonable
    choice.

    You can easily define your own measure function based on what you feel
    should represent the "size" or "complexity" of the input expression.  Note
    that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
    good metrics, but have other problems (in this case, the measure function
    may slow down simplify too much for very large expressions).  If you don't
    know what a good metric would be, the default, ``count_ops``, is a good
    one.

    For example:

    >>> from sympy import symbols, log
    >>> a, b = symbols('a b', positive=True)
    >>> g = log(a) + log(b) + log(a)*log(1/b)
    >>> h = simplify(g)
    >>> h
    log(a*b**(-log(a) + 1))
    >>> count_ops(g)
    8
    >>> count_ops(h)
    5

    So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
    However, we may not like how ``simplify`` (in this case, using
    ``logcombine``) has created the ``b**(log(1/a) + 1)`` term.  A simple way
    to reduce this would be to give more weight to powers as operations in
    ``count_ops``.  We can do this by using the ``visual=True`` option:

    >>> print(count_ops(g, visual=True))
    2*ADD + DIV + 4*LOG + MUL
    >>> print(count_ops(h, visual=True))
    2*LOG + MUL + POW + SUB
#.........这里部分代码省略.........

#.........这里部分代码省略.........
                            terms.add(erf(sqrt(-M[a])*x))

                        M = g.args[0].match(a*log(x)**2)

                        if M is not None:
                            if M[a].is_positive:
                                terms.add(-I*erf(I*(sqrt(M[a])*log(x)+1/(2*sqrt(M[a])))))
                            if M[a].is_negative:
                                terms.add(erf(sqrt(-M[a])*log(x)-1/(2*sqrt(-M[a]))))

                elif g.is_Pow:
                    if g.exp.is_Rational and g.exp.q == 2:
                        M = g.base.match(a*x**2 + b)

                        if M is not None and M[b].is_positive:
                            if M[a].is_positive:
                                terms.add(asinh(sqrt(M[a]/M[b])*x))
                            elif M[a].is_negative:
                                terms.add(asin(sqrt(-M[a]/M[b])*x))

                        M = g.base.match(a*x**2 - b)

                        if M is not None and M[b].is_positive:
                            if M[a].is_positive:
                                terms.add(acosh(sqrt(M[a]/M[b])*x))
                            elif M[a].is_negative:
                                terms.add((-M[b]/2*sqrt(-M[a])*\
                                           atan(sqrt(-M[a])*x/sqrt(M[a]*x**2-M[b]))))

        else:
            terms |= set(hints)

    for g in set(terms):
        terms |= components(cancel(g.diff(x)), x)

    V = _symbols('x', len(terms))

    mapping = dict(zip(terms, V))

    rev_mapping = {}

    for k, v in mapping.iteritems():
        rev_mapping[v] = k

    def substitute(expr):
        return expr.subs(mapping)

    diffs = [ substitute(cancel(g.diff(x))) for g in terms ]

    denoms = [ g.as_numer_denom()[1] for g in diffs ]
    try:
        denom = reduce(lambda p, q: lcm(p, q, *V), denoms)
    except PolynomialError:
        # lcm can fail with this. See issue 1418.
        return None

    numers = [ cancel(denom * g) for g in diffs ]

    def derivation(h):
        return Add(*[ d * h.diff(v) for d, v in zip(numers, V) ])

    def deflation(p):
        for y in V:
            if not p.has(y):
                continue

def ratsimpmodprime(expr, G, *gens, **args):
    """
    Simplifies a rational expression ``expr`` modulo the prime ideal
    generated by ``G``.  ``G`` should be a Groebner basis of the
    ideal.

    >>> from sympy.simplify.ratsimp import ratsimpmodprime
    >>> from sympy.abc import x, y
    >>> eq = (x + y**5 + y)/(x - y)
    >>> ratsimpmodprime(eq, [x*y**5 - x - y], x, y, order='lex')
    (x**2 + x*y + x + y)/(x**2 - x*y)

    If ``polynomial`` is False, the algorithm computes a rational
    simplification which minimizes the sum of the total degrees of
    the numerator and the denominator.

    If ``polynomial`` is True, this function just brings numerator and
    denominator into a canonical form. This is much faster, but has
    potentially worse results.

    References
    ==========

    .. [1] M. Monagan, R. Pearce, Rational Simplification Modulo a Polynomial
    Ideal,
    http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.163.6984
    (specifically, the second algorithm)
    """
    from sympy import solve

    quick = args.pop('quick', True)
    polynomial = args.pop('polynomial', False)
    debug('ratsimpmodprime', expr)

    # usual preparation of polynomials:

    num, denom = cancel(expr).as_numer_denom()

    try:
        polys, opt = parallel_poly_from_expr([num, denom] + G, *gens, **args)
    except PolificationFailed:
        return expr

    domain = opt.domain

    if domain.has_assoc_Field:
        opt.domain = domain.get_field()
    else:
        raise DomainError(
            "can't compute rational simplification over %s" % domain)

    # compute only once
    leading_monomials = [g.LM(opt.order) for g in polys[2:]]
    tested = set()

    def staircase(n):
        """
        Compute all monomials with degree less than ``n`` that are
        not divisible by any element of ``leading_monomials``.
        """
        if n == 0:
            return [1]
        S = []
        for mi in combinations_with_replacement(range(len(opt.gens)), n):
            m = [0]*len(opt.gens)
            for i in mi:
                m[i] += 1
            if all([monomial_div(m, lmg) is None for lmg in
                    leading_monomials]):
                S.append(m)

        return [Monomial(s).as_expr(*opt.gens) for s in S] + staircase(n - 1)

    def _ratsimpmodprime(a, b, allsol, N=0, D=0):
        r"""
        Computes a rational simplification of ``a/b`` which minimizes
        the sum of the total degrees of the numerator and the denominator.

        The algorithm proceeds by looking at ``a * d - b * c`` modulo
        the ideal generated by ``G`` for some ``c`` and ``d`` with degree
        less than ``a`` and ``b`` respectively.
        The coefficients of ``c`` and ``d`` are indeterminates and thus
        the coefficients of the normalform of ``a * d - b * c`` are
        linear polynomials in these indeterminates.
        If these linear polynomials, considered as system of
        equations, have a nontrivial solution, then `\frac{a}{b}
        \equiv \frac{c}{d}` modulo the ideal generated by ``G``. So,
        by construction, the degree of ``c`` and ``d`` is less than
        the degree of ``a`` and ``b``, so a simpler representation
        has been found.
        After a simpler representation has been found, the algorithm
        tries to reduce the degree of the numerator and denominator
        and returns the result afterwards.

        As an extension, if quick=False, we look at all possible degrees such
        that the total degree is less than *or equal to* the best current
        solution. We retain a list of all solutions of minimal degree, and try
        to find the best one at the end.
        """
        c, d = a, b
#.........这里部分代码省略.........

#.........这里部分代码省略.........
        else:
            result[root] = k

    def _try_decompose(f):
        """Find roots using functional decomposition. """
        factors = f.decompose()
        result, g = {}, factors[0]

        for h, i in g.sqf_list()[1]:
            for r in _try_heuristics(h):
                _update_dict(result, r, i)

        for factor in factors[1:]:
            last, result = result.copy(), {}

            for last_r, i in last.iteritems():
                g = factor - Poly(last_r, x)

                for h, j in g.sqf_list()[1]:
                    for r in _try_heuristics(h):
                        _update_dict(result, r, i*j)

        return result

    def _try_heuristics(f):
        """Find roots using formulas and some tricks. """
        if f.is_ground:
            return []
        if f.is_monomial:
            return [S(0)]*f.degree()

        if f.length() == 2:
            if f.degree() == 1:
                return map(cancel, roots_linear(f))
            else:
                return roots_binomial(f)

        result = []

        for i in [S(-1), S(1)]:
            if f.eval(i).expand().is_zero:
                f = f.exquo(Poly(x-1, x))
                result.append(i)
                break

        n = f.degree()

        if n == 1:
            result += map(cancel, roots_linear(f))
        elif n == 2:
            result += map(cancel, roots_quadratic(f))
        elif n == 3 and flags.get('cubics', True):
            result += roots_cubic(f)
        elif n == 4 and flags.get('quartics', True):
            result += roots_quartic(f)

        return result

    if f.is_monomial == 1:
        if f.is_ground:
            if multiple:
                return []
            else:
                return {}
        else:
            result = { S(0) : f.degree() }

#.........这里部分代码省略.........

        if symbol_swapped:
            swap_back_dict = dict(zip(symbols_new, symbols))
    # End code for handling of Function and Derivative instances

    if bare_f:
        f = f[0]

        # Create a swap dictionary for storing the passed symbols to be solved
        # for, so that they may be swapped back.
        if symbol_swapped:
            swap_dict = zip(symbols, symbols_new)
            f = f.subs(swap_dict)
            symbols = symbols_new

        # Any embedded piecewise functions need to be brought out to the
        # top level so that the appropriate strategy gets selected.
        f = piecewise_fold(f)

        if len(symbols) != 1:
            soln = None
            free = f.free_symbols
            ex = free - set(symbols)
            if len(ex) == 1:
                ex = ex.pop()
                try:
                    # may come back as dict or list (if non-linear)
                    soln = solve_undetermined_coeffs(f, symbols, ex)
                except NotImplementedError:
                    pass
            if soln is None:
                n, d = solve_linear(f, x=symbols)
                if n.is_Symbol:
                    soln = {n: cancel(d)}
            if soln:
                if symbol_swapped and isinstance(soln, dict):
                    return dict([(swap_back_dict[k],
                                  v.subs(swap_back_dict))
                                  for k, v in soln.iteritems()])
                return soln

        symbol = symbols[0]

        # first see if it really depends on symbol and whether there
        # is a linear solution
        f_num, sol = solve_linear(f, x=symbols)
        if not symbol in f_num.free_symbols:
            return []
        elif f_num.is_Symbol:
            return [cancel(sol)]

        strategy = guess_solve_strategy(f, symbol)
        result = False # no solution was obtained

        if strategy == GS_POLY:
            poly = f.as_poly(symbol)
            if poly is None:
                msg = "Cannot solve equation %s for %s" % (f, symbol)
            else:
                # for cubics and quartics, if the flag wasn't set, DON'T do it
                # by default since the results are quite long. Perhaps one could
                # base this decision on a certain crtical length of the roots.
                if poly.degree() > 2:
                    flags['simplified'] = flags.get('simplified', False)
                result = roots(poly, cubics=True, quartics=True).keys()

def is_log_deriv_k_t_radical_in_field(fa, fd, DE, case='auto', z=None):
    """
    Checks if f can be written as the logarithmic derivative of a k(t)-radical.

    f in k(t) can be written as the logarithmic derivative of a k(t) radical if
    there exist n in ZZ and u in k(t) with n, u != 0 such that n*f == Du/u.
    Either returns (n, u) or None, which means that f cannot be written as the
    logarithmic derivative of a k(t)-radical.

    case is one of {'primitive', 'exp', 'tan', 'auto'} for the primitive,
    hyperexponential, and hypertangent cases, respectively.  If case it 'auto',
    it will attempt to determine the type of the derivation automatically.
    """
    fa, fd = fa.cancel(fd, include=True)

    # f must be simple
    n, s = splitfactor(fd, DE)
    if not s.is_one:
        pass
        #return None

    z = z or Dummy('z')
    H, b = residue_reduce(fa, fd, DE, z=z)
    if not b:
        # I will have to verify, but I believe that the answer should be
        # None in this case. This should never happen for the
        # functions given when solving the parametric logarithmic
        # derivative problem when integration elementary functions (see
        # Bronstein's book, page 255), so most likely this indicates a bug.
        return None

    roots = [(i, i.real_roots()) for i, _ in H]
    if not all(len(j) == i.degree() and all(k.is_Rational for k in j) for
               i, j in roots):
        # If f is the logarithmic derivative of a k(t)-radical, then all the
        # roots of the resultant must be rational numbers.
        return None

    # [(a, i), ...], where i*log(a) is a term in the log-part of the integral
    # of f
    respolys, residues = zip(*roots) or [[], []]
    # Note: this might be empty, but everything below should work find in that
    # case (it should be the same as if it were [[1, 1]])
    residueterms = [(H[j][1].subs(z, i), i) for j in xrange(len(H)) for
        i in residues[j]]

    # TODO: finish writing this and write tests

    p = cancel(fa.as_expr()/fd.as_expr() - residue_reduce_derivation(H, DE, z))

    p = p.as_poly(DE.t)
    if p is None:
        # f - Dg will be in k[t] if f is the logarithmic derivative of a k(t)-radical
        return None

    if p.degree(DE.t) >= max(1, DE.d.degree(DE.t)):
        return None

    if case == 'auto':
        case = DE.case

    if case == 'exp':
        wa, wd = derivation(DE.t, DE).cancel(Poly(DE.t, DE.t), include=True)
        with DecrementLevel(DE):
            pa, pd = frac_in(p, DE.t, cancel=True)
            wa, wd = frac_in((wa, wd), DE.t)
            A = parametric_log_deriv(pa, pd, wa, wd, DE)
        if A is None:
            return None
        n, e, u = A
        u *= DE.t**e
#        raise NotImplementedError("The hyperexponential case is "
#            "not yet completely implemented for is_log_deriv_k_t_radical_in_field().")

    elif case == 'primitive':
        with DecrementLevel(DE):
            pa, pd = frac_in(p, DE.t)
            A = is_log_deriv_k_t_radical_in_field(pa, pd, DE, case='auto')
        if A is None:
            return None
        n, u = A

    elif case == 'base':
        # TODO: we can use more efficient residue reduction from ratint()
        if not fd.is_sqf or fa.degree() >= fd.degree():
            # f is the logarithmic derivative in the base case if and only if
            # f = fa/fd, fd is square-free, deg(fa) < deg(fd), and
            # gcd(fa, fd) == 1.  The last condition is handled by cancel() above.
            return None
        # Note: if residueterms = [], returns (1, 1)
        # f had better be 0 in that case.
        n = reduce(ilcm, [i.as_numer_denom()[1] for _, i in residueterms], S(1))
        u = Mul(*[Pow(i, j*n) for i, j in residueterms])
        return (n, u)

    elif case == 'tan':
        raise NotImplementedError("The hypertangent case is "
        "not yet implemented for is_log_deriv_k_t_radical_in_field()")

    elif case in ['other_linear', 'other_nonlinear']:
#.........这里部分代码省略.........