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SolveInTheImaginarySumOfAConstantAndATangent​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

 Method (no overloads)

Summary

A rational function of tan(z) beside a power of S = a + c tan(z) with
c = ±i a, below the bar or not whole, integrated in S: tan(z) is
(S - a)/c, and since c^2 = -a^2, dS/dz = c sec(z)^2 = S (S - 2a)/c, so
dz = c dS/(S (S - 2a)).

Remarks

Rubi's cot(c + d x)^3 (A + B tan(c + d x))/(a + i a tan(c + d x))^2 and the rest of
its 4.3.3.1 with a power of the tangent beside (A + B tan) over a power of
a + i a tan ran past the budget, whole powers and half-odd ones alike: under
u = tan(z) each is a rational function over 1 + i u and 1 - i u with
symbols in every coefficient. In S the factors are S, S - 2a and,
from a cotangent, S - a, with no imaginary root among them, and the same
integrands are a second or two; a half-odd power of S is a root of a linear.
 Exact wherever the integrand is defined: the substitution is the identity
sec(z)^2 = 1 + tan(z)^2 read in S, and the power of S is the
integrand's own. Only whole powers of the tangent and cotangent of z beside it,
and nothing else in x; a base standing only to positive whole powers is a polynomial in
the tangent, which the rules for those answer.
 Beside a second such sum of the same argument, q - i q tan(z), which is linear in
the first, the one under a power that is not whole is the variable and the other a whole
power of a linear in it: (a + i a tan(z))/(q - i q tan(z))^(3/2), Rubi's 4.3.2.1,
ran past the budget too. A symbol for the power is not whole either:
(a + i a tan(z))^m (q - i q tan(z))^4 is S^(m - 1) times a polynomial in
S, and was declined or past the budget with the rest of 4.3.2.1 and 4.3.3.1 that
write a symbolic power of one sum beside a whole one of the other.
https://github.com/asc-community/AngouriMath/issues/718

























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