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

 Method (no overloads)

Summary

A power of A + i A tan(z) beside a power of the secant of the same argument, one
of them not whole and the two adding up to a whole number k, integrated as the
exponential it is: A + i A tan(z) is A sec(z) e^(i z) on the real line, so
(A + i A tan(z))^n (c sec(z))^m is a constant on every interval where it is
continuous times sec(z)^k e^(i n z), which in w = e^(i z) is
(2 w/(w^2 + 1))^k w^n, and dz is dw/(i w).

Remarks

sqrt(a + i a tan(x))/sqrt(c sec(x)) is sqrt(a/c) e^(i x/2) wherever the
cosine is positive, and was declined, with every other power of the pair whose exponents
add up to a whole number: a root of a sum with the imaginary unit in it beside a root of the
secant is read by no rule, and e^(i x/2)/cos(x), what the positive sums leave, is
declined as well. Rubi's 4.3.1.2. A symbol for the powers is read too:
(pe sec(z))^(-4 - n) (a + i a tan(z))^n adds up to -4, and the integral in w is w^(n - 5) (w^2 + 1)^4, a sum of powers.
 The constant is not written: the answer is the integrand times the antiderivative in
w over what that antiderivative differentiates back to, sec(z)^k (e^(i z))^n times i and the slope, a quotient whose logarithmic derivative is zero -- both
halves have n (tan(z) + i) z' -- so it is constant wherever it is continuous, and
the answer holds on every interval where the integrand is.
https://github.com/asc-community/AngouriMath/issues/718

























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