Half-odd powers of two conjugate tangent sums are integrated as the exponential they make

[Rafael-SOWNet]

Part of #718.

(a + i a tan(pe + f x))^(7/2) (A + B tan(pe + f x))/(c - i c tan(pe + f x))^(9/2) ran past the budget, with 43 more rows of Rubi's 4.3.3.1 that put both conjugate sums beside A + B tan:

integrand	2.5.0	master a5ce1c55	this
(a + i a tan(x))^(7/2) (A + B tan(x))/(c - i c tan(x))^(9/2)	declined	2,075 characters	271 characters
(a + i a tan(x))^(3/2) (A + B tan(x))/(c - i c tan(x))^(3/2)	declined	531 characters	243 characters
(A + B tan(x))/((a + i a tan(x))^(3/2) (c - i c tan(x))^(3/2))	declined	645 characters	388 characters

With x alone master answers these since #1795. With the corpus's pe + f x, master a5ce1c55 runs past the 5-second budget on all 44 problems this answers, each asked alone, and this branch answers every one of them in under a second.

What changes. a + i a tan(z) is a sec(z) e^(i z) on the real line and c - i c tan(z) is c sec(z) e^(-i z), so two powers of them adding up to a whole number k are a constant on every interval where they are continuous times sec(z)^k e^(i (p - q) z), and in w = e^(i z), beside any function of the tangent, the secant, the cosine or the sine of z, the integrand is a rational function of w times a power of it. SolveAConjugatePairOfImaginaryTangentSums, asked after the rule for one sum beside the secant, integrates there, and returns the integrand times the antiderivative in w over what that antiderivative differentiates back to, as the rule for one sum does; #1791's rule integrates in one of the sums beside a whole power of the other, and with both half-odd neither is. The antiderivative in w is checked in w, where it is the only thing computed: checked at sampled x, the quotient's constant need not be the same on both sides of the points, and right answers were declined that way.

Tests: ConjugatePairOfImaginaryTangentSumsIntegralTest, five rows, with and without A + B tan, compared as complex numbers at real points.

Measured on the 338 corpus problems with both conjugate sums in the integrand, at the corpus's 5-second budget, against master 2aae852a, the base the branch was measured on (bc6de839):

	master	this
solved	287	331
unverifiable on the reals	1	1
past the budget	50	6

The 44 problems the two builds disagreed on, all from 4.3.3.1, run again one build at a time: master answers none inside the budget, and this branch answers all 44, in 0.7 seconds on average with the check and 2.1 at most. The 161 rows of 4.3.2.1 in the pocket are answered alike by both, 158 of them.

The suite on the commit measured, bc6de839, passes, 15,056 tests, and so does the allocation gate: every gated benchmark allocates what the baseline says. The head here, 649b3b8f, merges master e604369d; the calculus and corpus tests pass on it, 4,355 tests, and the library builds for netstandard2.0.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura