A symbolic power of a + i a tan beside a power of the secant is integrated as an exponential

[Rafael-SOWNet]

Part of #718.

(k sec(c + d x))^(-4 - n) (a + i a tan(c + d x))^n and the rest of Rubi's 4.3.1.2 whose two powers are symbols adding up to a whole number were declined:

integrand	2.5.0	master a5ce1c55	this
(a + i a tan(c + d x))^n/(pe sec(c + d x))^n	declined	declined	172 characters
(pe sec(c + d x))^(-1 - n) (a + i a tan(c + d x))^n	declined	declined	232 characters
(pe sec(c + d x))^(-2 - n) (a + i a tan(c + d x))^n	declined	declined	273 characters
(pe sec(c + d x))^(-3 - n) (a + i a tan(c + d x))^n	declined	declined	335 characters
(pe sec(c + d x))^(-4 - n) (a + i a tan(c + d x))^n	declined	declined	374 characters

pe stands for Rubi's e, which the parser reads as Euler's number. Each answer is differentiated back at six points with the symbols pinned, compared as complex numbers.

What changes. SolveAPowerOfAnImaginaryTangentBesideAPowerOfTheSecant integrates (A + i A tan(z))^n (c sec(z))^m with n + m a whole number k as the exponential it is, A + i A tan(z) being A sec(z) e^(i z) on the real line: in w = e^(i z) it is (2 w/(w^2 + 1))^k w^n dw/(i w). It read the two powers as numbers only. A symbol is read now, the sum simplified to its number, so that (-4 - n) + n is -4: the integral in w is then w^(n - 5) (w^2 + 1)^4, a sum of powers. At a sum of 0 the power of w is written alone, since (w^2 + 1)^0 is 1 only where w^2 + 1 is not 0, and the integral with that condition beside it was declined. A short sum of symbols in the answer is simplified, since the power rule writes w^(n + -4 - 1 + 1)/(n + -4 - 1 + 1) for w^(n - 4)/(n - 4).

The rest of 4.3.1.2's symbolic rows, (k sec(z))^(6 - 2n) (a + i a tan(z))^n and the like, add up to j - n and are not read by this rule: there sec(z)^2 is the product of the two conjugate sums, which leaves a symbolic power of one beside a whole power of the other.

Tests: ImaginaryTangentBesideTheSecantIntegralTest, three rows with a symbolic power, adding up to -4, -1 and 0, compared as complex numbers on both sides of 0 and where the cosine is negative.

Measured first on all of Rubi's 4.3.1.2, at the corpus's 5-second budget, against master b1af529d, the branch's base:

	master	this
solved	328	332
unevaluated	75	70
wrong	0	0
past the budget	29	29

Measured then on the Rubi corpus against master b1af529d:

	master	this
family 0, independent suites (1814)	1775	1775
family 1, 40 a file (1381)	1316	1318
families 2 to 8, sampled (2410)	2315	2314

The harness counts no answer wrong in either.

The 16 problems the two builds disagreed on, pocket and sample together, run again one build at a time: this answers four that master does not, (pe sec(c + d x))^(-k - n) (a + i a tan(c + d x))^n for k from 0 to 3, and a fifth, k = 4, which the harness cannot verify. One that master answered alone ran past the budget here, 6.6.1:24, which takes 23 seconds on either build asked on its own. The other ten are answered by both or by neither.

The suite on the commit measured, 7e709fc3, passes, 14,988 tests, and so does the allocation gate: every gated benchmark allocates what the baseline says. The head here, 360de735, merges master a5ce1c55; the calculus and corpus tests pass on it, 4,350 tests, and the library builds for netstandard2.0.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura