a + i a tan under a root beside the tangent's differential is lowered over its conjugate

[Rafael-SOWNet]

Part of #718.

sqrt(a + i a tan(e + f x))/(c + d tan(e + f x))^(3/2) ran past thirty seconds, with sixteen more of Rubi's 4.3.2.1: a root of a + i a tan beside a root of c + d tan. Under u = tan(x) it is sqrt(a + i a u)/((c + d u)^(3/2) (1 + u^2)), and the quotient by 1 + u^2 sent it into a search. But

(a + i a u)(a - i a u) = a^2 (1 + u^2)

so the sum's power is lowered by one over its conjugate, a^2 (a + i a u)^(-1/2)/((c + d u)^(3/2) (a - i a u)): roots of two linears over a linear, which is answered in a second. SolveByTangentSubstitution writes the integrand in u so where a factor is a root of A +- i A u. A constant taken out of the root first leaves the integrand as one reciprocal, (sqrt(1 + i u) (c + d u)^(3/2))^(-1), and that is read as the quotient it is. A whole power of the sum is left as it was: beside the cosine it is a polynomial in the sum, which the rules for the sine and the cosine answer sooner.

integrand	2.5.0	master f8e9734d	this
sqrt(a + i a tan(x))/(c + d tan(x))^(3/2)	declined	past 30 seconds	1,893 characters, 0.4 s
(a + i a tan(x))^(3/2) (c + d tan(x))^(5/2)	declined	past 30 seconds	670 characters, 0.2 s
1/(sqrt(a + i a tan(x)) (c + d tan(x))^(3/2))	declined	past 30 seconds	2,217 characters, 0.8 s
1/((a + i a tan(x))^(3/2) (c + d tan(x))^(5/2))	declined	past 30 seconds	3,333 characters, 1.0 s
sqrt(a + i a tan(x))/sqrt(c + d tan(x))	declined	declined after 14 s	169 characters, 0.1 s

Tests: ImaginaryTangentSumBesideARootOfALinearInTheTangentIntegralTest, the five rows above, each differentiated back and compared as a complex number at six real points.

Measured first on every corpus problem with i in its integrand, 2,253 of them, at the corpus's 5-second budget, against master f8e9734d:

	master	this
solved	2006	2023
unevaluated	51	50
wrong	1	1
past the budget	125	108

The one counted wrong on both is the known 6.1.5 1/(a + i a sinh(c + d x))^(1/2), the harness's own. Seventeen problems are answered here and not on master, all of 4.3.2.1, and none the other way; one of 4.3.3.1 goes from past the budget to an answer the harness cannot check on the reals. On the 2,006 both answer the time goes from 1,784 seconds to 1,685.

Measured then on the Rubi corpus:

	master	this
family 0, independent suites (1814)	1782	1782
family 1, 40 a file (1381)	1341	1341
families 2 to 8, sampled (2410)	2329	2329

The harness counts no answer wrong in either. The eighteen problems the builds disagreed on, run again one build at a time: master answers none, this answers seventeen and the eighteenth past the budget no longer.

The suite passes: 15,153 passed, 13 skipped, none failed. The allocation gate passes: every gated benchmark allocates what the baseline says. The library builds for every target.

Left in 4.3.2.1: a whole power of the sum beside a root, 1/((a + i a tan)^3 (c + d tan)^(3/2)), where the time goes to splitting a fourth power of a linear with symbolic complex coefficients into partial fractions.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura