An odd power of the secant over a whole power of a + i a tan is integrated in the sum

[Rafael-SOWNet]

Part of #718.

Nine more of Rubi's 4.3.1.2 rows, which #1812's rule should have taken and did not:

integrand	master b3ac1e8d	this
sec(c + d x)^5/(a + i a tan(c + d x))^2	declined	2,819 characters
sec(c + d x)^9/(a + i a tan(c + d x))^4	past the budget	6,166 characters
(d sec(e + f x))^(2/3)/(a + i a tan(e + f x))^(4/3)	declined	1,033 characters
1/((k cos(c + d x))^(5/2) sqrt(a + i a tan(c + d x)))	declined	1,057 characters

Each is differentiated back and compared with the integrand as a complex number at real points with the symbols pinned; the nine answers run from 888 to 6,166 characters.

What changes. Two things in SolveAPowerOfTheSecantBesideAPowerOfAnImaginaryTangentSumInTheSum.

• It checked each formula it found by differentiating the answer in x, which carries the factor constant on intervals and an antiderivative in u of thousands of characters: large enough that the sampled evaluations could not decide, and right answers were declined after seconds -- (d sec)^(2/3)/(a + i a tan)^(4/3) and the cosine's rows. It checks the formula in u now, on the path itself, u = A + i A t for a real t, which is the claim the answer rests on.
• It left whole powers on both to the rules for the sine and the cosine. They answer cos(x)^5/(a + i a tan(x))^3 and sec(x)^3/(a + i a tan(x))^4 in milliseconds, and not sec(x)^5/(a + i a tan(x))^2. An odd power s of the secant over the sum's n-th with s + 2 n = 1 is taken now: in u it is a whole power of 2 A - u over the root of u, answered in a second or two. Every other pair of whole powers is still left alone, since in u it is answered more slowly than they answer it, or not at all; letting them all through lost five rows the sine and cosine rules answer.

Tests: SecantBesideAnImaginaryTangentSumIntegralTest gains sec(x)^5/(a + i a tan(x))^2.

Measured first on every corpus problem with i in its integrand, 2,253 of them, at the corpus's 5-second budget, against master b3ac1e8d:

	master	this
solved	1978	1987
unevaluated	72	63
wrong	1	1
past the budget	137	137

The one counted wrong on both is the 6.1.5 1/(a + i a sinh(c + d x))^(1/2) the harness counts wrong on every build. Nine problems are answered here and not on master, and none the other way; on the 1,978 both answer the time is 1,839 seconds against 1,838.

Measured then on the Rubi corpus against master b3ac1e8d:

	master	this
family 0, independent suites (1814)	1782	1782
family 1, 40 a file (1381)	1341	1341
families 2 to 8, sampled (2410)	2328	2328

The harness counts no answer wrong in either.

The ten problems the two builds disagreed on, run again one build at a time: master answers none, this answers nine, in 1.8 to 10.6 seconds with the check. The tenth, sqrt(a + i a tan(e + f x))/sqrt(c + d tan(e + f x)), is declined by both, master after 24 seconds.

The suite passes on the head here, c4ad9d2f, which is master b3ac1e8d and this change: 15,131 tests, every one reported. The allocation gate passes: every gated benchmark allocates what the baseline says. The library builds for every target.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura