Pair the half-odd powers of two linears whose product is an arsinh radicand

[Rafael-SOWNet]

Beside (a + b arsinh(c x))^n, the product (d + i c d x)^(3/2) (f - i c f x)^(5/2) is K^3 (f - i c f x) M^(3/2), where M is the radicand arsinh holds and K is constant: a polynomial beside the root, integrated in about half a second. Paired the other way, as K^5 M^(5/2)/(d + i c d x), it ran past the budget. With both factors above the bar, the smaller power is the one paired, so the whole power left over stays above the bar too.

With the first factor above the bar and the second below it at a power of -3/2 or less, the whole power is written out as a sum of monomials, each over the root. sqrt(f - i c f x)/(d + i c d x)^(3/2) is K^(-3) (f - i c f x)^2/M^(3/2), a product the rules for a power of a quadratic did not read.

Measured on 8a1ef454, branch against master:

Run	Problems	Master	Branch
Pocket: Rubi's 5.1 and 7.1	1,304	1,281 solved	1,293 solved
Family 1, 40 a file	1,381	1,362	1,362
Family 0, whole	1,814	1,789	1,789
Families 2 to 8, sampled	2,410	2,362	2,363

Re-run alone on both builds, 12 of Rubi's 7.1.4 go from timed out to answered. None goes the other way. No run had a wrong answer.

APairOfLinearsOverARadicandAsMonomialsIntegralTest differentiates the answers back. BREAKING-CHANGES.md has the entry, with the 2.5.0 column measured on v2.5.0. The library builds for every target: a KeyValuePair deconstruction that netstandard2.0 lacks was written out.

Part of #233.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura