[Rafael-SOWNet]

A sum of terms, each a power of one function K whose derivative is free of it (an inverse trigonometric function or an error function) times a cofactor free of it, is now integrated by parts down the powers. The rule uses ∫ A K^p = F K^p - ∫ p F K^(p - 1) K', with F = ∫ A for the cofactor of the top power, and the last integral joins the terms one power below. Where every power cancels that way, the parts alone are the antiderivative: x^3/((1 + a^2 x^2) atan(a x)^3) - 3 x^2/(2 a atan(a x)^2) is (-x^3/(2 a atan(a x)^2))', found in one step.

Measured on 8a1ef454, branch against master:

Run	Problems	Master	Branch
Pocket: Rubi's 5.x and 8.1–8.5	2,194	2,167 solved	2,171 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	2,362

Re-run alone on both builds, 5 problems go from declined or timed out to answered, in 25–430 ms: Rubi's 5.1.4:534, 5.3.4:768 and 8.1:122/251/382. None goes the other way. One problem that master answered at 23.6 s timed out on the branch in the pocket (5.3.6:115); it sits at the edge of the budget and is not among the moves re-run alone. No run had a wrong answer.

ASumOfPowersOfAnInverseOrErrorFunctionByPartsDownThePowersIntegralTest differentiates the answers back. BREAKING-CHANGES.md has the entry, with the 2.5.0 column measured on v2.5.0.

Part of #233.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura