Integrate a power of a linear beside a quadratic's derivative over its power by parts

[Rafael-SOWNet]

Part of #718

Rubi's 1.2.1.3 (b + 2 c x)(d + e x)^m/(a + b x + c x^2)^n ran past the 5 s budget for every one of its 11 rows: the partial fractions took the symbolic quadratic apart twice over. The numerator is Q', so one step of parts lowers the power of the quadratic. With u = (d + e x)^m and dv = Q'/Q^n dx:

-(d + e x)^m/((n - 1) Q^(n - 1)) + m e/(n - 1) * integral (d + e x)^(m - 1)/Q^(n - 1)

Over Q^1, what is left is answered in a third of a second.

SolveAPowerOfALinearBesideTheDerivativeOfAQuadraticOverItsPower does that step. It reads the numerator up to a constant multiple of Q', checks the derivative of what it returns, and runs right after SolveByWritingEachLinearOnce, beside #1889's split. A negative whole power of the linear is left to the partial fractions: parts raises that power, and (b + 2 c x)/((d + e x) Q^3) went from 6,220 characters to 14,351.

Measured

intbench, this branch against master 954a9cd, 5 s budget. The pocket was 1.2.1.2–1.2.1.4, 1.2.1.9, 1.3.1 and 1.3.2 in full (6,664 rows), plus the usual family sample:

	master	this branch
pocket	6,550 solved, 42 timeouts	6,561 solved, 31 timeouts
time on the rows both solve	1,366 s	1,282 s
families 0–8 sample		identical
wrong / error	0 / 0	0 / 0

All 11 moves held when re-run alone (master solves none of them); none went the other way. They are 1.2.1.3:1723 and 1821–1830. The Q^2 rows take about 50 ms. The Q^3 rows take 14 s, because what is left, sqrt(d + e x)/Q^2, is still slow on master.

One row is slower: 1.3.2:871, 1/(x^3 sqrt(a + b sqrt(c + d x))), went from 1.4 s to 4.3 s. The rule meets this shape under a substitution, and the answer is shorter, 3,025 characters against 4,465.

Exact checks (exact pins, 50 and 2,000 digits, four points, three pin sets): 72 of 72 diff 0. The gate and the suite pass (15,395) on the branch before merging master. After the merge, the dispatch puts this rule just before #1889's split, and every target of both gives the same answer as when each was measured alone.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura