A sum of two squares below the bar, beside a repeated factor, is split over its conjugates

[Rafael-SOWNet]

Part of #718.

(c + d tan(x))^(3/2)/(a + b tan(x))^2 ran past two minutes, with the rest of Rubi's 4.3.2.1 that put one root, or two, over a whole power of a tangent sum:

integrand	master 15aae3d3	this
(c + d tan(x))^(3/2)/(a + b tan(x))^2	past 60 s	1.4 s, 47,209 characters
(c + d tan(x))^(5/2)/(a + b tan(x))^2	past 60 s	0.8 s, 55,579 characters
(c + d tan(x))^(3/2)/(a + b tan(x))^3	past 60 s	1.2 s, 71,293 characters
x^4/((a d + (x^2 - c) b)^2 ((x^2 - c)^2 + d^2))	past 60 s	4.2 s, 42,592 characters
1/((a + b tan(x))^(3/2) (c + d tan(x))^(3/2))	past 60 s	0.5 s, 18,658 characters
1/((a + b tan(x))^(5/2) (c + d tan(x))^(3/2))	past 60 s	0.7 s, 28,342 characters

What changes. Under u = tan(x) and t = sqrt(c + d u), the 1 + u^2 the tangent leaves is (t^2 - c)^2 + d^2, a quartic beside the square of a d + b (t^2 - c). The split by residues reads factors of the first and second degree only, so the quotient went to the Hermite reduction, which solved for its numerators with the symbols in every entry. SolveByPartialFractions writes a factor that is a sum of two squares, P^2 + Q^2 with the greater of their degrees two, as (P - i Q)(P + i Q) now, beside a repeated factor, and asks the quotient again; over those two every factor is one the split reads. Under the root of the quotient of two such linears the sum is (b - d t^2)^2 + (c t^2 - a)^2, with both parts in t, and the same holds. Not of the first degree: (a + b x)^2 + k^2 is a quadratic every rule reads, by an arctangent.

The constants of that split hold i, and it is short only with them in lowest terms, which #1795 does; this branch is measured on top of it.

Tests: SumOfTwoSquaresBelowTheBarIntegralTest, six rows, one root and two, differentiated back where the roots are real.

Measured on Rubi's 4.3 rows with a root and no i, 1,095 problems, at the corpus's 5-second budget, against #1795's commit 714fb609:

	#1795	this
solved	965	1002
unevaluated	11	3
unverifiable on the reals	1	1
past the budget	118	89

The harness counts no answer wrong in either.

The 104 problems the two builds disagreed on, run again one build at a time: #1795's commit answers 32, declines 8 and runs past the budget on 64; this answers 69 and runs past the budget on 35.

The suite on the commit measured, 9e81e67c, passes, 15,047 tests, and so does the allocation gate: every gated benchmark allocates what the baseline says. The head here, b986f4e9, merges master 15aae3d3; the calculus and corpus tests pass on it, 4,364 tests, and the library builds for netstandard2.0.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura