A rational function of a sine over a symbolic quadratic in it is integrated over the two roots

[Rafael-SOWNet]

Rubi's trig^m (a + b sin^n + c sin^(2n))^p file and its cosine sibling lost every even numerator: sin(x)/(a + b sin(x) + c sin(x)^2) came back unevaluated, sin(x)^4/(…) was still running after three minutes. Under the half-angle the quadratic in the sine is a quartic in t with a symbol in every coefficient, which nothing factors; the substitution u = sin(x) wants a cosine above the bar that an even power does not give.

The rule. Written in s = sin(x) — the other four functions as quotients of the two, the cosine's even powers as 1 - s^2 — the integrand is a rational function of s. It is divided down where improper, split over the written factors (s^k, (1 - s)^m (1 + s)^m from a secant, the quadratic) by the existing symbolic partial fractions, and the piece over the quadratic is taken by its two roots, exactly as SolveAnEvenPolynomialOverASymbolicBiquadratic does for x^2: with q = sqrt(b^2 - 4ac) and r = (-b ± q)/(2c), (d + e s)/(a + b s + c s^2) is (d + e r_1)/(q (s - r_1)) - (d + e r_2)/(q (s - r_2)), and

• ∫ dx/(sin(x) - r) = -2 atan((r tan(x/2) - 1)/sqrt(r^2 - 1))/sqrt(r^2 - 1),
• ∫ dx/(cos(x) - r) = -2 atan(tan(x/2)/σ)/((1 + r) σ) with σ = sqrt((r - 1)/(r + 1)),

each for any complex r but ±1, each checked by differentiation (d/dx atan(u/s)/s = u'/(s^2 + u^2) whatever s is). That matters because the chain's own answer for 1/(sin(x) - r) is a piecewise on the sign of r^2 - 1, which has no value for the conjugate pair the ordinary coefficients give — 1/(sin(x) - ((-b + sqrt(b^2 - 4ac))/(2c))) is unverifiable through the chain and exact here. The two forms are the ones the answers land on. A linear argument k x + m divides by k. Before the substitution search, which spends its budget on the quartic; it declines in a few reads for anything else — a tangent, an odd cosine, a second quadratic, a repeated one, a zero discriminant.

Measured. All eleven rows of SineOverASymbolicQuadraticTest differentiate back with the coefficients pinned to real roots (1, 5, 2) and to a conjugate pair (3, 1, 2), at six points; sin(x)^4/(a + b sin(x) + c sin(x)^2) answers in 71 ms where master had not returned after 190 s. All four test suites green (11,493). The 1,774-problem independent suite: 1706/1774, 0 wrong, 0 timeouts, per-problem verdicts identical to master, 89.6 s → 85.2 s in total. Rubi's family 4 sample (422, six per file): seven rows go from unevaluated or timeout to solved — sin(x), csc(x)^2, cos(x)^2, sec(x)^2 and sin(x)^4 over a + b sin(x) + c sin(x)^2, sin(x)^2 over the cosine quadratic, and 1/(a + b csc(c + d x)^2), which is sin^2/(b + a sin^2) — and no row goes the other way (one (a + b tan)^2 (…)/(c + d tan)^3 reads Solved at 24.7 s on one run and Timeout on the other, both past the 5 s budget). New SineOverASymbolicQuadraticTest with its .editorconfig section; BREAKING-CHANGES entry.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

[Rafael-SOWNet]

Merged master (#1399, same anchor in IndefiniteIntegralSolver.cs, .editorconfig and BREAKING-CHANGES; both kept). On the merge: both new test classes and the corpus gate green, the 1,774-problem suite 1706/1774 with 0 wrong and 0 timeouts, verdicts identical to master.