Split a power of x over a sum of two cubes over the cube roots of unity

[Rafael-SOWNet]

Part of #718.

Rubi's 1.3.1 rows 199–205 are x^k/((3a + bx^2)^3 + 27a^2 c x^3) for k = 4 down to −2. They were declined: the denominator is symbolic, and the integrator's resultant needs numbers for coefficients. Their numeric twins (rows 206–220) were already answered as sums over the roots.

SolveAPowerOfXOverASumOfTwoCubes reads the sextic as d6 x^6 + d4 x^4 + d3 x^3 + d2 x^2 + d0. When d4^2 = 3 d6 d2 and d2^2 = 3 d4 d0, that is d6 (U^3 + V^3) with U = x^2 + s, V = m x, s = d4/(3 d6) and m^3 = d3/d6. It splits by

$$\frac{1}{U^3 + V^3} = \sum_j \frac{w_j}{3V^2 (U + w_j V)}$$

over the cube roots of unity w_j, into three quadratics x^2 + w_j m x + s. The rule asks one integral, x^n/(x^2 + B x + S), with B and S named, and substitutes each w_j m for B. The integrator answers it split on the sign of 4S − B^2, which means nothing for a complex B. The rule takes the arctangent arm, which differentiates back for any B with 4S ≠ B^2, and the whole answer is checked by differentiation at sampled points. Two of the quadratics have complex coefficients, so the answer carries i, as Rubi's does.

It is guarded to symbolic coefficients, the question asked, and no imaginary unit in the sextic.

Measured

Pocket of 7,779 Rubi rows, 5 s per row, with both arms on master cba4232:

	master	this branch
solved	7,701	7,708
wrong	0	0
timeout	13	13

All seven held when each row was rerun alone, in 0.23–0.68 s. Answers run from 849 to 3,820 characters, and 2.5.0 declined all seven. Rows both arms solve take 690 s against 700 s on master. Families 0–8 have no losses and no row more than twice as slow. Exact checks at 50–2000 digits pass 336 of 336, across the seven rows and four pin sets, including negative a, b and c, where the principal cube root is complex. The gate passes and the suite passes 15,397. Since then the branch has merged master b4f5fbe; it builds for every target, and the unit test passes on the merge.

BREAKING-CHANGES.md records the new answers.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura