Integrate a linear below a cube root of a quadratic in closed form

[Rafael-SOWNet]

Part of #718.

1/((d + e x) (a + b x + c x^2)^(1/3)) has a closed form in two cases, and both were declined. With y = x + b/(2c), d' = d - e b/(2c) and a' = a - b^2/(4c):

• c d'^2 = 3 a' e^2: Rubi's arctangent and two logarithms, with q = (6 c^2 e^2/d'^2)^(1/3).
• c d'^2 = -9 a' e^2: with t = e y/d', u = 1 - 3t, κ = a'^(1/3) and R the cube root,
  [-(3/8) ln(2R + κu) + (1/4) ln(d + e x) + (1/8) ln u - (√3/4) arctan(√3 κu/(4R - κu))]/(e κ).

SolveALinearBelowACubeRootOfAQuadraticInClosedForm reads the shape, checks which relation holds, and checks the answer by differentiating it back at sampled points before returning it. It runs before the rule that writes two linears to one power as one quadratic.

Checks

• Exact checks at 50–2000 digits, every symbol pinned (including negative values): 108/108 for the first case, 84/84 for the second.
• ALinearBelowACubeRootOfAQuadraticIntegralTest: 9 cases, including a negative control, 1/((1 + x) (1 + 2x^2)^(1/3)), where neither relation holds and no closed form is claimed.

Corpus, measured against master 35c9ba5 (this branch then merged 55a8044):

	master	this branch
pocket (7,779 rows)	7,733	7,740
families 0–8	unchanged	unchanged
rows moved		+7: 1.2.1.2:806, 816, 819, 820, 825, 2819, 2834

No losses and no row slower than 2× master + 200 ms. The seven answer in 70–450 ms. Gate passed; suite 15,419 passed, 0 failed.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura