Reduce a rational function over a power of a symbolic quadratic a power at a time

[Rafael-SOWNet]

Part of #718

1/((d + e x)(a + b x + c x^2)^4), x^8/(a + b x + c x^2)^4 and their kin in Rubi's 1.2.1 ran past a minute. The partial fractions take a symbolic quadratic apart over its roots, and its powers with it. The new SolveARationalFunctionOverAPowerOfASymbolicQuadratic never takes the quadratic apart:

• The linear comes off at its root p. Using Q - Q(p) = (x - p)(c (x + p) + b), N/(L^j Q^k) becomes N(p)/(Q(p)^k L^j), plus short polynomials over lower powers of L and Q. The terms are kept in a worklist keyed by those two powers, so each stays of small degree.
• The power of the quadratic is lowered one at a time. P = S Q + alpha Q' + beta, alpha Q'/Q^k integrates directly, and beta/Q^k follows the recurrence integral 1/Q^k = (2 c x + b)/((k - 1) D Q^(k - 1)) + 2 c (2 k - 3)/((k - 1) D) integral 1/Q^(k - 1), with D = 4 a c - b^2.
• The first power is written in closed form, as alpha ln Q plus beta times the table's integral 1/Q.
• D, Q(p) and p are named atoms while it runs, and the answer is checked by its derivative.

It applies to the question asked only, with one linear at most. It declines in three cases, each of which a chain caught:

Declines when	What happened without it
b^2 - 4 a c is a polynomial square, so the quadratic factors over linears	1.2.1.2:2145 was lost; 1/(x^2 (d + e x)^4 (d^2 - e^2 x^2)^(7/2)) went from 943 characters to 22K
the coefficients hold the imaginary unit	TrigonometricOfAnImaginaryLogarithmIntegralTest failed
Q has no x term and there is no linear (a binomial in x^2)	x^22/(a + b x^2)^10 ran past the budget

Measured

intbench, this branch against master 37b4798, 5 s budget. The pocket was 1.2.1.2–1.2.1.4, 1.2.1.9, 1.3.1, 1.2.2.2 and 1.2.2.3 in full (7,020 rows), plus the usual family sample:

	master	this branch
pocket	6,940 solved, 28 timeouts	6,947 solved, 21 timeouts
time on the rows both solve	1,379 s	1,339 s
rows more than twice as slow		none
families 0–8 sample		identical
wrong / error	0 / 0	0 / 0

Seven rows moved, all to solved, and all held when re-run alone (master solves none of them):

• 1.2.1.2: 611, 2505, 2514, 2522 and 2523
• 1.2.1.3:2704
• 1.2.1.9:84

Every new answer was checked exactly with exactcheck (exact pins, 50 and 2,000 digits, four points, three pin sets): 192 of 192 diff 0. The gate and the suite pass (15,401).

The answers are 8.5K to 26K characters, in under 0.4 s each. Much of the length is the table's integral 1/Q, which is a piecewise on the sign of the discriminant.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura