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Integrate a power of x that is not whole beside binomials in a power of it, and thirteen more algebraic shapes - #2027
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…of it, and thirteen more algebraic shapes Fourteen shapes of Rubi's family 1 and the Welz suite, each listed with its rule and an example in the BREAKING-CHANGES entry: among them a half-odd power of x beside binomials in x^2, a rational function of a quadratic beside a half-odd power of it under t = Q'/sqrt(Q), two equal half-odd powers of linears beside a power of a quadratic as one power of their product, cube roots of two linears whose product is a quadratic, and a power below -1 of a cube root of a binomial divided out first. The scaling of the variable to numbers declines where two roots scale to one polynomial: it multiplied them into that polynomial, and the answer to 1/(sqrt(-2 - b x) sqrt(2 + b x)) had the wrong sign where 1 + b x/2 < 0, which the existing test's points reach and the chain's rows do not. And the rule under t = Q'/sqrt(Q) runs after the reduction over a power of the quadratic, which answers the third and fourth powers in a third of the characters (1,379 where this wrote 4,428); it answers the fifth, which that declines. Chain against master f7f2a7a: pocket of 988 rows, 921 -> 953 solved, 0 wrong; family 0 whole 1,805 -> 1,807 and family 1's sample 1,375 -> 1,376, the other samples unchanged; of the 34 rows that moved, alone, master solved 1 and this solves all 34; 6,453 rows compared for allocation, none grew past 1.3x. Part of #233. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Part of #233.
Fourteen shapes of Rubi's family 1 and the Welz suite. The BREAKING-CHANGES entry lists each with its rule and an example. Among them:
x^2;t = Q'/sqrt(Q);Chain against master f7f2a7a, with the harness at its current revision on both arms:
The entry's "Now" lengths and its comparison with the unreleased master are re-measured on this build and on f7f2a7a. One answer is longer than master's:
x^(7/2)/((a + b x^2)(c + d x^2)^2)is 1,246 characters here against 799 there. 56 tests pass locally across its test classes and the corpus gate.Fixed after the first CI run, which failed two existing tests that this branch's local run had not included:
SolveByScalingTheVariableToNumbersmultiplied two roots that scale to one polynomial into that polynomial.1/(sqrt(-2 - b x) sqrt(2 + b x))then came back with the opposite sign where1 + b x/2 < 0, which is wherePowersOfOneLinearAsOnePowerIntegralTestsamples. The rule now declines that case, and master's answer stands.t = Q'/sqrt(Q)ran ahead of master's reduction over a power of the quadratic, and wrote 4,428 characters for1/((d + b x + c x^2)^3 sqrt(a + b x + c x^2))where the reduction writes 1,379. It now runs after the reduction, and answers the fifth power, which the reduction declines.Re-chained against master f7f2a7a:
The full suite passes locally: 16321 passed, 0 failed. The branch is rebased onto 47ab958.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura