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Integrate five shapes of the independent suites - #2024
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…- T^3), a root of a quartic's numerator square, an Ei derivative, roots of sec ± 1, and Euler's substitution on a nested root - A quotient beside the cube root of a cubic with a linear factor, where it is T'/(1 - T^3) for T the root over the linear: Welz 94-96. - A linear over a linear beside the root of a quartic, where it is a function of the numerator's square over the root: Welz 179, 180, and Rubi's 1.2.2.3:557-560. - An exponential of a rational function beside the derivative of its exponential integral: Hebisch 12. - A function of sqrt(sec(u) + 1) and sqrt(sec(u) - 1), in their difference: Charlwood 140. - A half-odd power of a linear plus the root of a quadratic, in Euler's substitution where their squares differ by a square: Bondarenko 24, 25. Chain against master f7f2a7a: pocket of 1,774 rows, 1,738 -> 1,742 solved, 0 wrong; family 0 whole 1,805 -> 1,814 of 1,814; other samples unchanged; the 13 rows that moved, alone, 0 -> 13 solved; no row's allocation grew past 1.3x. 2.5.0 declined all thirteen. CubeRootLogarithmAnsatzTest held `1/(x ((x - 1)(q - 2 q x + x^2))^(1/3))` as declined, the ansatz's field not holding q^(1/3). The first rule above answers it, with T = q^(1/3) (x - 1)/root, and the answer differentiates back for q = 0.37, 2.3, -0.7 and -3.1 at eleven points on both sides of the radicand's zeros; the row moves to a theory of its own that checks the derivative. 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.
Five shapes from Rubi's independent test suites (family 0) and its 1.2.2.3, one test class and one BREAKING-CHANGES entry each:
T'/(1 - T^3)forTthe root over the linear: Welz 94–96.(2 + 2x + 3x^2 - x^3 + 2x^4) e^(x/(2 + x^2))/(2x + x^3), 61 characters.sqrt(sec(u) + 1)andsqrt(sec(u) - 1), in their difference: Charlwood 140.Chain against master f7f2a7a, with the harness at its current revision on both arms:
2.5.0 declined all thirteen. The entries' "Now" lengths are re-measured on this build after the rebase onto f7f2a7a; most answers came out shorter than when the branch was written (Charlwood 140: 3,179 → 2,635). 24 tests pass locally across the five classes and the corpus gate.
Test change.
CubeRootLogarithmAnsatzTest.DeclinedWhereTheFieldIsNotHeldexpected1/(x ((x - 1)(q - 2 q x + x^2))^(1/3))to be declined, because the ansatz's field does not holdq^(1/3). The first rule here now answers it, withT = q^(1/3) (x - 1)/root. The answer differentiates back forq = 0.37, 2.3, -0.7, -3.1at eleven points on both sides of the radicand's zeros. The row moves to a theory of its own that checks the derivative, and the Q(sqrt(3)) pole pair stays declined. The full suite passes locally: 16297 passed, 0 failed.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura