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Raise a negative power of x beside a power of a binomial by the reduction that takes out its pole, read a derivative's constant at a point, and rationalise the roots inside a root - #2033
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Rafael-SOWNet merged 1 commit intoOct 21, 2026
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…tion that takes out its pole, read a derivative's constant at a point, and rationalise the roots inside a root - x^m (a + b x^n)^p R(x^n) with m <= -2: a multiple of the derivative of x^(m + 1) (a + b x^n)^(p + 1) takes the pole at zero out and raises the power of x by n, the rest asked on its own. Beside a root, last and at the question asked, and not where m + 1 is a multiple of n, where the substitution reads the integrand whole: there the rest was asked and declined at twenty-two gigabytes before master's answer. Over a whole power, with a symbol in the binomial, past the twenty-fourth degree. - A product of powers whose brackets are proportional at sampled points, the constant read at x = 1 where the quotient cannot be shown free of x. - A root of a radicand that holds the root of a quadratic or the roots of two linears with one slope, in Euler's substitution or their sum, for integrands that are roots of x and nothing else with a radicand that is a polynomial in x and those roots: the tangent's substitution hands (1 + coth(x))^(7/2) on as a power of (u + 1)/u + sqrt(1 - u^2)/u, which this answered in 1,661 characters where the rules after it give 287. Chain against master f7f2a7a: pocket of 814 rows, 772 -> 779 solved, 0 wrong; the samples unchanged; the 7 rows that moved, alone, 0 -> 7 solved; 6,304 rows compared for allocation, none grew past 23 MB. 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.
Seven of Rubi's algebraic rows that master f7f2a7a declines or runs out of time on:
x^m (a + b x^n)^p R(x^n)withm <= -2, a multiple of the derivative ofx^(m + 1) (a + b x^n)^(p + 1)takes the pole at zero out and raises the power of x by n. The rest is asked on its own. This answers 1.1.3.4's1/(x^3 (1 - x^3)^(2/3) (1 + x^3)), and 1.1.2.2's1/(x^6 (a + b x^2)^10)and its neighbours, where the partial fractions ran past the budget.x = 1. This answers 1.1.3.3:544.sqrt(1 - x^2 + x sqrt(x^2 - 1))andsqrt(-x + sqrt(x) sqrt(1 + x))/sqrt(1 + x), which Rubi itself does not integrate.Two scopes, found by the first chain's allocation check.
m + 1is a multiple ofnbeside a root. There the substitution reads the integrand whole, and the rule asked what was left, declined it, and spent 22 GB before master's identical answer. That was1/(x^5 (1 - x^2)^(1/3) (3 + x^2)^2): 5.75 GB on master, 27 GB before the scope.(1 + coth(x))^(7/2)became a power of(u + 1)/u + sqrt(1 - u^2)/u, and the rule answered it in 1,661 characters where the rules after it give 287.Chain against master f7f2a7a, after both scopes:
The full unit suite passes locally: 16296 passed, 0 failed. Before the scopes it failed two coth tests, the same respelling. The BREAKING-CHANGES lengths were re-measured on the branch, and the v2.5.0 column on a v2.5.0 build.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura