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Lower the power of a binomial beside x below a sine before splitting over its roots - #2051

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lower-the-power-of-a-binomial-beside-x-below-a-sine
Oct 22, 2026
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Part of #233.

  • A sine over x and a power of a binomial, with the power lowered first. Rubi's 4.1.11 sin(c + d x)/(x (a + b x^3)^3) was declined. SolveATrigonometricOfALinearOverAPowerOfALinear splits over the linears of the denominator, which here are x and the cube's three roots, each to the third power. That split came out past the size it hands on. Over the square of the binomial it answers, in 6,403 characters.
  • The change. Since a + b x^n - b x^n is a, 1/(x Q^k) is 1/(a x Q^(k - 1)) - (b/a) x^(n - 1)/Q^k, for Q = a + b x^n with n three or four. Each of the two is a question the rule answers: the first is lower in k, until the split takes it, and the second is over Q^k without x. Only where the direct split declines are the two asked as questions of their own.
Input master ef4150e this branch
4.1.11:136 sin(c + d x)/(x (a + b x^3)^3) declined, 1.6 s 13,110 characters, 2.7 s
cos(c + d x)/(x (a + b x^3)^3) declined 13,132 characters

Rubi's own answer is a sum over the cube's roots of sine and cosine integrals, as this one is.

Chain against master ef4150e:

  • The pocket of every 4.1.10 to 4.1.12 and 4.2.10 to 4.2.12 row, 493 of them, went from 492 to 493 solved.
  • Every family's sample is unchanged.
  • The one row that moved is 4.1.11:136, run alone: declined on master, solved here.
  • 6,097 rows were compared for allocation, and none grew past 1.3x and 300 MB.

The library builds for netstandard2.0, net8.0 and net10.0. The suite is left to CI. The BREAKING-CHANGES lengths were measured on the branch, and the 2.5.0 column on a v2.5.0 build. Rebased onto de11aaf.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

…over its roots

Rubi's 4.1.11 sin(c + d x)/(x (a + b x^3)^3) was declined: the split over x
and the cube's three roots, each to the third power, came out past the size
it hands on. Since a + b x^n - b x^n is a, 1/(x Q^k) is
1/(a x Q^(k - 1)) - (b/a) x^(n - 1)/Q^k, and each of the two is a question
SolveATrigonometricOfALinearOverAPowerOfALinear answers: the first lower in
k until the split takes it, the second over Q^k without x. Where the direct
split declines, the two are asked as questions of their own.

Part of #233.

Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
@Rafael-SOWNet Rafael-SOWNet added this to the 2.6.0 milestone Oct 22, 2026
@Rafael-SOWNet
Rafael-SOWNet merged commit 890a032 into master Oct 22, 2026
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@github-project-automation github-project-automation Bot moved this from Todo to Done in AngouriMath Oct 22, 2026
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