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Read the derivative of a sum of exponentials over another off the integrand over the other's square - #2040
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Rafael-SOWNet merged 1 commit intoOct 22, 2026
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…egrand over the other's square Rubi's 6.7.1 x (b - a sinh(x))/(a + b sinh(x))^2 ran past the budget: the split took the bracket's terms apart, the substitution in e^x answered each over the square of a quadratic with a piecewise on its discriminant, and parts beside x integrated that piecewise again. For a sum D of exponentials of x, the integrand over D^2 is now read as (N/D)', N a constant plus a constant times each exponential in D. The exponentials' span is closed under the derivative, so N' D - N D' is linear in N's constants, and with each exponential a whole power of the first, w, the identity read at six values of w is a linear system. N and N + c D share a derivative over D; the solution with no constant is tried first, which is -cosh(x)/(a + b sinh(x)). Beside x, by parts, the integral of N/D is written as a constant times ln(D) where N is a constant times D', and asked otherwise. The trigonometric twin is SolveAsTheDerivativeOfASineRaisedFromNothingBesidePowersOfTheCosine's; there are no hyperbolic nodes, since sinh is parsed into exponentials. 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.
The derivative of a sum of exponentials over another, read off the integrand over the other's square. Rubi's 6.7.1
x (b - a sinh(x))/(a + b sinh(x))^2ran past the budget, in three steps:e^xanswered each over the square of a quadratic, with a piecewise on its discriminant;For a sum
Dof exponentials of x, the integrand overD^2is now read as(N/D)', whereNis a constant plus a constant times each exponential inD.How N is found.
N' D - N D'is linear in N's constants. With each exponential a whole power of the first,w, the identity read at six values of w is a linear system.NandN + c Dshare a derivative overD. The solution with no constant is tried first, and that is-cosh(x)/(a + b sinh(x)).N/Dis written as a constant timesln(D)whereNis a constant timesD', and asked as a question otherwise.Where it sits. It is the hyperbolic twin of
SolveAsTheDerivativeOfASineRaisedFromNothingBesidePowersOfTheCosine, dispatched beside it. There are no hyperbolic nodes, since the parser writessinhas exponentials.x (b - a sinh(x))/(a + b sinh(x))^2x (b + a cosh(x))/(a + b cosh(x))^2sech(x)^2 tanh(x)^2/(2 + tanh(x)^3)^2-1/3/(2 + tanh(x)^3), 60 characters, 55 ms(b - a sinh(x))/(a + b sinh(x))^2Chain against master 78cd31b:
The full unit suite passes locally: 16,554 passed, 0 failed. The BREAKING-CHANGES lengths were measured on the branch. Rebased onto 37e2b1a.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura