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Integrate an exponential of a quadratic over the linear through its vertex - #1951
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Rafael-SOWNet merged 1 commit intoOct 19, 2026
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…ertex f^(a + b x + c x^2)/(b + 2 c x)^n was declined. Under u = b + 2 c x the exponent is u^2/(4 c) + a - b^2/(4 c), with no linear part, and the integrand is a moment about 0 of the Gaussian, which the integrator answers: the exponential integral of the square for an odd n, the error function for an even one. The exponent in u is written from the coefficients, and the vertex condition 2 A p = q B is decided as a value. Rubi's 2.3: 6 rows go from unevaluated to solved, 0 wrong. Families 0 and 1 and samples of 2-8 otherwise identical; gate and suite pass. 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.
f^(a + b x + c x^2)/(b + 2 c x)^nwas declined. Underu = b + 2 c xthe exponent isu^2/(4 c) + a - b^2/(4 c), with no linear part, and the integrand is a moment about 0 of the Gaussian, which the integrator answers. An oddngives the exponential integral of the square, and an even one the error function. The new rule,SolveAnExponentialOfAQuadraticOverTheLinearThroughItsVertex, takes only a linear whose root is the vertex (2 A p = q B, decided as a value). Elsewhere the shift leaves a linear part:e^(x^2 + 2 x)/(x + 2)is still declined, and is pinned by a test.Measured against master
8a1ef454:Rebased onto #1950 without conflict. Both together put family 2 at 635 of 650. All six rows were declined by 2.5.0 too, and
BREAKING-CHANGES.mdhas the entry.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura