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Integrate an exponential over a quadratic with a double root over that linear twice - #2008
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Rafael-SOWNet merged 1 commit intoOct 21, 2026
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…t linear twice SolveAnExponentialOverSeveralLinears' quadratic branch (#1708) applied the quadratic formula and never asked whether the discriminant is zero as a value: e^(-t)/(a^2 + 2 a b t + b^2 t^2), which t = ln(c x) makes of 1/(x^2 (a + b ln(c x))^2), came back over two "different" roots carrying sqrt((2 a b)^2 - 4 b^2 a^2), merged by TrySplitOverWrittenFactors with the root still inside -- an answer whose derivative had no value at any point. A discriminant zero by PartialFractions.IsZeroAsAValue now takes the repeated root, the linear x + m/(2 l) twice, as the trigonometric twin does since #1880; a lone double root goes to the one-linear rules as a polynomial over its power. 1/(x^2 (a + b ln(c x))^2) is c e^(a/b) (-e^(-L)/L - Ei(-L))/b^2, L = ln(c x) + a/b, Rubi's 3.1.2:95 at n = 1; 23 integrals whose answer had no value anywhere -- logarithm squares for m = 2..4 and n = 1, 2, and exponentials and hyperbolic functions over a perfect-square quadratic -- now differentiate back at four sign patterns. All 169 rows of 3.1.2 and 1,247 neighbours of families 2, 3 and 6 answer as before, character for character. 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 exponential twin of #1879, which #1880 fixed for the sine and cosine.
SolveAnExponentialOverSeveralLinears' quadratic branch (#1708) applied the quadratic formula and never asked whether the discriminant is zero as a value.e^(-t)/(a^2 + 2 a b t + b^2 t^2)is whatt = ln(c x)makes of1/(x^2 (a + b ln(c x))^2). It came back over two "different" roots carryingsqrt((2 a b)^2 - 4 b^2 a^2). That gave an answer whose derivative had no value at any point.A discriminant that
PartialFractions.IsZeroAsAValuefinds zero now takes the repeated root: the linearx + m/(2 l), twice, as the trigonometric twin has done since #1880. So1/(x^2 (a + b ln(c x))^2)is-c e^(a/b) Ei(-(a + b ln(c x))/b)/b^2 - 1/(b x (a + b ln(c x))), Rubi's 3.1.2 atn = 1.Twenty-three integrals whose answer had no value anywhere now differentiate back at four sign patterns: logarithm squares for m = 2..4 and n = 1, 2, and exponentials and hyperbolic functions over a perfect-square quadratic. 2.5.0 declined them. The entry in BREAKING-CHANGES.md has 2.5.0's column measured.
The chain ran against master a567c8b, before the rebase onto a4e7656:
n.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura