An exponential of a polynomial beside the polynomial's derivative is integrated under u = P - #1525
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…integrated under u = P G^P k P' f(P), with P a polynomial of degree two or more and P' a factor of its own up to a constant, is k G^u f(u) under u = P. The substitution search reads this u and does not reach it: it writes an exponential of a sum as a product of exponentials for its own search, and e^(a + b x + c x^2) is then e^a e^(b x) e^(c x^2), with no P in it to replace; it took 24 s to decline. Rubi's 2.3, e^(a + b x + c x^2) (b + 2 c x) (a + b x + c x^2)^(n/2). Part of #1501. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
… one it follows Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Part of #1501.
G^P k P' f(P), wherePis a polynomial of degree two or more andP'is a factor of its own up to a constant, is now integrated underu = P, ask G^u f(u). Rubi's 2.3 has nine of these,e^(a + b x + c x^2) (b + 2 c x) (a + b x + c x^2)^(n/2). Beside the exponential,u^(n/2)is answered already, asF^(a + b x) x^(n/2)is.The substitution search reads this
uand does not reach it. For its own search it writes an exponential of a sum as a product of exponentials, so thate^(e^x) e^xbecomese^u du. Bute^(a + b x + c x^2)then becomese^a e^(b x) e^(c x^2), with noPleft in it to replace. On master the search declinese^(a + b x + c x^2) (b + 2 c x) sqrt(a + b x + c x^2)after 24 s, where 2.5.0 declined it in under 2. The new rule answers it in 270 ms, from before the search.Only a derivative written as a sum is taken, so
x^2 e^(x^3)stays with the rules that answer it already.e^(a + b x + c x^2) (b + 2 c x) (a + b x + c x^2)^(n/2),n= 1, 3, 5, 7erfi(a + b x + c x^2)^(n/2),n= 1, 3, 5, 7, 9erfiMeasured
Measured with the change on
12ee9809(#1520), before the rebase onto #1521 and #1522. After the rebase, the library builds for every target, and the Gaussian, exponentials-of-quadratics and exponential-substitution test classes pass (175). The erf baseline is22021b1c, one merge earlier, and the 29 problems #1520 gained are counted apart, by name.erf,erfcorerfi: 811 across families 0 to 7, at a 3 s budget. This PR solves 631, where22021b1csolves 593 and An exponential of a quadratic in the reciprocal of a linear is integrated onto the Gaussian #1520 adds 29, with 0 wrong and nothing lost. The 9 this rule gains are the 9 of its shape, each answered in 3 to 8 ms, and 3 of them were timeouts, so timeouts go from 14 to 11. The total time goes from 574 s to 544 s.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura