An exponential of a quadratic in a logarithm beside a power of x is integrated onto the Gaussian - #1522
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…ntegrated onto the Gaussian x^p G^(Q(L)) with L = ln(c x^r) and Q a quadratic with a square term: under t = L, with L' = s/x, dx = x dt/s and x^(p + 1) e^(-(p + 1) L/s) is a constant, so the integral is that constant over s times the integral of e^(Q(t) ln G + (p + 1) t/s), the Gaussian with a linear term, which the table answers. A polynomial beside it is summed a power at a time, and the logarithm of a power of a linear is the same question under u = d + e x, with a constant multiple of the linear read as a multiple of u. Rubi's 2.3, F^(f (a + b ln(c (d + e x)^n))^2) (g + h x)^m and F^(f (a + b ln(c (d + e x)^n)^2)) (g + h x)^m. Part of #1501. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…er templates kept Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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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.
x^p G^(Q(L)), withL = ln(c x^r)andQa quadratic with a square term, is now integrated. Undert = L, withL' = s/x,dx = x dt/s, andx^(p + 1) e^(-(p + 1) L/s)is constant. So the integral is that constant overstimes the integral ofe^(Q(t) ln G + (p + 1) t/s): the Gaussian with a linear term, which the table answers (#1512). This is #1518's substitution witht = Lin place oft = sqrt(F), and the constant is written as #1518 writes it, and as Rubi does.A polynomial beside it is summed a power at a time. The logarithm of a power of a linear is the same question under
u = d + e x. A constant multiple of that linear, as Rubi's(d g + e g x)^mwrites one, is read asg u. It is a power only inu, so the logarithm is found, and the substitution made, before any factor is read.e^(ln(x)^2),x e^(ln(x)^2),2^(ln(x)^2 + ln(x))erfix^m F^(f (a + b ln(c x^n))^2)F^(f (a + b ln(c (d + e x)^n))^2) (d g + e g x)^m,msymbolic, 2, 1, -2, -3(g + h x)^3,^2,^1F^(f (a + b ln(c (d + e x)^n)^2)), the square insideMeasured
Measured with the change on
22021b1c, before the rebase onto #1520. After the rebase, the library builds for every target, and the logarithm and Gaussian test classes pass (86).erf,erfcorerfi: 811 across families 0 to 7, at a 3 s budget. This PR solves 614, where master (22021b1c) solves 593, with 0 wrong and nothing lost. Both have the same 14 timeouts, and the total time goes from 574 s to 568 s. The 21 gained are all in 2.3, and every one of its 23 exponentials of a quadratic in a logarithm is now solved; the two belowd g + e g xalone were solved already.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura