A power of x times a half-odd power of a logarithm is integrated onto the Gaussian's moments

[Rafael-SOWNet]

Part of #1501.

x^p F^n, where F is ln x or A + B ln(c x^r) and n is half an odd number, is now integrated onto the Gaussian's moments, which #1512 answers.

With t = sqrt(F) and F' = s/x, dx = 2 x t dt/s, and x^(p + 1) e^(-(p + 1) F/s) is constant. So the integral is that constant times 2/s times the integral of t^(2n + 1) e^((p + 1) t^2/s). That is an even moment of the Gaussian, so the table answers it. Beside 1/x it is elementary: F^(n + 1)/(s (n + 1)).

The constant is written x^(p + 1) e^(-(p + 1) F/s), as Rubi writes it, rather than with ln(c x^r) split into ln c + r ln x. That split holds only where c is positive. For ln x itself the constant is 1 wherever the logarithm is real, so it's left out.

integral	master	this PR
sqrt(ln(x)), 1/sqrt(ln(x)), ln(x)^(3/2)	answered	answered
x sqrt(ln(x)), x^2/sqrt(ln(x)), sqrt(ln(x))/x^2	unevaluated	answered
sqrt(ln(x^2)), sqrt(ln(a x^n)), (a + b ln(c x^n))^(1/2)	unevaluated	answered
x^m/sqrt(ln(a x^n)), x^3 ln(a x^n)^(3/2), x^2/ln(a x^n)^(3/2)	unevaluated	answered
(f + g x)^2 sqrt(a + b ln(c (d + k x)^n))	unevaluated, after 21 s	answered, in 20 ms

A polynomial beside the logarithm is summed a power at a time. The logarithm of a power of a linear, Rubi's 3.3, is the same question under u = d + e x. The first version had neither. The integrator then found the half-odd power integrable, took (f + g x) (a + b ln(c (d + e x)^n))^(3/2) by parts against that antiderivative, and gave up after up to 55 s. On Rubi's corpus, 18 problems went from a 50 ms decline to the 3 s budget, until the rule read those shapes itself.

Measured

Measured with the change on ca85d0c4, before the rebase onto #1514 to #1517. After the rebase, the logarithm, Gaussian, downcasting and definite-integral test classes pass (96).

• Rubi's problems whose answer uses erf, erfc or erfi: 811 across families 0 to 7, at a 3 s budget. This PR solves 565, where The Gaussian, its even moments and the error functions are integrated #1512 alone solves 473, with 0 wrong and nothing lost. Both have 14 timeouts, and the total time goes from 609 s to 614 s. The 92 gained are 36 in 3.1.2, 50 in 3.3, 3 in 3.1.4 and 3 in 3.5, and every erf problem in those four sections is now solved: 36/36, 57/57, 3/3 and 7/7.
• The textbook suites (family 0) are unchanged: 1723/1778, 0 wrong, 0 timeouts.
• The unit suite passes on net10.0: 12896 passed, 14 skipped, none failed, of 12910.
• The allocation gate passes on all 19 gated benchmarks, with no allocation moved.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura