[Rafael-SOWNet]

Answers ∫ x^m F^(a + b x^n) dx and ∫ (c + d x)^m F^(a + b x) dx with the upper incomplete gamma function, where that is what the integral is. Part of #2075; answers #180's item 22, t^(-1/2) e^(-t^(-2)).

The rule. With z = -b ln(F) x^n and s = (m + 1)/n, x^(m + 1) z^(-s) is constant between the cuts and gamma(s, z) differentiates to -z^(s - 1) e^(-z) z', so the integral is -F^a x^(m + 1) z^(-s) gamma(s, z)/n. The second shape is the first with n = 1 under t = c + d x.

• Where it applies: where s is neither whole nor half a whole number. Those integrals are elementary, exponential integrals or error functions, and keep the answers they had.
• Past 30: a whole or half-whole s is the gamma function, where parts would take s steps. x^100 e^(-a x) is -a^(-101) gamma(101, a x).
• Where the constant factor is one number on both half-lines, it is written as that number, and the answer has a value at 0: e^(x^3) is gamma(1/3, -x^3)/3.

Two costs it exposed, each fixed here.

• Slow declines. By parts integrated (a + b ln(c x^n))^p into a gamma function of a logarithm and handed on a remainder that nothing integrates. The searches then sampled it three thousand times before declining, and 29 declines in the 3.x files became timeouts (one went from 2.7 s to 116 s). Three by-parts paths now decline a gamma function of an argument that is not linear. Evaluate the incomplete gamma function's series faster #2079 made each evaluation ten times cheaper too.
• A definite integral through 0. For x^2 e^(x^4) from 0, the answer is 0 times an infinity at the bound. Its limit went through l'Hôpital's rule on the whole product for over a minute, and it is left as written there, as master leaves it. Γ's own limit is taken by continuity where its value is finite. limit(gamma(-1, x), x, 0) stays undecided, since the function is about 1/x there.

Measured on Rubi's suite, read with the harness that maps GAMMA and Psi, against master e62a56d6, the whole corpus each side:

	master	this
solved	48,502 of 49,052	48,634
wrong	0	0
timeouts	5	2
rows lost		0

The gains are 68 in 2.3, 24 in 3.1.2, 12 in 8.6, 7 each in 3.3 and 3.5, 6 in 3.1.5, and 8 elsewhere. Time and allocation per stream are within 3%. Two rows' allocation grew in the run, 3.3:246 and 7.3.6:563, and alone they are identical on both builds: 2.46 GB and 1.43 GB.

Checked by value, not only by differentiating back, since a derivative that matches is what hid #2080's wrong values:

	this	quadrature
integral(e^(x^3), x, 0, 1)	1.3419044179774…	1.34190441797742
integral(e^(x^2)*x^(1/3), x, 0, 1)	1.1717955466629…	1.17179554666298
integral(x^2*e^(x^4), x, 1, 2)	1129720.502623949…	1129720.50262395
integral(floor(x)*e^(x^3), x, 0, 2)	275.5109837633…	275.510983763312

Four tests changed, each because an integral they pinned as declined is answered now. Each new answer is checked by its value. A fifth, (pe x)^(n - 1) e^(x^n) with its pe > 0, failed on the first version and passes unchanged since the order is read through Simplify:

• e^(x^3) in the ansatz's decline list is replaced by e^(x^3 + x), which is neither elementary nor a gamma function.
• floor(x) e^(x^3) over [0, 2] moves to a test of its value. e^(x^3 + x) takes its place among the integrals left as written.
• x^m ln(x)^n (1 + ln x) is differentiated back.
• x^n e^(x^n) is differentiated back.

The suite passes (16,923) and so do the MCP tests (329).

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura