Powers of a logarithm's two linears beside it are integrated by t = L1/L2, a power of x beside a power of A + B ln(c x^n) by the closed reduction, and the principal root of a negative real is exactly imaginary

[Rafael-SOWNet]

Rubi's logarithm families (family 3): 180 → 206 of the 225-problem sample, 0 wrong, 0 error, both sides measured with the same harness. The independent suites: 1707/1774 (from 1706), 0 wrong, 0 timeouts. Cut from master; independent of #1377.

What it does

Powers of a logarithm's two linears beside a power of it, by t = L1/L2. Rubi's (f + g x)^m (h + i x)^q (A + B ln(e ((a + b x)/(c + d x))^n))^p with f + g x and h + i x proportional to the logarithm's linears took the rational-beside-a-logarithm rule one step of parts at a time, with the coefficients of its partial fractions growing to b^205, and then the substitution search past its budget — 21 of the 27 problems left in the family. Under t = (a + b x)/(c + d x), with D = b c − a d: a + b x = D t/(b − d t), c + d x = D/(b − d t), dx = D dt/(b − d t)^2, so the integrand is a power of t times a power of b − d t times a power of A + B ln(K t^n), which the same rule answers in one step (SolveByTheQuotientOfTheLogarithmsLinears, before that rule in the chain). The powers of t and of b − d t are summed so that each stands once — with (b − d t)^2 above and (b − d t)^(−4) below as they came, the rule expanded its rational part to a page. Exponents may be symbols ((g L1)^m taken as g^m L1^m, K L1^n/L2^n as K t^n, the generic case) and the answer is checked against the integrand at sampled points. A polynomial beside the logarithm is left to parts in x. The proportionality of a linear to the logarithm's is decided at sampled points and its ratio read exactly, because the chain hands the linear on as a g^0 b^(−1) + x for (a g + b g x) made monic, which is not read as a polynomial; a linear in that spelling is read by its derivative.

x^p (A + B ln(c x^r))^n by the closed reduction of x^p ln(x)^n. The affine logarithm's derivative is s/x for s = B r, each step of parts brings the factor s; a lone power of it is p = 0, and (d x)^m is d^m x^m. Rubi's (e x)^q (a + b ln(c x^n))^3, (a + b ln(c (d x^m)^n))^4, and t^(−2 − m) (A + B ln(e t^n))^2 in the variable of the substitution above.

#1378, found on the way. That reduction wrote u ln(u^2) for an integrand's ln(−1 + a x) under u = sqrt(−1 + a x), and the corpus harness read the answer as wrong at 3 of 6 points where its derivative is the integrand: at −1 + a x < 0 the root came back from the polar form with a real part of 6e-102 (cos(pi/2) at a hundred digits), and its square lay a hair below the negative axis, where ln reads −i pi for the +i pi the number has. Fixed in Number.Pow: a negative real to a real power that is a whole number of halves is the modulus to that power times i^k, exactly; and for a base off the positive axis a real power that is a whole number (how 2 arrives with the downcasting off) is the integer it is, so the square is the exact multiplication. A positive real keeps its exp(p ln b) — the first version applied the integer path there too, and a sum with 3.66^120 in every term stopped cancelling, because every power of the base shares the logarithm's digits and not a repeated product's.

Measured

Family 3, 25 problems a file, 5 s budget, master 2b4a5256 against this branch, same harness on both (the 185 recorded for master in #1363 was an older verification — a right answer with b^205 coefficients cannot be verified at a hundred digits by either, and two problems master "solved" then are timeouts for both now):

Source	Run	master	branch
3.1.2 (d x)^m (a+b log(c x^n))^p	25	24	25
3.1.4 (f x)^m (d+e x^r)^q (a+b log(c x^n))^p	25	23	24
3.1.5 u (a+b log(c x^n))^p	25	22	24
3.2.1 (f+g x)^m (A+B log(e ((a+b x) over (c+d x))^n))^p	25	15	22
3.2.2 (f+g x)^m (h+i x)^q (A+B log(e ((a+b x) over (c+d x))^n))^p	25	9	24
3.2.3 u log(e (f (a+b x)^p (c+d x)^q)^r)^s	25	23	23
3.3 u (a+b log(c (d+e x)^n))^p	25	21	22
3.4 u (a+b log(c (d+e x^m)^n))^p	25	23	23
3.5 Logarithm functions	25	20	19
total	225	180	206

The one loss, ln(d (a + b x + c x^2)^n)/(d + e x)^5, is 19 s alone on both builds and was answered in master's run from a warm memo at 24 s of wall time — budget-edge, not a change. Every answer is verified by differentiating back with the parameters pinned, in the harness and in the library.

Left in the family: polynomial × log with the coefficients growing ((f + g x)^3 (A + B ln(...)), 8 s on both), (a + b ln(c (d (e + f x)^p)^q))/(g + h x)^2 (a different linear below; the step of parts declines its symbolic partial fractions), and the 3.5 miscellany.

Four suites green. The gate moved, downwards: SolveEasy allocates 783,622 → 453,364 B (−42%), 390 → 222 µs — the solve of x^2 + x + 1 evaluates sqrt(−3) four times, each of which was a hundred-digit Arctan2, cos and sin of pi/2 through the polar form and is now the decimal square root times i. The first push took the modulus through the exact-base path, whose search over every root of unity of 3 cost the same row +17.4% and failed the gate on this PR, which is what the gate is for; the 2077th entry in version_performance_control.md records both, and the baseline is the run of commit 6e5986a7.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

[Rafael-SOWNet]

Composition with #1377 (which carries #1374): merge-tree reports one conflict, Sources/Tests/UnitTests/Core/Numeric.cs — both append a test class at the end of the file, same anchor, keep both. The library files merge clean. Whichever lands second gets that one mechanical round from me.