A partial-fraction coefficient with symbols in it is in lowest terms, its rational content included, and a constant factor is not a symbolic coefficient

[Rafael-SOWNet]

Two rows of the Rubi hyperbolic sample that were answered in under two seconds became timeouts on today's master: csch(x)^5/(a + b cosh(x)) with #1389 (1.4 s → >60 s) and tanh(x)^6/(a + a sech(x)) with #1392 (0.66 s → >60 s). Bisected with a solo probe at each merge; both are the symbolic partial-fraction decomposition, and both are coefficients that were not in lowest terms.

The rational content. InLowestTermsOverTheSymbols put a coefficient in lowest terms by the polynomial gcd, which normalizes its divisor to whole coprime coefficients and so never cancels a number, and calls two sides without a variable in common coprime before it looks: -1024/(1024 a) stayed as it was. Newton's iteration for the inverse modulo (1 + u^2)^6 squares its iterate every round, so the second round carried 1024^2, the third integers of twelve hundred digits ([newton 2] maxchars=1190), and the minute went into EInteger.HalfGCD on them (dotnet-stack). The rational content of the denominator now goes up into the numerator — the denominator's coefficients whole and coprime with a positive leading one, by the Normalized the gcd already has, with the scale it used handed back — on every path out of the function.

The elimination's values. Where symbols are among them they are put in lowest terms before the terms are built. csch^5/(a + b cosh) under u = e^x, after the Hermite reduction, is a numerator with a page of a and b over (u^2 - 1)(b + 2 a u + b u^2); its values came out at 2233 nodes each as quotients of determinants, and each term went round the chain with that in front and did not return. In lowest terms they are 1/256 (3 a^3 - 7 a b^2)/(a^4 - 2 a^2 b^2 + b^4) and the like; the answer is 27 kB where it was 59, and the row is 1.3 s.

A constant factor is not a symbolic coefficient. The content of a written sum comes out before the symbolic split rather than after it, and the split's gate asks for a symbol among the coefficients of the factors in the variable rather than anywhere in the denominator: a (1 + u^2) + 2 a u is a (1 + u)^2, a repeated linear over the rationals, which the refactoring and the Hermite reduction read, where the split computed its digits with a in every coefficient for an answer twice the length. tanh^6/(a + a sech) is 6.7 kB and 2 s, from 7.8 kB before the regression.

row	2.5.0	master e780ce9a	this branch
csch(x)^5/(a + b cosh(x))	unevaluated	>60 s	1.3 s, 27 kB
tanh(x)^6/(a + a sech(x))	unevaluated after 115 s	>60 s	2.1 s, 6.7 kB
1/(u (u^2 + 1)^6 (a (u^2 + 1) + 2 a u))	unevaluated	0.1 s, 1066 chars	50 ms, 499 chars
u^5/((u^2 - 1)^5 (2 a u + b (u^2 + 1)))	unevaluated	>60 s	0.4 s

Measured: the four rows under IntegrationDecline.Guard in SymbolicFactorsPartialFractionsTest, differentiated back and bounded in length; the suite green locally (11523 passed, 6 m 40 s); BREAKING-CHANGES entry with the 2.5.0 values from a build of the tag. The independent Rubi suite and the family-6 sample are running and will be posted here.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

[Rafael-SOWNet]

Corpus on d5ce8276, both runs detached and alone on the machine:

• Independent Rubi suite: 1706/1774, 0 wrong, 0 timeouts — identical verdicts to master e780ce9a (0 changed; 89.6 → 86.0 s).
• Family 6 (hyperbolic, 417 problems): 315/417 from 312, 0 wrong, 0 error. Three rows changed, all Timeout → Solved: csch(x)^5/(a+b cosh(x)) 1.1 s, tanh(x)^6/(a+a sech(x)) 0.7 s, and coth(x)^4/(a+b csch(x)) (14.8 s in the run, answered from a warm memo — over the budget solo). Nothing lost.

The second commit is what the first run of family 6 asked for: it lost cosh(c+d x)^6/(a+b sinh(c+d x)^2)^2 (6.4 s on master) and 1/(a+b csch(c+d x)^2)^3 (3.4 s), confirmed solo, because their elimination hands back values of 12–28 thousand nodes and putting one in lowest terms did not return in three seconds where the answer with them as they come is six. A value past complexity 4096 is now left as the elimination gave it (the ones that needed the lowest terms were ~2 thousand nodes); both rows are back at master's timings.