Take a linear factor common to a sum below the bar out of it

[Rafael-SOWNet]

Part of #718.

1/(a x + b x/log(c x^n)) was left unevaluated, while 1/(x (a + b/log(c x^n))), the same integrand with x taken out of the sum below the bar, is answered in a tenth of a second as a function of the logarithm beside its differential. The existing SolveByGatheringACommonPolynomialFactorOfASum already takes a common power of a linear out of a sum that is the whole integrand. This does the same for a sum below the bar, alone or to a whole power, dispatched right after it.

Rubi corpus, against master 081b56f (budget 5 s):

	master	this branch
3.5:323 1/(a x + b x/log(c x^n))	unsolved	488 ms
3.5:324 … /log(c x^n)^2	unsolved	199 ms
3.5:325 … /log(c x^n)^3	unsolved	311 ms
3.5:326 … /log(c x^n)^4	unsolved	270 ms
3.5:37 (a x + 2 b n log(c x^n))/(a x^2 + b x log(c x^n)^2)	unsolved after 9 s	solved in 22 s (37 characters, log(a x + b log(c x^n)^2))

The other rows that moved:

• 6.3.2:218 tanh(a + 2 log(x))^2/x^3 is still answered, now in 0.5 s rather than 4.4 s. Its answer is 7,796 characters against master's 4,078: the tanh's exponential form has a sum with a common x^2 below the bar. 2.5.0 left it unevaluated, and it is recorded with the others.
• 3.5:38 goes from unsolved after 24 s on master to a timeout.
• 3.2.1:258 moved in the sampled run, but run alone it is solved on both arms in about 20 s.

Family 3 whole is the pocket. Family 0 whole and families 1–8 sampled show no losses and no wrong answers. The allocation gate and the suite pass.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura