Integrate the reciprocal of a binomial in a high power of a sine, cosine or hyperbolic one over its roots

[Rafael-SOWNet]

Part of #718.

1/(a + b s^n) for s a sine, cosine, hyperbolic sine or hyperbolic cosine of a linear, n from 5 to 12, and a symbol in a or b, is split over the roots of the binomial. With q = 2 for an even n and 1 for an odd one, m = n/q and r_k the m-th roots of -a/b:

1/(a + b s^n) = (1/(m a)) sum_k r_k/(r_k - s^q)

Each term is an arctangent of tan(y) (even n) or of tan(y/2) (odd n), or the inverse hyperbolic tangent of tanh for the hyperbolic functions. It is written F(K u)/K with only K^2 fixed, so its derivative does not depend on which root K is. (atan(K tan(y))/K)' is 1/(cos(y)^2 + K^2 sin(y)^2) for either. A root that makes K^2 zero or infinite declines. The hyperbolic functions are read back from (e^y -+ e^(-y))/2, which is how the parser writes them.

Two things the first version got wrong:

• The roots of unity. (-1)^(2/3) simplifies to 1 here, the real cube root of 1. It means the principal e^(2 pi i/3) in Rubi's answers. Written that way, every term was the same. They are written as cos + i sin now.
• The check. DerivativeHoldsAtSampledPoints could not settle the arctangents of complex arguments at enough points, and declined 1/(a + b sin(x)^5), which is right at every point checked exactly. The rule refuses an answer only where it differs, through DerivativeDiffersWhereReal.

n up to 4 is left to the rules before it. They answer 1/(a + b cosh(x)^3) through e^x in 655 characters, and asked first this rule wrote 1491. That costs Rubi's 6.2.7 1/(a - b cosh(x)^3), which those rules decline on the sign of b alone; that is a separate defect.

Checks

• Exact checks at 50–2000 digits, a and b of every sign pattern: 768/768 for the sine and cosine (twelve shapes, a linear argument, n = 7 and 10), 624/624 for the hyperbolic ones (eleven shapes).
• AReciprocalOfABinomialInAPowerOfASineIntegralTest: 11 cases.

Corpus, measured against master 55a8044 (this branch then merged 5b5252b):

	master	this branch
pocket: every row of 4.1.7, 4.2.7, 6.1.7, 6.2.7 (966 statable)	924	945
families 0–8 samples	4.1.7:323, 4.2.7:147	+2 (the same rows)
rows moved		+21: 4.1.7:320–325, 4.2.7:145–150, 6.1.7:310–312, 6.2.7:111–116

No losses and no slower rows once narrowed; the pocket was re-run on both arms after the narrowing. Gate passed. The chain's suite was killed by the local memory guard at 7.8 GB, as on four earlier branches. Re-run under a 4 GB heap limit: 15,424 passed, 0 failed.

The answers are long: 0.6–3.6 K characters, against about 1 K for Rubi's. The hyperbolic tangent and its inverse are written out in exponentials and a logarithm.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura