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Integrate the reciprocal of a binomial in a high power of a sine, cosine or hyperbolic one over its roots (#1905)
* Integrate the reciprocal of a binomial in a power of a sine over its roots
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 r_k/(r_k - s^q), and each term is an arctangent of tan(y) or of
tan(y/2), written F(K u)/K so that only K^2 is fixed. The roots of unity are a cosine and a sine:
(-1)^(2/3) is the real cube root of 1 here. Rubi's 4.1.7 and 4.2.7 1/(a + b sin(x)^5),
1/(a - b cos(x)^6) and the like, with a symbol in a or b, were declined.
Part of #718
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
* Integrate the reciprocal of a binomial in a power of a hyperbolic sine or cosine over its roots
The same split as for the sine and cosine, each term an inverse hyperbolic tangent of tanh(y)
or of tanh(y/2), the functions read back from (e^y -+ e^(-y))/2 as the parser writes them.
Rubi's 6.1.7 and 6.2.7 1/(a + b sinh(x)^5), 1/(a - b cosh(x)^3) and the like were declined.
Part of #718
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
* Leave a hyperbolic cube to the rules before the split over roots
They write 1/(a + b cosh(x)^3) through e^x in 655 characters, where the split writes 1491.
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
* Record the split over a binomial's roots in the breaking changes
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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