Integrate half-odd powers of a ± a sin beside a fractional power of the cosine

[Rafael-SOWNet]

Rubi's (e cos(c + d x))^p (a + a sin(c + d x))^m with half-odd p and m ran past the budget: the substitution search spent it on the two radicals. Under u = sin(y) they are (1 + u)^A (1 - u)^B with A and B quarter-odd, which by Chebyshev's theorem is elementary only where A + B = m + p - 1 is whole.

With every power a number, the rule asks only where A, B or A + B is whole and the rest is a polynomial in u. (a + a sin(y)) sqrt(e cos(y)), which is elliptic, is declined before anything is asked. Numeric powers are read one level below the question too, where the polynomial term passes 1/(A B) on as (A B)^(-1), and a whole power of a product is read as the powers of its factors.

Measured on 8a1ef454, branch against master:

Run	Problems	Master	Branch
Pocket: Rubi's 4.1 and 4.2, every third solved problem	2,000	1,910 solved	1,934 solved
Family 1, 40 a file	1,381	1,362	1,362
Family 0, whole	1,814	1,789	1,789
Families 2 to 8, sampled	2,410	2,362	2,356

Re-run alone on both builds, 24 of Rubi's 4.1.1.2 go from timed out to answered. None goes the other way. The drop in families 3, 6 and 7 is problems near the budget: re-run alone, they time out on both builds or answer on both. No run had a wrong answer. Three problems in the file that master answered in 6–11 s, under a condition, now take about 0.05 s and need no condition.

FractionalPowersOfACosineBesideOnePlusASineIntegralTest differentiates the answers back, at both signs of the constants. BREAKING-CHANGES.md has the entry, with the 2.5.0 column measured on v2.5.0.

Part of #233.

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