Split a sum of two powers of even quadratics over the roots of its binomial

[Rafael-SOWNet]

Part of #718.

(1 + x^2)^2/(a x^6 + b (1 + x^2)^3) (Rubi's 1.3.1:542) was left unevaluated. SolveAPowerOfAnEvenQuadraticOverABinomialInIt splits w^m/(a + b w^n) over the n roots of the binomial, where w is an even quadratic. It took only a constant beside the power.

It now reads a second power of an even quadratic v as well. In t = w/v, w^m v^(n-1-m)/(a v^n + b w^n) is the same sum, and each root's 1/(w - r v) is again an even quadratic. With w = p + q x^2 and v = pV + qV x^2, each root contributes arctan(sqrt(Q) x/sqrt(P))/(sqrt(P) sqrt(Q)), where P = p - r pV and Q = q - r qV. An x^(2n) beside v^n is read as (x^2)^n.

	master 0f840669	this branch
1.3.1:542 (1+x^2)^2/(a x^6 + b (1+x^2)^3)	Unsolved	Solved, 878 chars
x^2 (1+x^2)/(a x^6 + b (1+x^2)^3)	Unsolved	Solved, 896 chars
(1+x^2)/(a x^4 + b (1+x^2)^2)	Unsolved	Solved, 289 chars

The answers are exact at four sign cases of a and b.

Measured:

• Pocket. 1,111 rows: all of 1.3.1, plus 6.1.7 and 6.2.7. Branch against master 0f840669. 542 is the one move.
• Chain. No moves on the per-family samples. The allocation gate passed. Suite: 15,568 passed, 0 failed. Calculus after merging master (e04410a5): 4,826 passed.
• Unit test. ASumOfTwoPowersOfEvenQuadraticsIntegralTest, five rows, each at two sets of constants.

BREAKING entry: A quotient over a sum of two powers of even quadratics is split over the roots of its binomial. The 2.5.0 column was measured on the v2.5.0 package.

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