SolveAPowerOfAnEvenQuadraticOverABinomialInIt​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

The residue of w^m/(a + b w^n) at r_j is r_j^m/(n b r_j^(n - 1)), and
b r_j^n = -a makes it -r_j^(m + 1)/(n a). The roots are
(-a/b)^(1/n) times the n-th roots of unity. The one integral 1/(P + q x^2) is asked with
P named, and splits on the sign of P q, which means nothing for a complex
P: its first arm, the arctangent, differentiates back for any P q != 0, and
the answer is checked below. Rubi's 1.3.1 (1 - x^2)/(a + b (1 - x^2)^4) was declined.
Two powers of even quadratics, w^m v^(n - 1 - m)/(a v^n + b w^n), are the same sum in
t = w/v: the integrand is t^m/(v (a + b t^n)), and 1/(v (t - r_j)) is
1/(w - r_j v), again an even quadratic P + Q x^2, whose integral
arctan(sqrt(Q) x/sqrt(P))/(sqrt(P) sqrt(Q)) differentiates back for any P and
Q off zero. Rubi's 1.3.1 (1 + x^2)^2/(a x^6 + b (1 + x^2)^3), a sum of two cubes,
was declined.
https://github.com/asc-community/AngouriMath/issues/718