SolveAPowerOfAnInverseFunctionOfAQuadraticByTheRecurrence​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

A whole power of a + b u, where u is the inverse sine or cosine of
w = c + d x^2 with c^2 = 1, the inverse hyperbolic cosine of one, or the
inverse hyperbolic sine of one with c^2 = -1, by the recurrence
integral (a + b u)^n = x (a + b u)^n + 2 e n b H (a + b u)^(n - 1) - 4 e b^2 n (n - 1) integral (a + b u)^(n - 2),
with H = 1/(d x u'(w)), e = 1 for the circular functions and -1 for the
hyperbolic ones. It closes at n = 0 and n = 1, so nothing is asked below. A
negative power goes through u: w - c is a constant times g(u/2 + beta)^2 for a sine or a cosine g, so x/g is constant on each side of 0, and the
powers -1 and -2 are sine and cosine integrals of (a + b u)/(2 b) (hyperbolic ones for the hyperbolic cosine and sine), the rest the recurrence run backwards.
Exact on the real domain; off it x/g can change sign across a cut. For the inverse
hyperbolic sine g is cosh(u/2 + c pi/4): cosh(u/2) + c sinh(u/2) is
sqrt(2) times it, and its square is 1 + c sinh(u) = c d x^2.