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

 Method (no overloads)

Summary

A polynomial over a power of a quadratic beside the square root of another quadratic,
P/(A^k sqrt(B)) with k at least two and P of degree below
2k, one power of A at a time down to the first, which
SolveALinearOverAQuadraticBesideTheRootOfAnother(AngouriMath.Entity,AngouriMath.Entity.Variable) closes.

Remarks

Over A^j, P is L + A R with L linear, and R goes a
power down as it is. For L = g + h x, the derivative of
(p + q x) sqrt(B)/A^(j - 1) is
(2 q A B + (p + q x) W)/(2 A^j sqrt(B)) with W = B' A - 2 (j - 1) B A',
so L/(A^j sqrt(B)) is that derivative and S/(A^(j - 1) sqrt(B)) wherever
2 L = 2 q A B + (p + q x) W + 2 A S. That is five linear equations, one for each
power of x up to the fourth, in p, q and the three coefficients of a
quadratic S, and they have a solution where A and B have no root
in common: Hermite's reduction, as Rubi's 1.2.1.6 takes these. Over A itself,
what is left is sigma A + L, sigma over the root alone and L over
A beside it.
 Beside a half-odd power of the secant or the cosecant, the half-angle tangent leaves
these: sqrt(a + a sec(x))/(c + d sec(x))^2 has (c + d) + (d - c) t^2 squared below the bar beside sqrt(1 - t^2).
https://github.com/asc-community/AngouriMath/issues/718

























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