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

A polynomial over a power of a linear beside the square root of a quadratic,
P/((x - p)^k sqrt(Q)) with P of degree below k, by the reciprocal
of the linear as SolveALinearBesideTheRootOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable) takes it for a
constant over the first power: under t = 1/(x - p) it is
-sgn(t) P(p + 1/t) t^(k - 1)/sqrt(R(t)) with
R(t) = Q(p) t^2 + Q'(p) t + a, a polynomial over the root of a quadratic, which
the reduction for those closes. The pieces the partial fractions leave over
t^3 when cot(x)^3 sqrt(a + b tan(x) + c tan(x)^2) is taken under the
tangent.
https://github.com/asc-community/AngouriMath/issues/718