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

 Method (no overloads)

Summary

A rational function times sqrt(c x^2 + d sqrt(P))/sqrt(P), P = a + b x^4 with
c^2 = b d^2, written over two conjugate roots: on the real line it is
(d/sqrt(2)) (1/sqrt(c x^2 - i d sqrt(a)) + 1/sqrt(c x^2 + i d sqrt(a))), and each of
the two is a question the rules for a root of a quadratic answer.

Remarks

With alpha = (c x^2 + i d sqrt(a))/2, d^2 P = c^2 x^4 + d^2 a = 4 |alpha|^2, so
d sqrt(P) = 2 |alpha| for d > 0, and (sqrt(alpha) + sqrt(conj alpha))^2 =
c x^2 + 2 |alpha|: the sum is twice the real part of a principal root, not negative, and
is the outer root. And sqrt(P) = 2 sqrt(alpha) sqrt(conj alpha)/d. Both for real x
and real c, with a and d positive, which is where this is asked, and the
answer is differentiated back at real points before it is returned.
 After SolveANestedRootByItsClosedForm(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), which answers a constant beside the
roots in closed form. Welz's sqrt(x^2 + sqrt(1 + x^4))/((1 + x)^2 sqrt(1 + x^4)) was
declined, with numbers only, as here.
https://github.com/asc-community/AngouriMath/issues/233

























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