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

 Method (no overloads)

Summary

A root whose radicand holds roots one substitution rationalises -- the root of a
quadratic, by Euler's, or the roots of two linears with one slope, by their sum --
integrated in the substitution's variable, where the radicand is a rational function:
sqrt(1 - x^2 + x sqrt(x^2 - 1)) is sqrt((t^2 - 1)/(2 t^2)) (t^2 - 1)/(2 t^2) at
t = x + sqrt(x^2 - 1), and sqrt(-x + sqrt(x) sqrt(1 + x))/sqrt(1 + x) is
sqrt((v^2 - 1)/(2 v^2)) (v^2 - 1)/v^2 at v = sqrt(x) + sqrt(1 + x).

Remarks

For Q = a x^2 + b x + c, Euler's first substitution t = sqrt(Q) + sqrt(a) x has
x = (t^2 - c)/(2 sqrt(a) t + b) and sqrt(Q) = t - sqrt(a) x, and his second,
t = (sqrt(Q) - sqrt(c))/x, has x = (2 sqrt(c) t - b)/(a - t^2) and
sqrt(Q) = x t + sqrt(c). For L1 - L2 = k with one slope m,
v = sqrt(L1) + sqrt(L2) has sqrt(L1) = (v + k/v)/2, sqrt(L2) = (v - k/v)/2 and dx = (v^4 - k^2)/(2 m v^3) dv. Each is an identity in the values of the roots
wherever they are defined, whatever their branches, so the radicand written in the new
variable is the radicand, and the integrand there a rational function beside roots of
rational functions, which the chain is asked.
 Rubi's 1.3.2 has both above without an answer; the Euler substitution and the sum of two
roots read only a rational function of the roots, and declined. Numbers only, roots
nested two deep, at the question asked or one below it, and late; the answer is
differentiated back at the sampled points where the integrand is real.
https://github.com/asc-community/AngouriMath/issues/233

























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