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

 Method (no overloads)

Summary

Four nested roots with a closed form, Rubi's 1.3.3: sqrt(a + b sqrt(c + d x^2)) with a^2 = b^2 c; sqrt(c x^2 + d sqrt(a + b x^4))/sqrt(a + b x^4) with
c^2 = b d^2; 1/((a + b x^n) sqrt(c x^2 + d (a + b x^n)^(2/n))); and
sqrt(a x^2 + b x sqrt(c + d x^2))/(x sqrt(c + d x^2)) with a^2 = b^2 d and
b^2 c + a = 0. And the second over a power of a linear, for numbers.

Remarks

The first is 2 b^2 d x^3/(3 S^(3/2)) + 2 a x/sqrt(S) for the root S. The
second and third are the derivative of w = x/sqrt(S) over 1 - 2c w^2 and
1 - c w^2: sqrt(S)/R for the inner root R is d w'/(1 - 2c w^2), since
S' = 2 c x S/(d R) where c^2 = b d^2, and 1 - 2c w^2 = d R/S. The fourth is
sqrt(2) b/a times 1/sqrt(1 + t^2/a) in t = a x + b sqrt(c + d x^2). The
integral in w or t is asked of the integrator with its coefficient named, and
the answer is differentiated back before it is returned. Rubi's 1.3.2 and the Welz and
Timofeev suites declined all of them.
https://github.com/asc-community/AngouriMath/issues/718
Over a power of a linear L the substitution leaves x as a fourth root in
w; there the quartic is split instead. With beta = sqrt(a/b),
a + b x^4 = b (x^2 + i beta)(x^2 - i beta), and on the real line the two roots of
the factors are conjugates whose sum is sqrt(2 x^2 + 2 sqrt(x^4 + beta^2)), so
sqrt(S)/R is sqrt(d/(2 sqrt(b))) times
1/sqrt(x^2 + i beta) + 1/sqrt(x^2 - i beta), each over L^k an inverse
hyperbolic tangent and roots, and the two together real. For positive numbers
a, b, c, d: Rubi's 1.3.2
sqrt(2 x^2 + sqrt(3 + 4 x^4))/((c + d x) sqrt(3 + 4 x^4)) and its square, and
Welz's 47 and 48, declined before.
https://github.com/asc-community/AngouriMath/issues/233

























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