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

 Method (no overloads)

Summary

A linear over a linear beside the reciprocal of the square root of an even quartic,
N/(M sqrt(P)) with P = a + b x^2 + c x^4, at the coefficients where its
integral is K artanh(N^2/(q sqrt(P))) or K atan(N^2/(q sqrt(P))): Rubi's 1.2.2.3
(1 + x - sqrt(3))/((1 + x + sqrt(3)) sqrt(-4 + x^4 + 4 sqrt(3) x^2)).

Remarks

The derivative of artanh(N^2/(q sqrt(P))) is
q N (2 N' P - N P'/2)/(sqrt(P) (q^2 P - N^4)), and that of the arctangent is the same
with -q^2 for q^2 and -N^4 for N^4. So the integrand is K times
one of them where sigma T = rho P - N^4 as polynomials, for T = (2 N' P - N P'/2) M,
with sigma = K q and rho = q^2, or sigma = -K q and rho = -q^2.
The top terms of 2 N' P - N P'/2 cancel, so both sides have degree four, and the
five coefficients are a linear system in sigma and rho: solved from the first and the last,
it holds on the rest exactly where Rubi's three conditions on the coefficients do, and
otherwise the integral is elliptic. The answer is differentiated back at sampled points,
which is where the rest of the system is checked. These were declined: no rule read the
square of the numerator over the root.
https://github.com/asc-community/AngouriMath/issues/233

























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