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

 Method (no overloads)

Summary

The root of a trinomial in the square over a binomial in the fourth power,
sqrt(a + b x^2 + c x^4)/(d + e x^4), where c d + a e = 0 and a c < 0:
with w = sqrt(b^2 - 4 a c), k = sqrt(2 (w + b)) and l = sqrt(2 (w - b)),
its integral is
(a/d) (atanh(x (b + 2 c x^2 + w)/(k R))/k - atan(x (b + 2 c x^2 - w)/(l R))/l),
R the root. Welz's sqrt(1 - x^4)/(1 + x^4) and sqrt(1 + p x^2 - x^4)/(1 + x^4).

Remarks

The binomial is (d/a) (a - c x^4), and u = x/R has
u' = (a - c x^4)/R^3, so the integrand is (a/d) du/(1 - 2 b u^2 + w^2 u^4), a
rational function of u -- Rubi's 1.2.2.6. Where a c is negative, w is
real and greater than |b|, and k and l are real; the two terms are
Rubi's, written for any a. k^2 R^2 - x^2 (b + 2 c x^2 + w)^2 is
2 (a - c x^4) (b + 2 c x^2 + w) and l^2 R^2 + x^2 (b + 2 c x^2 - w)^2 is
-2 (a - c x^4) (b + 2 c x^2 - w), so the inverse hyperbolic tangent's argument is
1 or -1, and the arctangent's i or -i, only at a zero of the binomial, none of them real,
or of the trinomial: the answer has no jump between two zeros of the trinomial. Each term
uses no more about the root than R^2 = a + b x^2 + c x^4, and so holds where the
trinomial is negative as well. Where a c is positive the substitution that writes
the even integrand in x - 1/x already answers it, and is left to:
sqrt(1 + x^4)/(1 - x^4). Declined before, since x/R is no substitution any
rule tries. The answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























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