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

 Method (no overloads)

Summary

A rational function beside the reciprocal of the square root of a palindromic quartic
P = p0 + p1 x + p2 x^2 + p1 x^3 + p0 x^4, where it is a function of
u = x + 1/x: P/x^2 = p0 (u^2 - 2) + p1 u + p2 = Q(u), and with
S = sqrt(P)/x, a root of Q, R(x) dx/sqrt(P) = h(u) du/S for
h = R(x) x/(x^2 - 1). Rubi's
(e f - e f x^2)/((a d + b d x + a d x^2) sqrt(a + b x + c x^2 + b x^3 + a x^4)) is
-(e f/d) int du/((a u + b) sqrt(a u^2 + b u + c - 2 a)).

Remarks

h is a function of u where it is the same at x and at 1/x, the two roots of
X^2 - u X + 1; that is read by writing its numerator and denominator modulo that
quadratic, each as beta x + alpha with alpha and beta polynomials in u, the quotient
free of x exactly where beta_N alpha_D = alpha_N beta_D, decided as a value.
 The integral in u is written back with each power of the root of Q as the same power of
S = sqrt(P)/x rather than of sqrt(Q(x + 1/x)): S is that root for a positive x
and its negative for a negative one, and the answer holds for any root whose square is Q,
so with S it holds on both sides of zero, where |x| would have to be taken out
otherwise. These were declined: no rule read the quartic's symmetry. The answer is
differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























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