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

 Method (no overloads)

Summary

A polynomial in x^2 over a + b x^2 + c x^4 with a symbol in it, by the
two roots in x^2: with q = sqrt(b^2 - 4 a c) and
r = (-b ± q)/(2c), (d + e x^2)/(a + b x^2 + c x^4) is
(d + e r_1)/(q (x^2 - r_1)) - (d + e r_2)/(q (x^2 - r_2)), and
1/(x^2 - r) is atan(x/sqrt(-r))/sqrt(-r) for any complex r. A
higher even degree is divided down first, and an odd numerator is left to the
substitution u = x^2. And over x^(2k) times the biquadratic, a power of
x on both sides cancelled first: in w = x^2 the remainder R over
w^k Q is the expansion of R/Q at w = 0 up to w^(k-1), over
w^k, plus (d + e w)/Q, so the terms in x^(-2j) go by the power
rule and the rest as above. Split in x, the repeated factor x^(2k) beside
the quartic went to the Hermite reduction or to the conjugates of a sum of two squares,
and 2 a (1 - b t^2)^4/(t^4 (1 - 2 b t^2 + (a^2 + b^2) t^4)), which the root of the
quotient makes of Rubi's cot(c + d x)^(5/2)/(a + b tan(c + d x))^(3/2), was
declined after seconds in one spelling and answered at length in the other.

Remarks

The partial fractions read written factors, and a + b x^2 + c x^4 is written
as one; over the rationals it is factored, over symbols it was declined, and Rubi's
quartic files -- (d x)^m (a + b x^2 + c x^4)^p and the next four -- lost the
even numerators to it: x^2/(a + b x^2 + c x^4), (d + e x^2)/(...).
Exact wherever the two roots differ, which is the generic case; a discriminant
that is zero as written declines, the square of a quadratic being another shape.
https://github.com/asc-community/AngouriMath/issues/718

























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