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TrySplitBiquadraticOverTheReals​(AngouriMath.​Entity,​AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​AngouriMath.​Entity@,​AngouriMath.​Entity@)

 Method (no overloads)

Summary

N/D written as two fractions over the quadratic factors of a biquadraticdenominator — one with no odd power in it — or
false where it is not one, or does not split into two distinct
quadratics.

Remarks

Why this exists next to the step above. That one factors over the rationals,
and stops where the rationals do: x^4 + 1 is irreducible over Q, so it
is left whole and x^2/(x^4 + 1) has no antiderivative — the case
#233 names as
wanting "partial fractioning". Over the reals it is
(x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1), and both halves are read by the rule
for a linear numerator over a quadratic. Nothing was missing but a factorisation the
rational one is right to refuse.
Biquadratic only, and that is a real boundary rather than a first cut. A
general quartic factors into real quadratics through its resolvent cubic, whose roots
carry Cardano's nested radicals; a biquadratic x^4 + px^2 + q is the case where
the resolvent is solvable by inspection, and the two factors stay in one square root.
Two shapes come out of it, by the sign of p^2 - 4q:
Negative — no real root in x^2. Matching
(x^2 + ax + b)(x^2 - ax + b) = x^4 + (2b - a^2)x^2 + b^2 gives
b = sqrt(q) and a = sqrt(2b - p), both real because q > 0 and
p^2 < 4q forces p < 2sqrt(q). This is x^4 + 1, at
a = sqrt(2), b = 1.
Positive — two distinct real roots in x^2, so
(x^2 + M)(x^2 + N) with M, N = (p +- sqrt(p^2 - 4q))/2. Both factors are
even, and the split is two independent pairs of equations rather than four.
Zero — (x^2 + p/2)^2, a repeated quadratic, declined for the same reason
the guard above declines one: there is no rule for a numerator over
(x^2 + c)^k, so decomposing it ends in the integral it started from.
No condition is owed, on the same argument as the step above: the two factors
are distinct and coprime, so their product vanishes exactly where the original
denominator does. q > 0 is required rather than assumed, which is what keeps
b real; a negative q puts a real root in x^2 of either sign and
is left to the rational step, which reaches it whenever the root is rational.
 Reached only after the rational split has declined, so a biquadratic that factors over
Q — x^4 + 3x^2 + 2 — is decomposed there, in exact arithmetic, and never
arrives here to be given a square root it does not need.

























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