A biquadratic denominator is decomposed over the reals (#233)

[Rafael-SOWNet]

Both existing partial fraction steps factor over Q and stop where Q does, so x^4 + 1 —
irreducible over the rationals — was left whole and x^2/(x^4 + 1) came back unevaluated. Over
the reals it is (x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1), and both halves are read by the
existing rule for a linear numerator over a quadratic. Nothing was missing but a factorisation
the rational step is right to refuse.

This is the fourth of the five integrals #233 lists, and the first of them to need a
factorisation rather than a rule.

"x^2/(x^4 + 1)".Integrate("x")

was  integral(x ^ 2 / (x ^ 4 + 1), x)
is   -1/2 * sqrt(2) * 1/2 / 2 * ln(x ^ 2 + sqrt(2) * x + 1)
     + 1/2 * arctan((2 * x + sqrt(2)) * 1/2 * sqrt(2)) * 1/2 * sqrt(2)
     + 1/2 * sqrt(2) * 1/2 / 2 * ln(x ^ 2 - sqrt(2) * x + 1)
     + 1/2 * arctan((2 * x + -sqrt(2)) * 1/2 * sqrt(2)) * 1/2 * sqrt(2) + C

1/(x^4 + 1), 1/(x^4 - 2), 1/(x^4 + 3x^2 + 1) and (x^3 + 1)/(x^4 + 1) come with it.

Biquadratic only, and why that is a 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 both factors stay inside a single square root. The sign of
p^2 - 4q picks the shape:

p^2 - 4q	factors	example
negative	(x^2 + ax + b)(x^2 - ax + b), b = sqrt(q), a = sqrt(2b - p)	x^4 + 1 at a = sqrt(2), b = 1
positive	(x^2 + u)(x^2 + v), u, v = (p -+ sqrt(p^2 - 4q))/2	x^4 - 2, x^4 + 3x^2 + 1
zero	(x^2 + p/2)^2 — a repeated quadratic, declined	x^4 + 2x^2 + 1

Each split is closed form. Matching coefficients gives two sums and two differences in the
first branch, and two independent pairs in the second, so neither needs a symbolic linear
solve. The zero case is declined for the reason the step above declines a repeated quadratic:
there is no rule for a numerator over (x^2 + c)^k, so decomposing it ends in the integral it
started from.

A quartic with an odd power in it is still declined — x^4 + x^3 + 1, x^4 + x + 1 — and
so is anything of degree five and up that does not factor over Q. Both are pinned by tests.

Where it sits, and what it costs

Tried last, after both rational steps, so a denominator that factors over Q is still taken
apart in exact arithmetic and never given a square root it does not need: x^4 + 3x^2 + 2 is
decomposed by the step above and never reaches this one.

Declining stays as cheap as it was. Every guard is rational arithmetic on coefficients already
read, and the entity work happens only for a genuine biquadratic. (1 - x^4)/(1 + x^4 + x^8) —
the case the existing guard's comment records as 18s before that guard and 203ms after —
returns the same unevaluated integral in the same fraction of a second, and there is now a test
asserting it.

Two identity checks rather than one

The step above checks one identity because its factorisation is exact by construction. Here the
factors are built by matching coefficients through a square root, so the factorisation is a
claim of its own and is checked too.

They do not imply each other, which is worth stating because it is not obvious: a wrong term
present in both factors cancels between the two halves of the numerator identity and passes
it, while the denominator identity sees it immediately.

Three tests changed their verdict

x^2/(x^4 + 1) was recorded as declined in three places. It is now answered, so all three
change — but two of them are about other code and keep their subject rather than being
weakened:

• PowerSubstitutionIntegralTest.AnIntegralThisDoesNotReachIsStillDeclined asserts the
  u = x^2 rewrite rejects the integrand. That is still true; the integrand simply stopped
  being a witness once a neighbouring step began answering it. It now uses
  x^2/(x^4 + x + 1) and x^2/(x^4 + x^3 + 1), which carry an odd power and so are out of
  reach of this step as well.
• RationalIntegralsTest.ADenominatorThatDoesNotFactorIsStillDeclined likewise.
• PartialFractionsTest.WhatCannotBeSplitIsLeftAlone drops it and gains it as an answer.

The general point, recorded in the tests: a test that something is declined stops testing its
own subject the moment any other capability answers the case. Only one of the three files was
about the code this changes.

Not in this change

sqrt(tan(x)), the last entry on #233's list, is not answered. It reduces under
u = sqrt(tan x) to 2 * integral(u^2/(u^4 + 1), u), which is now integrable — but the
substitution that gets there is a separate capability and is not built here.

Verification

Full suite: 9324 passed, 0 failed, 14 skipped. Every new answer is checked by
differentiating it back and comparing at four points, through the file's existing helper, which
asserts the antiderivative contains no integral( before comparing — without that guard the
check passes vacuously, since d/dx of an unevaluated integral(f, x) is f.

BREAKING-CHANGES.md carries the entry, and corrects the #919 entry in place: it names
x^4 + 1 as still declined, which this makes false.

Part of #233.