A constant over a linear beside the root of a quadratic is answered by the reciprocal of the linear, and Euler's substitution reaches a radical that is real only between its roots

[Rafael-SOWNet]

K/((x - p) sqrt(Q)) was the Euler substitution's, answered at length as a
partial-fraction decomposition in its t, and declined outright where the
leading coefficient and the constant are symbols it cannot take a real root
of: Hearn's 1/(r sqrt(-alpha^2 - epsilon^2 + 2h r^2 - 2k r^4)) is
1/(2u sqrt(-alpha^2 - epsilon^2 + 2h u - 2k u^2)) under u = r^2, with
-2k in front and -alpha^2 - epsilon^2 behind, and had no antiderivative.

With t = 1/(x - p), Q(p + 1/t) is (Q(p) t^2 + Q'(p) t + a)/t^2, so the
root is sqrt(R(t))/|t| for a quadratic R in t alone, dx is -dt/t^2,
and the integrand is -K sgn(t) dt/sqrt(R(t)) -- the table's, an arcsine
or a logarithm by the sign of Q(p), a piecewise where that sign is a
symbol's. The sign of t is the sign of x - p, and
-K sgn(x - p) G(1/(x - p)) is an antiderivative on both sides of p,
exactly. Closed, so volunteered at any depth, before Euler:
1/((x + a) sqrt(x^2 + b x + c)) is three arms of a line each where it was
two kilobytes of partial fractions in t, and Hearn's is Rubi's arcsine.

Three things stood in the way one level down.

A^(-1) written as one quotient was 1^1/A^1, and nothing reading the
factors of the denominator took A^1 for A; a first power is the base.

Euler's substitution declined a < 0 beside c < 0 as a radical nowhere
real, and it is real between the roots where there are two: shifted to the
vertex, x = y - b/(2a), the quadratic is a y^2 + c - b^2/(4a), the
constant positive exactly when the roots are, and that is the second
substitution's. 1/(x sqrt(-5 + 10x - 4x^2)) is one, Hearn's numerically.

And the third substitution can make a polynomial of the integrand --
1/(x^2 sqrt(3x - x^2)) is -2(t^2 + 1)/9 -- which none of the rational
readers behind it took; a polynomial in t goes to the chain, which
answers one in a step. Behind that, the table's rule for k/(a x^2 + b x + c)
answered NaN for a constant read as k/(0 x^2 + 0 x + 3): a zero a gives
a zero discriminant, and the perfect-square arm divided by 2 a x + b. A
decidably linear denominator is the linear case and nothing else.

u = r^2 reaches Hearn's only with the two bare r in 1/(r sqrt(P))
over 2r collected into r^2, which was done for numeric integrands of
modest size only -- a symbolic partial fraction written as one quotient is
a page, and its simplification took twelve minutes of one test. A symbolic
integrand of forty nodes or fewer is collected the same way now, and
Hearn's is one.

Rubi sample (463 problems, 5 s budget), against the #1313 master this branch is cut from:

master at #1313     405/463, 0 wrong, 0 timeouts, ~100 s
with this           405/463, 0 wrong, 0 timeouts, 102 s

Hearn 255 is answered in 0.5 s and the harness counts it as unverifiable rather than answered: its own pins for alpha, epsilon, h and k make the radical complex at its sample points. The test checks it at parameters where the radical is real, on both signs of Q(0)'s piecewise. No other problem moved by more than noise (re-measured solo; the first run had a build beside it).

Composes with #1314, merged meanwhile: this touches the solver, the table, the chain and SingleQuotient; that one the elimination and two other test files.

The five test suites are green.

Part of #718.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura