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OverTheRootsOfTheQuadraticBelow(AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity)

Method (no overloads)

Summary

(g + h x)/(A sqrt(B)) over the two roots of A, where its discriminant is
not a number: with s = sqrt(A1^2 - 4 A0 A2) and r = (-A1 +- s)/(2 A2),
(g + h x)/A is ((g + h r1)/(x - r1) - (g + h r2)/(x - r2))/s, and each
1/((x - r) sqrt(B)) is -ln((N + 2 sqrt(B(r)) sqrt(B))/(x - r))/sqrt(B(r)) with N = 2 B0 + B1 r + (B1 + 2 B2 r) x, Rubi's artanh of N/(2 sqrt(B(r)) sqrt(B)) written with N once. Null where the discriminant is a number.

Remarks

With y = x - r, N is 2 B(r) + B'(r) y and y N' - N is
-2 B(r) - sqrt(B(r)) N/sqrt(B), so the derivative of the logarithm is
-sqrt(B(r))/(y sqrt(B)) on either branch of sqrt(B(r)), which appears
twice: the piece holds where B(r) is negative or off the real line, and where
r is. Where the roots are real, the argument of the logarithm is real and of
one sign, or off the real line, on the whole of each interval between the roots of
A and B, since (N + 2 sqrt(B(r)) sqrt(B)) (N - 2 sqrt(B(r)) sqrt(B)) is (B1^2 - 4 B0 B2) y^2, zero only at the pole: continuous on each. Where they
are a conjugate pair that is not shown; at four pinnings with B positive
everywhere, F(b) - F(a) agreed with Simpson's rule on intervals up to
[-5, 5]. B(r) is written as the
remainder of B by A at r, (P + E r)/A2 with
P = A2 B0 - A0 B2 and E = A2 B1 - A1 B2, as Rubi writes it.
The roots, s, B(r) and a compound g or h are symbols of
their own while the answer is built, written back after. Over the linears for which
A is a sum of multiples of B and their squares, Rubi's 1.2.1.5
1/(sqrt(a + b x + c x^2) (d + e x + f x^2)) was 4,982 characters, the root of the
resultant of the two in every coefficient; it is 863.
https://github.com/asc-community/AngouriMath/issues/233

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