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SolveALinearOverAQuadraticBesideTheRootOfAnother​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

A linear over a quadratic beside the square root of another quadratic,
(g + h x)/(A sqrt(B)), in closed form, symbols in the coefficients included:
two arctangents of L/sqrt(B), each for a linear L for which A is
a sum of multiples of B and L^2.

Remarks

For L = lambda + mu x, u = L/sqrt(B) has du = N dx/(2 B^(3/2)) with N = (2 mu B0 - lambda B1) + (mu B1 - 2 lambda B2) x, and
alpha + beta u^2 is (alpha B + beta L^2)/B. Where
alpha B + beta L^2 = rho A, then, N dx/(A sqrt(B)) is
2 rho du/(alpha + beta u^2). The three equations that says, one for each power
of x, have a solution where their determinant is zero, which is a quadratic in
lambda : mu, E lambda^2 - 2 P lambda mu + F mu^2 with
P = A2 B0 - A0 B2, E = A2 B1 - A1 B2 and F = A1 B0 - A0 B1. Its two
roots give two such numerators, independent unless q = sqrt(P^2 - E F) is zero
or B is a square, and g + h x is a sum of the two.
 Each piece is 2 rho arctan(k u)/(alpha k) with k = sqrt(beta/alpha). Its
derivative is 2 rho/(alpha + alpha k^2 u^2), and alpha k^2 is beta on either branch of the root, so the piece holds where beta/alpha is negative
too, as the hyperbolic arctangent it is there.
 Rubi's 1.2.1.6 closes these the same way, and its 4.3.9 and 4.4.9 come to them under
the tangent: 1/sqrt(a + b tan(x) + c tan(x)^2) is
1/((1 + t^2) sqrt(a + b t + c t^2)), for which q is
sqrt((a - c)^2 + b^2). That is the nested surd the rotation of the tangent
takes only as a number, since the rotation hands the rational integrator a quotient
over it; here it is one root in a closed form.
https://github.com/asc-community/AngouriMath/issues/718

























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