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SolveAnOddHalfPowerOfALinearOverAPowerOfAQuadraticWithARootAtZero(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A polynomial times an odd half power of a linear over a whole power of a quadratic with
a root at zero, P(x) (d + e x)^(j/2)/(b x + c x^2)^m with j odd and
positive and m >= 2, under t = sqrt(d + e x) with the quadratic written
as its two linears, x (b + c x), and split over them.

Remarks

With s = t^2, x = (s - d)/e, b + c x = c (s + rho)/e for
rho = r/c and r = b e - c d, and dx = 2 t dt/e, the integrand is
2 e^(2m - 1)/c^m N(s)/((s - d)^m (s + rho)^m) with
N(s) = s^((j + 1)/2) P((s - d)/e), and exactly so: t^j is
(d + e x)^(j/2) for the principal root and every complex x. That is its
quotient by the monic denominator, a polynomial in s, and its partial fractions
over s - d and s + rho, whose coefficients are those of the Taylor series
of N over the other power at each root. The roots are b e/c apart, which
is not zero. Each linear's fractions are asked as one question in t. Each of
d, b e/c, e and the coefficients of P that is compound is
named by a symbol of its own and written back into the answer, as in
SolveCubeRootsOfTwoLinearsBesideAWholePowerOfAThird(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), and rho is
written in them, so that a power of x in P cancels.
Rubi's (A + B x) (d + e x)^(j/2)/(b x + c x^2)^3 ran past the budget: the
substitution that reaches it is late in the chain and multiplies the quadratic out.
Asked as one rational function in t, the second power came back in up to 940,000
characters and 11 GB, (d + e x)^(7/2)/(b x + c x^2)^2 among them, and the third
in 10,000 to 66,000; split, each is under 7,000 characters, in megabytes. The first
power and numbers for coefficients are answered compactly as they are, and left to
that, and so is the root below the bar, which puts a power of s below it too.
Declined where d or r is zero as a value, the linear then sharing a root
with the quadratic.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

Table of contents

AngouriMath

SolveAnOddHalfPowerOfALinearOverAPowerOfAQuadraticWithARootAtZero​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

SolveAnOddHalfPowerOfALinearOverAPowerOfAQuadraticWithARootAtZero(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)