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SolveAPolynomialTimesAnOddHalfPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A polynomial times an odd half power of a quadratic, P Q^(m/2), reduced to
R Q^(k + 1/2) + K/sqrt(Q) by one linear solve, with K/sqrt(Q) the
table's -- an arcsine or a logarithm by the sign of the leading coefficient, a
piecewise where that sign is a symbol's.

Remarks

For m = 2j - 1 at least -1 the ansatz is F = R sqrt(Q), and
F' = P Q^j/sqrt(Q) - K/sqrt(Q) is R' Q + R Q'/2 + K = P Q^j; for
m = -2j - 3 it is F = R Q^(-j - 1/2), and the identity
R' Q - (j + 1/2) R Q' + K Q^(j + 1) = P. Each is one equation per power of
x, linear in the coefficients of R and in K, exact; a solution is an
answer and its absence a decline.
Hearn's r/sqrt(-alpha^2 - 2k r + 2pe r^2), Stewart's
x^2/(a^2 - x^2)^(3/2) and Apostol's (a^2 - x^2)^(5/2) had no
antiderivative: with a symbol in the radicand the trigonometric substitution has
no sign to go on and Euler's takes a root of it, while 1/sqrt(Q) alone was
answered, as a piecewise on the sign of the leading coefficient, with
arcsin(x/sqrt(a^2)) right for either sign of a. The reduction is what
every table does before that line. Numeric coefficients reach this only from a
sub-integral, since the substitutions in front answer them.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

SolveAPolynomialTimesAnOddHalfPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)