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SolveABinomialDifferential(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A binomial differentialx^m (a + b x^n)^(p/q), in the two of Chebyshev's
three cases that are not a whole power: (m + 1)/n whole, or
(m + 1)/n + p/q whole.

Remarks

x^2 sqrt(1 + x^3) came out and x^5 sqrt(1 + x^3) did not, and the two are
the same substitution. Under u = (a + b x^n)^(1/q) the first leaves a monomial
in u, which the general substitution finds because x^(n-1) is du up to a constant; the second leaves a polynomial, which it does not, because
what multiplies du there is x^3 rather than a constant.
Writing s = (m + 1)/n and substituting:

x^n = (u^q - a)/b       x^m dx = (q/(n b)) ((u^q - a)/b)^(s-1) u^(q-1) du
int x^m (a + b x^n)^(p/q) dx = (q/(n b)) int ((u^q - a)/b)^(s-1) u^(p+q-1) du

which is a polynomial in u when s is a whole number of at least one,
expanded by the binomial theorem and integrated term by term; and a rational
function of u when s is a whole number of at most zero, which the
rational integrator answers — sqrt(1 + x^3)/x is (2/3) int u^2/(u^2 - 1) du,
and 1/(x sqrt(1 - x^3)) is (2/3) int 1/(u^2 - 1) du. Both were declined
for s = 0, which the rule read as "not a polynomial" and left at that.
The third case comes down to the second. Where s + p/q is whole instead,
x = 1/y turns x^m (a + b x^n)^(p/q) dx into
-y^m' (b + a y^n)^(p/q) dy with m' = -m - 2 - n p/q, a whole number, and
(m' + 1)/n = -(s + p/q), whole — so it is the second case in y, with the
roles of a and b exchanged, and y = 1/x put back afterwards.
x^6 (3 + 4x^4)^(1/4) and (x^3 - 1)/(2 + x^3)^(1/3) are this. Chebyshev
proved there is no fourth case: outside these the integrand has no elementary
antiderivative at all, which is worth knowing before anyone goes looking.
Closed in one step: the polynomial case is a sum of powers, and the rational case
goes to the rational integrator directly, not back into the chain -- which is what
lets it be volunteered at any depth rather than asked at the top only
(#1265):
tan(x)/sqrt(1 + sec(x)^3) is 1/(u sqrt(1 + u^3)) under u = sec(x),
one level down, and was declined there.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveABinomialDifferential(AngouriMath.Entity,AngouriMath.Entity.Variable)