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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 and
(m' + 1)/n = -(s + p/q), whole — so it is the second case in y, with the
roles of a and b exchanged. Its u, whose q-th power is
b + a/x^n, is put back as (a + b x^n)^(1/q)/x^(n/q), which has that power
at every x. (b + a/x^n)^(1/q) is the same number for a positive x,
and for a negative one only under an odd root, which is real here:
1/(1 + x^4)^(5/4) came out as 1/(1 + 1/x^4)^(1/4), an even function
whose derivative is the integrand's negative for every negative x, where
x/(1 + x^4)^(1/4) is right on the whole line.
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.
The theorem is about the exponents, and m and n are read as rationals and
a and b as anything free of the variable, since neither substitution
reads the coefficients: x^(7/3) (a + b x^2)^(1/3) is the third case. A power of
a multiple of the variable, (c x)^(5/2), is read as x^(5/2) times
(c x)^(5/2)/x^(5/2), whose derivative is zero: it is c^(5/2) for a
positive c x and constant on either side of zero whatever the signs, so it
stands outside the integral as written. Rubi's 1.1.2.2 and 1.1.3.2,
(c x)^m (a + b x^n)^p, are written so.
Where m + 1 + n (p/q + 1) is zero, a case of the third, the answer is one product
of powers, x^(m+1) (a + b x^n)^(p/q+1)/(a (m + 1)), and is given as that.
SolveAsTheDerivativeOfAProductOfPowers(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) finds it too, but is asked at the
top and one level below only, and below that the third case wrote it out through the
rational integrator, many terms where one will do.
Here with numbers only. With a symbol in a, b or the multiple it is
SolveABinomialDifferentialWithSymbols(AngouriMath.Entity,AngouriMath.Entity.Variable), asked after the rules written for
a root of a quadratic and for a rational function of x^n beside the root of its
binomial, which answer what they share with this more shortly: asked first,
1/(a - b x^4)^(1/4) came out as two logarithms and two arctangents where
SolveByDividingByTheRoot(AngouriMath.Entity,AngouriMath.Entity.Variable) gives two arctangents.
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

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AngouriMath

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

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