what's the best way to integrate $\int \frac{\sqrt{x}}{\sqrt[4] {x^3 +1}}dx$?

Question

I have to integrate $$\int \frac{\sqrt{x}}{\sqrt[4]{x^3+1}} \,dx$$

To solve this, I've tried various substitutions including a) let $x^{3/2}$=$e^{2u},$ which leads to the following integral $$\frac{4}{3} \int \frac{e^{2x}}{\sqrt [4] {e^{4x}+1}}du$$ which, according to integral calculator.net, has no closed form. I also tried the following $u$-sub let $u=x^{3/2}.$ However that didn't help much either, so what is the sub, if any that I must use here.

Edit:- My book does give the answer, but not the method, @greg Martin suggested I add that as it may help, so here it is

$\frac{4}{3} (\sqrt[4] {x^3} -\ln(\sqrt[4]{x^3} +1)) +C $

This problem is from N.Piskunov's Differential and integral calculus , Volume 1

Answer 1

The only solution is given in terms of Gauss hypergeometric function. $$\frac 1{\sqrt[4]{t+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} t^n$$ $$\frac{\sqrt{x}}{\sqrt[4]{x^3+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{3 n+\frac{1}{2}}$$ $$\int \frac{\sqrt{x}}{\sqrt[4]{x^3+1}} \,dx=\frac 2 3 x^{3/2}\sum_{n=0}^\infty \frac{\binom{-\frac{1}{4}}{n} }{2 n+1} x^{3 n}=\frac{2}{3} x^{3/2} \,\, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};-x^3\right)$$

Answer 2

Your integral is of a type of integrals known as the differential binomial:

$$ \int x^m(a+bx^n)^p\,dx\qquad a,b\in\mathbb{R}, \,m, n, p\in\mathbb{Q}$$ The change of variables $x=t^{1/n}$ yields $$ \int x^m(a+bx^n)^p\,dx = \frac1n\int (a+bt)^p t^{\tfrac{m+1}{n}-1}\,dt $$

P. L. Chebyshev studied these integrals and proved that there are only three cases in which they are solvable, in the sense that a primitive (antiderivative) can be found in terms of elementary functions:

1. $p\in\mathbb{Z}$: If $q=\frac{m+1}{n}-1=\frac{\alpha}{\beta}$, $\alpha,\beta\in\mathbb{Z}$ and $g.c.d(\alpha,\beta)=1$, then the substitution $z=t^{1/\beta}$ reduced the integral to one of a rational fraction.

2. $q\in\mathbb{Z}$: the substitution $z=(a+bt)^{1/\beta}$ reduced to integral to one of a rational fraction.

3. $p+q\in\mathbb{Z}$. The change of variable $z=\Big(\frac{a+bt}{t}\Big)^{1/\beta}$ reduces the integral to one of a rational fraction

In your case, the substitution $x=t^{1/3}$ ($\frac13 t^{-2/3} dt=dx$) yields to $$\int x^{1/2}(x^3+1)^{-1/4} \,dx=\frac13\int(1+t)^{-1/4}t^{-1/2}\,dt $$ None of the conditions (1)-(3) hold; hence there is no elementary primitive.