Solve an integral $\int\left(\frac{x^{1/3}-1}{x}\right)^{5/3}dx$

Solve an integral $$\int\left(\frac{x^{1/3}-1}{x}\right)^{5/3}dx$$

I am using Chebyshev’s criterion (integrals of differential binomials), but can’t figure out how to use it properly.

Integrand function is of the form $x^m(a+bx^n)^p$ where $a,b$ are coefficients and $m,n,p$ are rational numbers.


Here $a=-1,b=1,m=-5/3,n=1/3$.

Since $$\frac{m+1}{n}$$ is an integer, we use substitution $$a+bx^n=t$$

Using this substitution gives an integral
$$3\int t^{5/3}(1+t)^{-3}dt$$

Now $p=-3$ is an integer, so we use the first case in Chebyshev’s criterion.
I don’t understand how to use this case.

It says: Expanding $(a+bt^n)^p$ by the binomial formula we write the integrand as a rational function of the simple radicals $t^{j/k}$. Then the substitution $t=u^r$, where $r$ is the largest of all denominators $k$ will remove the radicals entirely.

How to use this case?

Reference: Chebyshe

Solutions Collecting From Web of "Solve an integral $\int\left(\frac{x^{1/3}-1}{x}\right)^{5/3}dx$"

If you set $u^3=x^\frac{1}{3}-1$ or equivalently $u^3=t$ you get $\int \frac{9u^7}{(u^3+1)^3}du$.


Happy integration…

Note: maybe there is a better method than partial fraction expansion at this stage, but you’ll have to read “Manuel Bronstein – Symbolic Integration I”.

Edit: I’ve now read the interesting part of this book, let’s apply the Horowitz-Ostrogradsky algorithm !








And let’s identify to the null polynom $\forall u, H(u)=A-B\,’D^*+BD^*{D^{-}}’/D^–CD^-=0$


This system solves to $(b_i)_i=(0,0,\frac{-5}{2},0,0,-4)$ and $(c_i)_i=(0,5,0)$

$$\int \frac{9u^7}{(u^3+1)^3}du=\frac{B}{D^-}+\int\frac{C}{D^*}du=\frac{-\frac52u^2-4u^5}{(u^3+1)^2}+\int\frac{5u}{u^3+1}du$$

The last part is solved classically:

$\int=-\frac53\ln(u+1)+\frac56\ln(u^2-u+1)+\frac{5}{\sqrt 3}\arctan(\frac{2u-1}{\sqrt 3})$