Evaluate the integral $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$ $$x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}=\frac{\sqrt{(-\sqrt[3]{x^2}+1)}}{\sqrt[3]{x^4}}$$
Is it necessary to simplify the function further? What substitution is useful?
$u=\sqrt[n]{\frac{ax+b}{cx+d}}$ doesn't work.

Following RecklessReckoner's comment:
We have $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$.
Now, let $u=x^{1/3}$ and $3u \mathrm du=\mathrm dx$
This gives $3\int u^{-4}(-u^{2}+1)^{\frac{1}{2}}u\mathrm du=3\int u^{-3}(-u^{2}+1)^{\frac{1}{2}}\mathrm du=3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du.$
Next, we perform another substitution: let $u=\sin t$ and $\mathrm du=\cos t \mathrm dt$
$3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du=3\int {\sin^{-3} t}(1-{\sin}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt=3\int {\sin^{-3} t}({\cos}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt$
$=3\int {\sin^{-3} t} \,{\cos}^{2} t\mathrm dt=3\int \frac{\cos^2t}{\sin^3 t} \mathrm dt=3\int {\cot^2t} \,{\csc t} \mathrm dt$
We can use the identity $\csc^2 t-1= \cot^2 t.$
$3\int {\cot^2t} \,{\csc t} \mathrm dt=3\int (\csc^2 t-1)\csc t \mathrm dt=3\int \csc^3 t-\csc t  \mathrm dt=3\int \csc^3 t \mathrm dt- \int \csc t  \mathrm dt.$
For the second term use the fact that $\int \csc t \mathrm dt =-\log|\csc t+\cot t| \mathrm dt+C$
For the first term, use the reduction formula $$\int \csc^m t \mathrm dt= \frac{(-\cos t)(\csc^{m-1} t)}{m-1}+\frac{m-2}{m-1}\int \csc^{m-2} t \mathrm dt$$
with m=3.
Then, substitute back for $t$ and $u$. Can you take it from here? 
