<s>[INST] Computing $I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$ I am trying to solve this integral and find its primitive, unfortunately I have not been successful. I made some variable changes, but I think it has become even more complicated.
My solution is as follows:
$$I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$$
$$u=x+\frac{1}{2}\iff u+\frac{1}{2}=x+1$$
$$du=dx$$
$$I=\int \frac{du}{(u+\frac{1}{2})(u^2+\frac{1}{12})^{1/3}}$$
$$\alpha=\frac{1}{2}$$
$$I(\alpha)=\int \frac{(u-\alpha)}{(u^2-\alpha^2)(u^2+\frac{\alpha^2}{3})^{1/3}}du$$
$$I(\alpha)=\frac{1}{3^{1/3}}\int \frac{(u-\alpha)}{(u^2-\alpha^2)(3u^2+\alpha^2)^{1/3}}du$$
I hope you can help me. Regards
 [/INST] \begin{gather*}
\int \frac{dx}{( x+1)\left( 3x^{2} +3x+1\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left( x^{3} +3x^{2} +3x+1-x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left(( x+1)^{3} -x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)^{2}\left( 1-\left(\frac{x}{x+1}\right)^{3}\right)^{\frac{1}{3}}}\\
Let\ \frac{x}{x+1} =t\\
\frac{dx}{( x+1)^{2}} =dt\\
The\ integral\ becomes\\
\int \left( 1-t^{3}\right)^{\frac{-1}{3}} dt
\end{gather*}
Does this help in some way?
Edit: For solving the last integral, just change it to
\begin{equation*}
-\int \frac{dt}{\left( t^{3} -1\right)^{\frac{1}{3}}}
\end{equation*}
and make a substitution
\begin{equation*}
u=\frac{\left( t^{3} -1\right)^{\frac{1}{3}}}{t}
\end{equation*}
After some manipulation, you will get an integrand that can be solved by partial fractions.
</s>