$\int \frac{d x}{\sqrt[3]{x^3-3 x^2-3 x-1}}$

hbghlyj

blog.wolfram.com/2021/08/18/new-methods-for-computing-algebraic-integrals/
For example, the following integral possesses an elementary (albeit enormous) solution:$$\int \frac{d x}{\sqrt[3]{x^3-3 x^2-3 x-1}}$$

ljh25252

$$
\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt[3]{x^3-3x^2-3x-1}}&=\frac{3}{2}\int\frac{1+2x}{3x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x-\frac{1}{2}\int\frac{1}{x\sqrt[3]{x^3-3x^2-3x-1}}\mathrm{d}x\\
&=\frac{3}{2}\int\frac{\mathrm{d}\left(\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)}{1-\left(\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)^3}-\frac{1}{2}\int\frac{\mathrm{d}\left(\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)}{1-\left(\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}\right)^3}\\
&=\frac{3}{2}\int\frac{\mathrm{d}p}{1-p^3}-\frac{1}{2}\int\frac{\mathrm{d}q}{1-q^3}\\
&\color{gray}{p=\frac{x-1}{\sqrt[3]{x^3-3x^2-3x-1}}}\quad \color{gray}{q=\frac{-x-1}{\sqrt[3]{x^3-3x^2-3x-1}}}
\\
\color{gray}{\int\frac{\mathrm{d}z}{1-z^3}}&\color{gray}{=\frac{1}{\sqrt{3}}\arctan\frac{1+2z}{\sqrt{z}}+\frac{1}{3}\ln\frac{\sqrt{1+z+z^2}}{1-z}}\\
\end{aligned}
$$