Integral of $$$\frac{x^{4}}{\sqrt{1 - x^{4}}}$$$

Find $$$\int \frac{x^{4}}{\sqrt{1 - x^{4}}}\, dx$$$.

This integral does not have a closed form:

$${\color{red}{\int{\frac{x^{4}}{\sqrt{1 - x^{4}}} d x}}} = {\color{red}{\left(\frac{x^{5} {{}_{2}F_{1}\left(\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle| {x^{4}} \right)}}{5}\right)}}$$

Therefore,

$$\int{\frac{x^{4}}{\sqrt{1 - x^{4}}} d x} = \frac{x^{5} {{}_{2}F_{1}\left(\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle| {x^{4}} \right)}}{5}$$

Add the constant of integration:

$$\int{\frac{x^{4}}{\sqrt{1 - x^{4}}} d x} = \frac{x^{5} {{}_{2}F_{1}\left(\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle| {x^{4}} \right)}}{5}+C$$

$$$\int \frac{x^{4}}{\sqrt{1 - x^{4}}}\, dx = \frac{x^{5} {{}_{2}F_{1}\left(\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle| {x^{4}} \right)}}{5} + C$$$A