$\int \frac{(1+x^4)}{{(1-x^4)}^{3/2}}dx=$

$=\int \frac{\left(1+x^4\right)}{{(1-x^4)}^{3/2}}dx$

$=\int \frac{\left(1+x^4\right)dx}{x^3{\left(\frac{1}{x^2}-x^2\right)}^{3/2}}$

$\begin{array}{l}=\int \frac{\left(\frac{1}{x^3}+x\right)dx}{{\left(\frac{1}{x^2}-x^2\right)}^{3/2}}dx\\ put\frac{1}{x^2}-x^2=t\\ (\frac{1}{x^3}+x)dx=\frac{-1}{2}dt\\ G.I=\int \frac{\frac{-1}{2}dt}{t^{{\displaystyle \frac{3}{2}}}}\\ =\frac{-1}{2}\int t^{-\frac{3}{2}} dt\\ =\frac{-1}{2}\frac{\left(t^{{\displaystyle \frac{-1}{2}}}\right)}{\frac{-1}{2}}+C\\ =\frac{1}{\sqrt{t}}+C\\ =\frac{1}{\sqrt{{\displaystyle \frac{1}{x^2}}-x^2}}+C\end{array}$