Evaluate $\int \dfrac{1}{\left(1+x^4\right)\left(\sqrt{\sqrt{1+x^4}}-x^2\right)}dx$
My attempt is as follows:-
$$x^2=\tan\theta$$ $$2xdx=\sec^2\theta d\theta$$ $$dx=\dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$$
$$\int \dfrac{1}{\left(1+\tan^2\theta\right)\left(\sqrt{\sec\theta}-\tan\theta\right)}\cdot\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta$$
$$\dfrac{1}{2}\cdot\int\dfrac{d\theta}{\sqrt{\tan\theta}\left(\sqrt{\sec\theta}-\tan\theta\right)}$$ $$\dfrac{1}{2}\cdot\int\dfrac{(\cos\theta)^\frac{3}{2} d\theta}{\sqrt{\sin\theta}\left(\sqrt{\cos\theta}-\sin\theta\right)}$$
$$\sqrt{\sin\theta}=y$$ $$\dfrac{\cos\theta}{2\sqrt{\sin\theta}}d\theta=dy$$ $$\int \dfrac{\sqrt{1-y^4}}{\sqrt{1-y^4}-y^2}dy$$
$$\int \dfrac{\sqrt{\dfrac{1}{y^4}-1}}{\sqrt{\dfrac{1}{y^4}-1}-1}dy$$
How to proceed from here or feel free to suggest shorter and clean approach
In the same spirit as Mariusz Iwaniuk's comment, a CAS gives $$\int \dfrac{\sqrt{1-y^4}}{\sqrt{1-y^4}-y^2}dy=\frac{1}{3} y^3 F_1\left(\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};y^4,2 y^4\right)+\frac{1}{8} \left(4 y+2^{3/4} \left(\tan ^{-1}\left(\sqrt[4]{2} y\right)+\tanh ^{-1}\left(\sqrt[4]{2} y\right)\right)\right)$$ where appears the Appell hypergeometric function of two variables.