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

  • 4
    $\begingroup$ Is it $$\sqrt{\sqrt{1+x^4}-x^2}$$ $\endgroup$ – lab bhattacharjee Dec 26 '19 at 7:27
  • $\begingroup$ no,its the same I wrote $\endgroup$ – user3290550 Dec 26 '19 at 7:33
  • 5
    $\begingroup$ The reason we ask is because $\sqrt{\sqrt{x}}$ is an atrocious notation that could've easily been simplified to $\sqrt[4]{x}$ so it seems suspicious that it would be there like that. Second, the square root over both makes it so that the powers match up with the $x^2$. Lastly, this integral has no elementary antiderivative, but the altered version does. $\endgroup$ – Ninad Munshi Dec 26 '19 at 7:39
  • $\begingroup$ Mathematica+Rubi can.Solution very large with Elliptic function and another elementary stuff. $\endgroup$ – Mariusz Iwaniuk Dec 26 '19 at 10:48
  • $\begingroup$ shouldn't the second-to-last line of your attempt have fourth-roots instead of square-roots, i.e., $\int \dfrac{\sqrt[4]{1-y^4}}{\sqrt[4]{1-y^4}-y^2}dy$ ? $\endgroup$ – David H Dec 28 '19 at 8:16

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.

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