Integration of $\frac{1}{(1+x^4)^\frac{1}{4}}$ This question has been puzzling me a lot. This is in the middle list of question(meaning its moderately tough). I have tried everything i could think of. 
My try: $$\int \frac{1}{x(1+\frac{1}{x^4})^\frac{1}{4}}dx$$
$$-\frac{1}{4}\int \frac{-4x^4}{x^5(1+ \frac{1}{x^4})^\frac{1}{4}}dx$$
putting $$1 + \frac{1}{x^4} = t \iff x^4=\frac{1}{t-1} \iff dt = -\frac{4}{x^5}dx$$ and substituting in the integral,
$$-\frac{1}{4}\int \frac{1}{(t-1)t^\frac{1}{4}}dt$$ 
And Now I am stuck here.

Let $$I = \int\frac{1}{(1+x^4)^{\frac{1}{4}}}dx\;,$$ Then Put 
$$x=\sqrt{\tan z }\;\;\;, \;\; dx = \frac{\sec^2 z}{2\sqrt{\tan z}}dz$$
$$I=\int \frac{1}{\sqrt[4]{1+\tan^2 z }} \frac{ \sec^2 z } {2 \sqrt{\tan z}}dz=\frac{1}{2} \int  \sec^{ \frac{3}{2}} z \tan^{-\frac{1}{2}} zdz$$
$$=\frac{1}{2} \int \frac{1}{\cos z \sqrt{\sin z}}dz=\frac{1}{2} \int \frac{1}{(1-\sin^2 z) \sqrt{\sin z}} \cos z dz$$
$$w=\sqrt{\sin z}, dw = \frac{\cos z}{2\sqrt{\sin z}}dz$$
$$I=  \int \frac{1}{1-w^4}dw=\frac{1}{2}\int \frac{1}{1-w^2} + \frac{1}{1+w^2} dw$$
$$=\frac{1}{2} \int \frac{1}{2} \left[ \frac{1}{1-w}+ \frac{1}{1+w} \right] + \frac{1}{1+w^2} dw$$
$$= \frac{1}{4} \ln \frac{1+w}{1-w} + \frac{1}{2} \tan^{-1} w +\mathcal{C}$$
$$= \frac{1}{4} \ln \left( \frac{1+ \sqrt{\sin z} }{1- \sqrt{\sin z}}\right) + \frac{1}{2} \tan^{-1} (\sqrt{\sin z}) +\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{1+ \sqrt{ \sin ( \tan^{-1} x^2 )}}{1-\sqrt{ \sin ( \tan^{-1} x^2 )}}\right) + \frac{1}{2} \tan^{-1} \left( \sqrt{ \sin ( \tan^{-1} x^2 )} \right) +\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{1+ \sqrt{ \frac{x^2}{\sqrt{1+x^4}}}}{1- \sqrt{ \frac{x^2}{\sqrt{1+x^4}}}}\right) + \frac{1}{2} \tan^{-1} \left(  \sqrt{ \frac{x^2}{\sqrt{1+x^4}}} \right)+\mathcal{C}$$
$$=\frac{1}{4} \ln \left( \frac{\sqrt[4]{1+x^4}+ x}{\sqrt[4]{1+x^4}-x}\right) + \frac{1}{2} \tan^{-1} \left( \frac{x}{\sqrt[4]{1+x^4}} \right)+\mathcal{C}$$
