Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intuitive substitution.

Rewrite the integral as
$$\int \frac{x\:dx}{x^2(1+x^4)\sqrt{\frac{\sqrt{1+x^2}-x^2}{x^2}}}$$
by pulling out an $x^2$ from the square root then multiplying the top and bottom by $x$.
Now let $x^2 = \sinh( t) \implies xdx = \frac{1}{2}\cosh(t) dt$:
$$\int \frac{\left(\frac{1}{2}\cosh(t)\right)\:dt}{\sinh(t)(1+\sinh^2(t))\sqrt{\frac{\sqrt{1+\sinh^2(t)}-\sinh(t)}{\sinh(t)}}} = \int \frac{\sqrt{e^t\sinh(t)}\:dt}{\sinh(2t)} = \sqrt{2}\int \frac{\sqrt{e^{2t}-1}}{e^{2t}-e^{-2t}}dt$$
Now let $e^t = \sec\theta \implies dt = \tan\theta \: d\theta$:
$$\sqrt{2}\int \frac{\sqrt{\sec^2\theta - 1}\tan\theta}{\sec^2\theta - \cos^2\theta}\:d\theta = \sqrt{2}\int \frac{\sin^2\theta}{1-\cos^4\theta}\: d\theta = \int \frac{\sqrt{2}\:d\theta}{1+\cos^2\theta}$$
$$= \int \frac{\sqrt{2}\:d\theta}{2\cos^2\theta+\sin^2\theta} = \int \frac{\sqrt{2}\sec^2\theta \:d\theta}{2 + \tan^2\theta} = \tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right) + C$$
Now undoing the substitutions, 
$$\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{e^{2t}-1} = \sqrt{2e^t \sinh(t)}$$
which can go in one of two directions, either
$$\sqrt{2(\cosh(t)+\sinh(t))\sinh(t)} = \sqrt{2(\sqrt{1+\sinh^2(t)}+\sinh(t))\sinh(t)} = x \sqrt{2(\sqrt{1+x^4}+x)}$$
$$\implies I = \tan^{-1}\left(x\sqrt{\sqrt{1+x^4}+x^2} \right) + C$$
or
$$\sqrt{\frac{2\sinh(t)}{\cosh(t)-\sinh(t)}} = \sqrt{\frac{2\sinh(t)}{\sqrt{1+\sinh^2(t)}-\sinh(t)}} = x\sqrt{\frac{2}{\sqrt{1+x^4}-x^2}}$$
$$\implies I = \tan^{-1}\left(\frac{x}{\sqrt{\sqrt{1+x^4}-x^2}} \right) + C$$
the latter of which matches wolfram's output, but both are equivalent.
