Definite integral with irrational function I wonder if it's possible solve this definite integral in terms of elemental or special functions:
$$\int_0^1\sqrt{\sqrt{x^4+a}-x^2}\,dx$$
with $a>0$. I've tried with Wolfram Mathematica but doesn't give me anything.
 A: Let's rewrite the integral as
$$\int_0^1 \sqrt{\sqrt{x^4+b^2}-x^2}\:dx = \int_0^1 \frac{1}{2}\sqrt{\sqrt{1+\frac{b^2}{y^2}}-1}\:dy$$
for $a = b^2$ and letting $x^2 = y$. Now use the substitution $y = b\operatorname{csch}t$ to get
$$\int_{\sinh^{-1}(b)}^\infty \frac{b\cosh t}{2\sinh^2 t}\sqrt{\cosh t - 1}\:dt$$ $$ = 
 \frac{b}{4\sqrt{2}}\int_{\sinh^{-1}(b)}^\infty \left[\frac{1}{\cosh\left(\frac{t}{2}\right)-1} - \frac{1}{\cosh\left(\frac{t}{2}\right)+1} + \frac{2}{\cosh^2\left(\frac{t}{2}\right)}\right]\frac{1}{2}\sinh\left(\frac{t}{2}\right)\:dt$$
$$ = \frac{b}{4\sqrt{2}}\left[\log\left(\frac{\cosh\left(\frac{t}{2}\right)-1}{\cosh\left(\frac{t}{2}\right)+1}\right) - \frac{2}{\cosh\left(\frac{t}{2}\right)} \right]_{\sinh^{-1}(b)}^\infty$$
where we went from the first line to the second with the hyperbolic double angle identities: $$\cosh t = \cosh^2\left(\frac{t}{2}\right) + \sinh^2\left(\frac{t}{2}\right) = 1+2\sinh^2\left(\frac{t}{2}\right)$$
$$\sinh t = 2\sinh\left(\frac{t}{2}\right)\cosh\left(\frac{t}{2}\right)$$
At infinity both terms vanish. To evaluate the lower bound, we'll have to use the double angle formulas in reverse.
$$\frac{b}{4\sqrt{2}}\left[\frac{2}{\cosh\left(\frac{t}{2}\right)} - \log\left(\frac{\cosh\left(\frac{t}{2}\right)-1}{\cosh\left(\frac{t}{2}\right)+1}\right) \right]$$ $$ = \frac{b}{4\sqrt{2}}\left[\frac{2}{\sqrt{\cosh^2\left(\frac{t}{2}\right)}} - \log\left(\frac{\sqrt{\cosh^2\left(\frac{t}{2}\right)}-1}{\sqrt{\cosh^2\left(\frac{t}{2}\right)}+1}\right) \right]$$
$$= \frac{b}{4\sqrt{2}}\left[\frac{2\sqrt{2}}{\sqrt{1+\cosh t}} - \log\left(\frac{\sqrt{1+\cosh t}-\sqrt{2}}{\sqrt{1+\cosh t}+\sqrt{2}}\right) \right]$$ $$ = \frac{b}{4\sqrt{2}}\left[\frac{2\sqrt{2}}{\sqrt{1+\sqrt{1+\sinh^2 t}}} - \log\left(\frac{\sqrt{1+\sqrt{1+\sinh^2 t}}-\sqrt{2}}{\sqrt{1+\sqrt{1+\sinh^2 t}}+\sqrt{2}}\right) \right]$$
Then plugging in our point we get the answer
$$\boxed{\frac{\sqrt{a}}{2\sqrt{1+\sqrt{1+a}}}+\frac{\sqrt{a}}{4\sqrt{2}}\log\left(\frac{\sqrt{1+\sqrt{1+a}}+\sqrt{2}}{\sqrt{1+\sqrt{1+a}}-\sqrt{2}}\right)}$$ $$ \longrightarrow \frac{1}{2\sqrt{1+\sqrt{2}}} + \frac{1}{2\sqrt{2}}\log\left(\sqrt{1+\sqrt{2}}+\sqrt{2}\right)\approx 0.8622$$
which agrees with the value of the integral when $a=1$
