$\int \sqrt{x^4+1}\ dx$ I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.
 A: With this antiderivative, you face elliptic integrals of the first kind.
If you are not aware of them, I prefer to just give the result
$$\int \sqrt{x^4+1}\,dx=\frac{1}{3} \left(x \sqrt{x^4+1}-(1+i) \sqrt{2} F\left(\left.i \sinh
   ^{-1}\left(\frac{(1+i) }{\sqrt{2}}x\right)\right|-1\right)\right)$$
Oherwise, use series expansions
$$\sqrt{x^4+1}=\frac 1{2\sqrt \pi}\sum_{n=0}^\infty (-1)^{n+1}\frac{ \Gamma \left(n-\frac{1}{2}\right)}{n!} x^{4n}$$ and integrate term wise.
A: By the reduction and transformation formulas mentioned in my Elliptic Integrals and Functions monograph
$$\int_0^y\sqrt{x^4+1}\,dx=\int_0^y\frac{x^4+1}{\sqrt{x^4+1}}\,dx=\frac{y\sqrt{y^4+1}}3+\frac23\int_0^y\frac1{\sqrt{x^4+1}}\,dx$$
$$=\frac{y\sqrt{y^4+1}}3+\frac23F\left(\sin^{-1}\sqrt{y^2+1-\sqrt{y^4+1}},m=\frac12\right)=\frac{y\sqrt{y^4+1}+F(2\tan^{-1}y,1/2)}3$$
In hypergeometric terms the integral is
$$y{_2F_1}(-1/2,1/4;5/4;-y^4)$$
A: I think you'll have to go to an integral table or software.  Regardless, here's the answer:
$$\frac{1}{3} \left(x \sqrt{x^4+1}-2 \sqrt[4]{-1} F\left(\left.i \sinh
   ^{-1}\left(\sqrt[4]{-1} x\right)\right|-1\right)\right) ,$$
where $F$ is an elliptic integral.
