<s>[INST] $\int \sqrt{x^4+1}\ dx$ I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.
 [/INST] 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
$$y{_2F_1}(-1/2,1/4;5/4;-y^4)$$
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