<s>[INST] Evaluate $\int {\sqrt{x^4+1}}dx$ Evaluate $$\int \sqrt{x^4+1}dx$$
My Try: I used parts as follows:
$$I=x{\sqrt{x^4+1}}-4 \int \frac{x^4 dx}{\sqrt{x^4+1}}$$ $\implies$
$$I=\frac{x}{\sqrt{x^4+1}}-4I+4 \int \frac{dx}{\sqrt{x^4+1}}$$ $\implies$
$$5I=\frac{x}{\sqrt{x^4+1}}+4J$$ where
$$J=\int \frac{dx}{\sqrt{x^4+1}}$$
any clue here?
 [/INST] To get
$$I = \int\sqrt{x^4+1}\,\mathrm{d}x=\int\frac{x^4+1}{\sqrt{x^4+1}}\,\mathrm{d}x.$$
First use partial fractions to get
$$
I=x{\sqrt{x^4+1}}-2\int x \mathrm{d}\sqrt{x^4+1}=x{\sqrt{x^4+1}}-2 \int \frac{x^4+1-1}{\sqrt{x^4+1}}\mathrm{d}x
$$
and rewrite as
$$
I=x{\sqrt{x^4+1}}-2I+2 J,\quad \mathrm{where}\quad J=\int \frac{1}{\sqrt{x^4+1}}\mathrm{d}x
$$

let $t^4=-x^4$, solved
$$
\begin{aligned}
t &= e^{\pi i / 4} x\\
x &= e^{-\pi i / 4} t\\
1-x^4&=1 - t^4= (1-t^2)(1- i^2t^2)\\
\end{aligned}
$$
So
$$
J=  \int \frac{1}{\sqrt{1+x^4}}\mathrm{d}x = \int \frac{e^{-\pi i/4}}{\sqrt{1-t^4}}\mathrm{d}x=e^{-\pi i/4}F(\arcsin t; i^2)
$$
where F(x;k) is Elliptic Integral of the First Kind.
$$
F(x;k^2)=\int _{0}^{\sin x}{\frac {{\rm {d}}t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}
$$

Then
$$
\begin{aligned}
I
&=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2}{3}J\\
&=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2e^{-\pi i/4}}{3}F(\arcsin e^{\pi i/4} x; -1)\\
\end{aligned}
$$
Finally, take the derivation and verify that the result is correct
D[x Sqrt[x^4 + 1] / 3 + 2 Exp[-Pi I / 4] EllipticF[ArcSin[Exp[Pi I / 4] x], -1] / 3, x] // FullSimplify

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