Evaluation of Integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ Calculate  integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ This was my exam question. I've tried many online math solvers and math programs but none were able to solve. If anybody has an answer would be helpful. Thanks
 A: $$I = \int\frac{x^2+1}{\sqrt{x^3+3}}dx=\underbrace{\int \frac{x^2}{\sqrt{x^3+3}}dx}_{I_1}+\underbrace{\int \frac{1}{\sqrt{x^3+3}}dx}_{I_2}$$

$$I_1 = \int \frac{x^2}{\sqrt{x^3+3}}dx$$
u-substitution $u=x^3$
$$\int \frac{1}{3\sqrt{u+3}}du  =\frac{1}{3}\int \frac{1}{\sqrt{u+3}}du$$
u-substitution $v=u+3$
$$I_1 = \frac{1}{3} \int \frac{1}{\sqrt{v}}dv = \frac{1}{3}\int v^{-\frac{1}{2}}dv = \frac{1}{3}\frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C$$
Revert the substitutions $v=u+3,u=x^3$
$$I_1 = \frac{2}{3}\sqrt{x^3+3} + C$$

Now
\begin{align}
I_2 &= \int \frac{1}{\sqrt{x^3+3}}dx\\
 &= \int \frac{1}{\sqrt{3\left(\frac{x^3}{3}+1\right)}}dx\\
 &= \int \frac{1}{\sqrt{3\left(\left(\frac{x}{\sqrt[3]{3}}\right)^3+1\right)}}dx
\end{align}
Set $u = \frac{x}{\sqrt[3]{3}} \implies dx = \sqrt[3]{3}\, du$, then 
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\int \frac{1}{\sqrt{u^3+1}}du\\
\end{align}
See : $\int\frac{1}{\sqrt{x^3+1}}$
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\left[\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+u}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)\right]+C\\
\end{align}
Revert the original substitution $u = \frac{x}{\sqrt[3]{3}} $:
\begin{align}
&= \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C\\
\end{align}

And

$$I = \frac{2}{3}\sqrt{x^3+3} + \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$

A: I have found a solution in terms of a hypergeometric function for positive $x$:
$$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{3}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{3}{x^3}\right)\right) +C
$$
and verified the derivative equals the integrand (for positive $x$), and in fact for general $a$,
$$
\int \frac{x^2+1}{\sqrt{x^3+a}}dx
$$
is given by $$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{a}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{a}{x^3}\right)\right) + C
$$
