Evaluation of Integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$

$$I = \int\frac{x^2+1}{\sqrt{x^3+3}}dx=\underbrace{\int \frac{x^2}{\sqrt{x^3+3}}dx}_{I_1}+\underbrace{\int \frac{1}{\sqrt{x^3+3}}dx}_{I_2}$$
$$I_1 = \int \frac{x^2}{\sqrt{x^3+3}}dx$$ 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

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