<s>[INST] 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
 [/INST] 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 $$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{a}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{a}{x^3}\right)\right) + C
$$
</s>