How do we integrate $x\sqrt{x^3 + 1}$? Tried to use integration by parts but just keep going round in circles. Any help appreciated, thanks.

Alternatively, and perhaps a little more transparent than just stating the result in terms of elliptic functions, we can find an antiderivative by expanding the square root and integrating the powers of x.
This leads to
$$a(x) = \sum _{k=0}^{\infty } \frac{\binom{\frac{1}{2}}{k} x^{3 k+2}}{3 k+2}$$
This infinite sum can be expressed by a hypergeometric function
$$a(x) =\frac{1}{2} x^2 \, _2F_1\left(-\frac{1}{2},\frac{2}{3};\frac{5}{3};-x^3\right)$$
Hence the function $a(x)$ is analytic in $x$ except for a branch cut from $x = -1$ to $x = -\infty$.
EDIT
We can obtain a possibly different antiderivative by writing 
$$\sqrt{x^3+1}=x^{3/2} \sqrt{\frac{1}{x^3}+1}$$
expanding into negative powers and integrating, viz.
$$b(x)=\sum _{k=0}^{\infty } \frac{\binom{\frac{1}{2}}{k} x^{\frac{7}{2}-3 k}}{\frac{7}{2}-3 k}$$
which can be expressed as
$$b(x) = \frac{2}{7} x^{7/2} \, _2F_1\left(-\frac{7}{6},-\frac{1}{2};-\frac{1}{6};-\frac{1}{x^3}\right)$$
This expression holds even for $x\gt 0$.
Corollary 
There should be a standard transformation of the hypergeometric function showing some equivalence of $a(x)$ and $b(x)$. Or, inverting the reasoning, we have just derived this transformation.
Not really, we must be careful here. In fact, $a(x)$ and $b(x)$ are not identical in the overlaping region of definition $x > 0$ but differ by a constant (which is permissible for antiderivatives):
$$b(x) = a(x) + c$$
where
$$ c = \lim_{x\to 0} \, b(x) = -\frac{2 \left(\Gamma \left(-\frac{1}{3}\right)-6 \Gamma \left(\frac{2}{3}\right)\right) \Gamma \left(\frac{5}{6}\right)}{21 \sqrt{\pi }} \simeq 0.739174$$
