Computing the integral $ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}$ Compute the following integral
$$ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}.$$
This integral obviously converges. However, is it computable?
 A: I have to share @RobertIsrael and @ProfessorVector's doubts.
Call it $I$. Following @WillJagy, take $2x^3+1=\sinh t$ so$$I=\frac{1}{3\sqrt[6]{2}}\int_{\operatorname{arsinh}3}^\infty(\sinh t-1)^{-1/3}\mathrm dt.$$Using this,$$I=2^{-5/3}[(\sinh t-1)^{2/3}\cosh t\cdot f(t)]_{\operatorname{arsinh}3}^\infty$$where$$f(t):=F_1(\tfrac23;\,\tfrac12,\,\tfrac12;\,\tfrac53;\,\tfrac12(1+i)(1-\sinh t),\,\tfrac12(1-i)(1-\sinh t)),$$with $F_1$ the Appell hypergeometric function. Assuming the bracketed expression vanishes in the $t\to\infty$ limit (hopefully an expert on $F_1$ can weigh in on that),$$I=-\sqrt{\tfrac52}F_1(\tfrac23;\,\tfrac12,\,\tfrac12;\,\tfrac53;\,-1-i,\,-1+i).$$If that has a nicer closed form, I'd love to see it.
A: A bit more Mathematica legerdemain: If we substitute $v=2x^3+1$, we obtain the form
$$I=\frac{1}{3 \sqrt[6]{2}}\int_3^\infty \frac{1}{\sqrt[3]{v-1}}\frac{dv}{\sqrt{1+v^2}}.$$
Writing the integration range as $\int_3^\infty = \int_1^\infty -\int_1^3$, Mathematica yields
$$I= \frac{2\pi}{3\sqrt{3}}{_2}F_1\left(\frac13,\frac56,1,-1\right)-  \frac12 F_1\left(\frac23;\frac12,\frac12;\frac53;-1+i,-1-i\right)$$
where ${_2}F_1(a,b;c;x)$ is the Gauss hypergeometric function and $F_1(a;b,c;d;x,y)$ is the Appell hyper-geometric function referenced in J.G.'s answer. Note that this means
\begin{align}
\int_0^\infty \frac{x\,dx}{\sqrt{2x^6+2x^3+1}} &= \frac{2\pi}{3\sqrt{3}}{_2}F_1\left(\frac13,\frac56,1,-1\right),\\
\int_0^1 \frac{x\,dx}{\sqrt{2x^6+2x^3+1}} &= \frac12 F_1\left(\frac23;\frac12,\frac12;\frac53;-1+i,-1-i\right)
\end{align}
Mathematica can't simplify the first term further, but there's so many identities for hypergeometric functions that further simplification seems plausible. The Appell F1 term, however, seems rather impenetrable.
