Integrate a quartic in a radical in the denominator of $\int_{1}^{\infty}\frac{1}{\sqrt{3x^{4}+6x^{2}-1}}dx$ Here is a rather tough integration. I think it looks like an Elliptic Integral of some sort.  
$$\int_{1}^{\infty}\frac{1}{\sqrt{3x^{4}+6x^{2}-1}}dx$$
Since there are no odd terms in the quartic, I thought maybe completing the square 
would be OK. But I got nowhere. I even factored it into:
$3x^{4}+6x^2-1=((2\sqrt{3}-3)x^{2}+1)((2\sqrt{3}+3)x^{2}-1)$ and got nowhere.
I think the solution will involve the Gamma function in some manner.
Does anyone have a good starting point for this... like a clever substitution?. 
Thanks for any input. 
 A: As has been mentioned many times on this site, anytime you have an algebraic function containing the square root of a cubic or a quartic, you are bound to bump into an elliptic integral.
Usually, such things are handled by using Jacobian elliptic functions for substitutions (in a manner similar to using substitution with trigonometric or hyperbolic functions when you have the square root of a quadratic in an integral).
I'll skip the tedious details of figuring out the proper substitution, since Byrd and Friedman give a formula for handling your integral (formula 212.00 in their handbook):
$$\int_y^\infty\frac{\mathrm dt}{\sqrt{(t^2+a^2)(t^2-b^2)}}=\frac1{\sqrt{a^2+b^2}}F\left(\arcsin\left(\sqrt{\frac{a^2+b^2}{a^2+y^2}}\right) \mid\frac{a^2}{a^2+b^2}\right)$$
where $F(\phi|m)$ is the incomplete elliptic integral of the first kind.
Coming back to your integral, we let $u=2\sqrt{3}-3$ and $v=2\sqrt{3}+3$ such that
$$\int_1^\infty\frac{\mathrm dx}{\sqrt{3x^4+6x^2-1}}=\frac1{\sqrt{uv}}\int_1^\infty \frac{\mathrm dx}{\sqrt{(x^2+1/u)(x^2-1/v)}}$$
Using the quoted formula, the integral reduces to
$$\frac1{\sqrt{u+v}}F\left(\arcsin\left(\sqrt{\frac{u+v}{v+uv}}\right)\mid\frac{v}{u+v}\right)$$
Substituting the values of $u$ and $v$ into this expression and simplifying, we have the result
$$\frac1{\sqrt[4]{48}}F\left(\arcsin\left(\sqrt{\sqrt{3}-1}\right)\mid\frac{2+\sqrt{3}}{4}\right)$$
which agrees with the numerical result in the comments.
As an aside, I consider it a capital annoyance that Mathematica often returns results with complex amplitudes even for real results...
