How to take the integral of the following $$\int\sqrt{16x^2+8+\frac 1{x^2}}\,\mathrm d x$$

I tried various substitutions, including trig functions (to see if I could do it the way I would do $\int\sqrt{1+x^2}\,\mathrm d x$), but didn't get an answer.  How would I do this?  

Thank you.
HINT, substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:

$$\mathcal{I}(x)=\int\sqrt{16x^2+8+\frac{1}{x^2}}\space\space\text{d}x=\int\frac{\sqrt{16x^4+8x^2+1}}{x}\space\text{d}x=$$
$$\frac{1}{2}\int\frac{\sqrt{16u^2+8u+1}}{u}\space\text{d}u=\frac{1}{2}\int\frac{\sqrt{(1+4u)^2}}{u}\space\text{d}u$$
Try the hyperbolic substitution $x=sinh(y)$