Difficulty evaluating $\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$ In evaluating
$$\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$$ I did not have a situation where the polynomial under the square root has lower degree. Neither trigonometric substitutions nor variable changes help, at least how I apply them.
I would like at least to know how to evaluate a similar integral, because I already spent a lot of time on this one.

The limits of integration and the form of the radical in the denominator strongly suggest $x = 2\sqrt{2}\sinh(u)$. Performing this substitution gives and expanding in partial fractions gives
$$
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\infty\frac{du}{1 + 2^{7/2}\sinh^3(u)} 
 =\frac{1}{6}\sum_{i=-1}^1\int_0^\infty\frac{du}{1 + 2^{7/6}\omega^i \sinh(u)}
$$
where $\omega = \exp(2\pi i/3)$. This last integral is doable, but it's not all that pretty. You get
$$
\int_0^\infty \frac{du}{1+a\sinh(u)} = \frac{\cosh^{-1}(1 + a + a^{-1})}{\sqrt{a}\sqrt{a + a^{-1}}},
$$
where the principal branch of the square root is used. The result is
$$
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{6}\sum_{i=-1}^1\frac{\cosh^{-1}(1 + 2^{7/6}\omega^i + 2^{-7/6}\omega^{-i})}{\sqrt{2^{7/6}\omega^i}\sqrt{2^{7/6}\omega^i + 2^{-7/6}\omega^{-i}}}
$$
And to check numerically, the integral equals the sum.
