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.
 A: 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.
A: Substitute $ x =2^{3/2}\tan y$, along with $a=2^{7/6}$
\begin{align}
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\frac\pi2\frac{\sec y}{1 + (a\tan y)^3} dy
=\frac12 (I_1 + I_2)
\end{align}
where
\begin{align}
I_1=& \int_0^\frac\pi2\frac{\sec y}{1 + a\tan y} dy
=\frac2{\sqrt{1+a^2}}\tanh^{-1} \frac{\sqrt{1+a^2}}{1+a}\\
I_2= &\int_0^\frac\pi2\frac{\sec y \ (2-a\tan y)}{1 -a\tan y+a^2\tan^2 y}\overset{y\to \frac\pi2 -y}{dy}\\
=& \ \frac12 \int_0^\frac\pi2\frac{\sec y \ (2-a\tan y)}{1 -a\tan y+a^2\tan^2 y}+\frac{\csc y \ (2-a\cot y)}{1 -a\cot y+a^2\cot^2 y}\ dy\\
 =& \int_0^\frac\pi2 \frac{2[a^3+a^2+a-2)\sin2y-4a^2+2a](\sin y +\cos y)}{(a^4-a^2+1)\cos4y+4a(a^2+1)\sin2y-(a^4+7a^2+1)} \ \overset{t=\sin y -\cos y}{dy}\\
=& \int_{-1}^1 \frac{(a^3+a^2+a-2)t^2-(a-2)(a^2-a+1)}
{(a^4-a^2+1)t^4 -2(a^4-a^3-a^2-a+1)t^2+(a^2-a+1)^2}dt\\
=& \ \frac{\sqrt2p_-}{\sqrt{q_-}}\bigg(\frac\pi2+\tan^{-1}\frac{s_-}{\sqrt{q_-}}\bigg)
- \frac{\sqrt2p_+}{\sqrt{q_+}}\coth^{-1}\frac{s_+}{\sqrt{q_+}}
\end{align}
with
\begin{align}
&s_\pm=\sqrt{a^4-a^2+1}\pm(a^2-a+1)\\
 &p_\pm=\frac12\bigg( \frac{a^3+a^2+a-2}{\sqrt{a^4-a^2+1}}\pm (a-2)\bigg)\\
&q_\pm= \sqrt{a^4-a^2+1}\ (a^2-a+1)\pm (a^4-a^3-a^2-a+1)
\end{align}
