Another two hard integrals Evaluate :
$$\begin{align}
  & \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\left( 2\cos x \right)}{{{\ln }^{2}}\left( 2\cos x \right)+{{x}^{2}}}}\text{d}x \\ 
 & \int_{0}^{1}{\frac{\arctan \left( {{x}^{3+\sqrt{8}}} \right)}{1+{{x}^{2}}}}\text{d}x \\ 
\end{align}$$
 A: For the second integral, consider the more general form
$$\int_0^1 dx \: \frac{\arctan{x^{\alpha}}}{1+x^2}$$
(I do not understand what is special about $3+\sqrt{8}$.)
Taylor expand the denominator and get
$$\begin{align} &=\int_0^1 dx \: \arctan{x^{\alpha}} \sum_{k=0}^{\infty} (-1)^k x^{2 k} \\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} \end{align}$$
Now we can simply evaluate these integrals in terms of polygamma functions:
$$\int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} = \frac{\psi\left(\frac{a+2 k+1}{4 a}\right)-\psi\left(\frac{3 a+2 k+1}{4 a}\right)+\pi }{8 k+4}$$
where
$$\psi(z) = \frac{d}{dz} \log{\Gamma{(z)}}$$
and we get that
$$\int_0^1 dx \: \frac{ \arctan{x^{\alpha}}}{1+x^2} = \frac{\pi^2}{16} - \frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \frac{\psi\left(\frac{3 \alpha+2 k+1}{4 \alpha}\right)-\psi\left(\frac{\alpha+2 k+1}{4 \alpha}\right) }{2 k+1} $$
This is about as close as I can get.  The sum agrees with the numerical integration out to 6 sig figs at about $10,000$ terms for $\alpha = 3+\sqrt{8}$.
