Evaluate $\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}$ How to prove
$$\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}=\frac{8}{5} \left(C-\frac{1}{8} \pi  \log \left(\frac{\sqrt{50-22 \sqrt{5}}+10}{10-\sqrt{50-22 \sqrt{5}}}\right)\right)$$
Where $L_k$ denotes Lucas number and $C$ Catalan? Any help will be appreciated.
 A: This is more of an extended comment at this point until I link to the final result, but numerically the above relation is plausible.
There are two observations to make to take the main step forward:
$$L_n = \phi^n + (1-\phi)^n $$
where $\phi = (\sqrt{5}+1)/2$ is the golden ratio, and
$$\sum_{k=0}^{\infty} \frac{x^{2 k+1}}{(2 k+1) \binom{2 k}{k}} = \frac{2 \arcsin{(x/2)}}{\sqrt{1-(x/2)^2}} $$
Then after some manipulation, one may show that the above sum is equal to the following definite integral:
$$2 \int_{\arcsin{(\phi-1)/2}}^{\arcsin{(\phi/2)}} d\theta \, \theta \, \csc{\theta} $$
Note that the lower limit is equal to $\pi/10$ and the upper limit is $3 \pi/10$.  This may be integrated by parts and manipulated a little further to produce
$$\frac{\pi}{5} \log{\left (\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} \right )} - 4 \int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2}$$
What remains for us to prove is that
$$\int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2} = -\frac25 C$$
which amazingly checks out numerically, and that the ratio
$$\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} = \frac{10-\sqrt{50-22 \sqrt{5}}}{10+\sqrt{50-22 \sqrt{5}}}$$
which also checks out numerically (and should be simple algebra).
