How can I find $\sum\limits_{n=0}^{\infty}\left(\frac{(-1)^n}{2n+1}\sum\limits_{k=0}^{2n}\frac{1}{2n+4k+3}\right)$?

Question

prove that$$\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{2n+1}\sum_{k=0}^{2n}\frac{1}{2n+4k+3}\right)=\frac{3\pi}{8}\log(\frac{1+\sqrt5}{2})-\frac{\pi}{16}\log5 $$

This problem, I think use $$\sum_{k=0}^{2n}\dfrac{1}{2n+4k+3}=H_{10n+3}-H_{2n+3}$$

Thank you everyone help