Evaluate $\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}$ I was given that
$$\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}=\frac{-9 \left(-\frac{2 C}{3}-1+\frac{1}{3} \pi  \log \left(\sqrt{3}+2\right)\right)}{2 \pi }$$
Here $\beta$ denotes Dirichlet Beta. Unfortunately, this result seems to be numerically incorrect. So what is the closed-form of this sum? Any help will be appreciated.

Here is a sketch. Using, say, the link with Euler numbers, one gets
$$\sum_{n=0}^{\infty}\beta(2n+1)x^{2n}=\frac{\pi}{4}\sec\frac{x\pi}{2},$$
which implies
$$\sum_{n=0}^{\infty}\frac{\beta(2n+1)}{2n+2}x^{2n+2}=\frac{1}{\pi}f\left(\frac{x\pi}{2}\right),\quad f(y)=\int_0^y\frac{x\,dx}{\cos x}.$$
The given sum is then equal to $\dfrac{9}{2\pi}f\left(\dfrac{\pi}{3}\right)-\dfrac{\pi}{4}$, with the known value of
$$f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}\ln(2+\sqrt{3})-\frac{2}{3}G,$$
where $G$ is Catalan's constant. The latter formula can be derived from
$$\int_0^y\ln\left|\tan\frac{x}{2}\right|\,dx=-2\sum_{n=0}^{\infty}\frac{\sin(2n+1)y}{(2n+1)^2}.$$
