The series arose in the calculation of Mean value of a function associated with continued fractions: $$C=\sum_{1\le b\le d<\infty}\frac{1}{b(b+d)d^2}.$$ Obviously $C=C_1-C_2,$ where $$C_1=\sum_{1\le b\le d<\infty}\frac{1}{bd^3},\quad C_2=\sum_{1\le b\le d<\infty}\frac{1}{(b+d)d^3}.$$ Mathematica gives $C_1=\frac{\pi^4}{ 72},$ and $C_2$ can be simplifyied to $$C_2=2\int_{0}^{1}\frac{\mathrm{Li}_2(t)}{t}\log(1+t)dt$$ using the following trick: $$\frac{1}{n+1}+\ldots+\frac{1}{2n}=1-\frac12+\frac13-\frac14+\ldots-\frac{1}{2n}=\int_0^1\frac{1-t^{2n}}{1+t}dt.$$

Q: Is it possible to find closed form for $C$?

The series arose in the calculation of Mean value of a function associated with continued fractions: $$C=\sum_{1\le b\le d<\infty}\frac{1}{b(b+d)d^2}.$$ Obviously $C=C_1-C_2,$ where $$C_1=\sum_{1\le b\le d<\infty}\frac{1}{bd^3},\quad C_2=\sum_{1\le b\le d<\infty}\frac{1}{(b+d)d^3}.$$ Mathematica gives $C_1=\frac{\pi^4}{ 72},$ and $C_2$ can be simplifyied to $$C_2=2\int_{0}^{1}\frac{\mathrm{Li}_2(t)}{t}\log(1+t)dt$$ using the following trick: $$\frac{1}{n+1}+\ldots+\frac{1}{2n}=1-\frac12+\frac13-\frac14+\ldots-\frac{1}{2n}=\int_0^1\frac{1-t^{2n}}{1+t}dt.$$

Q: Is it possible to find closed form for $C$?