I would like to know whether the sum \begin{align} \sum^\infty_{n=0}\binom{n+2}{n}\log(1+a\ q^n) \quad \text{ with } \quad 0<a<1 \text{ and } |q|<1 \end{align} has a closed form or not. The logarithm here is the natural logarithm. This is not quite the Euler function because of the $a$ and the positive sign in front of $q^n$. My hope is that someone could direct me to some known special function. Many thanks.
Update: Without the binomial coefficient it seems that we have what some called the q-Pochhammer symbol.
It could be interesting to compare with the integral since \begin{align*}&\frac{1}{2}\int (n+1) (n+2) \log \left(1+a q^n\right)\,dn\\ &= \frac{-\left(n+1)(n+2\right) \log ^2(q) \text{Li}_2\left(-a q^n\right)-2 \text{Li}_4\left(-a q^n\right)+(2 n+3) \log (q) \text{Li}_3\left(-a q^n\right)}{2 \log ^3(q)},\end{align*} \begin{align*}&\frac{1}{2}\int_0^\infty (n+1) (n+2) \log \left(1+a q^n\right)\,dn \\ & =\frac{2 \text{Li}_2(-a) \log ^2(q)-3 \text{Li}_3(-a) \log (q)+2 \text{Li}_4(-a)}{2 \log ^3(q)}.\end{align*}
For a quick check : $a=\frac 12$ and $q=\frac 34$ leads to $30.5249$ while the infinite summation gives $30.6846$