I see in some table pages sum like $$\sum _{k=1}^{\infty } \frac{1}{\left(q^k+2\right)^2 \left(q^k+1\right)^2 \left(q^k+3\right)}\text{==}$$ $$\frac{2 \log (q) \left(-9 \psi _q^{(0)}\left(-\frac{\log \left(-\frac{1}{q}\right)}{\log (q)}\right)+9 \psi _q^{(0)}\left(-\frac{\log \left(-\frac{2}{q}\right)}{\log (q)}\right)+\psi _q^{(0)}\left(-\frac{\log \left(-\frac{3}{q}\right)}{\log (q)}\right)\right)+12 \psi _q^{(1)}\left(-\frac{\log \left(-\frac{1}{q}\right)}{\log (q)}\right)+6 \psi _q^{(1)}\left(-\frac{\log \left(-\frac{2}{q}\right)}{\log (q)}\right)+\log (q) \left(2 \log (q-1)+2 \log \left(-\frac{1}{q}\right)+\log (q)-18+18 \log (2)+\log (9)\right)}{24 \log ^2(q)}$$ do you know how it is get.
Editorial note The series is more compactly presented as
\begin{align} & \sum _{k=1}^{\infty} \frac{1}{\left(q^k+2\right)^2 \left(q^k+1\right)^2 \left(q^k+3\right)} \\ &= \frac{1}{12 \ln(q)} \left(-9 \psi_q^{(0)}\left(-\frac{\ln\left(-\frac{1}{q}\right)}{\ln (q)}\right) + 9 \psi_q^{(0)}\left(-\frac{\ln\left(-\frac{2}{q}\right)}{\ln(q)}\right) + \psi_q^{(0)}\left(-\frac{\ln\left(-\frac{3}{q}\right)}{\ln(q)}\right)\right) \\ & \hspace{5mm} + \frac{1}{2 \ln^2(q)} \psi_q^{(1)}\left(-\frac{\ln\left(-\frac{1}{q}\right)}{\ln(q)}\right) + \frac{1}{4 \ln^2(q)} \psi _q^{(1)}\left(-\frac{\ln\left(-\frac{2}{q}\right)}{\ln(q)}\right) + \frac{\ln(q-1)}{12\ln (q)} \\ & \hspace{5mm} + \frac{ \ln\left(-\frac{1}{q}\right)}{12 \ln^2(q)} + \frac{1}{24\ln(q)} - \frac{9+9 \ln(2)+\ln(3)}{12 \ln^2(q)} \end{align}