The closed-form solution of the family $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)}$? (The results below extend this post.) Given the Clausen function $\operatorname{Cl}_n\left(z\right)$. And,
$$\begin{aligned}
\operatorname{Cl}_2\left(\frac\pi2\right) &= \text{Catalan's constant}\\
\operatorname{Cl}_2\left(\frac\pi3\right) &= \text{Gieseking's constant}\\
\operatorname{Cl}_2\left(\frac\pi4\right) &= \text{unnamed}\\
\operatorname{Cl}_2\left(\frac\pi6\right) &= \tfrac23\,\operatorname{Cl}_2\left(\frac\pi2\right)+\tfrac14\,\operatorname{Cl}_2\left(\frac\pi3\right)
\end{aligned}$$
Then we have the closed-forms,
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)} &=& \frac{11}{8} \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)} &=& \frac{5}{3} \zeta(3) -\frac{2}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(4n+m)} &=& \frac{67}{32} \zeta(3) -\frac{1}{2}\pi\, \operatorname{Cl}_2\left(\frac\pi2\right) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(6n+m)} &=& \frac{73}{24} \zeta(3) -\frac{8}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(8n+m)} &=& \frac{515}{128} \zeta(3) -\frac{3}{8}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{red}4}\right)\\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(12n+m)} &=& \frac{577}{96} \zeta(3) -\frac{7}{6}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\frac{19}{18}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\
\end{eqnarray*}
where for $p=12$ we could have used $\operatorname{Cl}_2\left(\frac\pi2\right)$ and $\operatorname{Cl}_2\left(\frac\pi6\right)$. As the OP from the other post points out, note that,
$$I(p)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)} =\int_0^1  \frac{\ln(1-z) \ln(1-z^p)}{z} dz$$

Q: The results above suggest a family. Can we find the closed-form of the integral $I(p)$ for $p=5$ and others?


$\color{red}{\text{Update July 24}}$: Thanks to Zacky's answer which provided the clue that more than one Clausen function with argument $\frac{m\,\pi}p$ may be needed, after some tinkering, I managed to find a closed-form for $I(p)$, namely,

$$I(p)= \frac{p^3+3}{2p^2}\zeta(3)-\frac{\pi}p\sum_{k=1}^{\lfloor(p-1)/2\rfloor}(p-2k)\operatorname{Cl}_2\left(\frac{2k\pi}p\right)$$

with floor function $\lfloor x\rfloor$. I found this using odd $p$, but it seems to work for even $p$ as well. However, a rigorous proof is needed to show it holds true for all $p$.
 A: We may apply a discrete Fourier transform to the following generating function
$$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$
since 
$$ I(p) = \sum_{n\geq 1}\frac{H_{p n}}{pn^2}. $$
The only term leading to a non-elementary contribution is the sum of $\operatorname{Li}_3(1-x)$ over the $p$-th roots of unity.
A: An additional note on how to derive the digamma (or harmonic numbers) series from the integral:
$$ p I(p) = \sum_{n=1}^\infty \frac{H_{p n}}{n^2}$$
$$I(p)= \int_0^1 x^{-1} \log (1-x) \log (1-x^p) dx= \\ = - \sum_{n=1}^\infty \frac{1}{n} \int_0^1 x^{pn-1} \log (1-x) dx$$
Now consider the following integral:
$$J(s)=-\int_0^1 x^s \log (1-x) dx$$
Let's integrate by parts with: $$u=x^s, \qquad du=s x^{s-1} dx \\ dv=- \log(1-x) dx, \qquad v=x+(1-x) \log(1-x)$$
We get:
$$J(s)=1-s\int_0^1 x^s dx-s\int_0^1 x^{s-1} \log (1-x) dx+s \int_0^1 x^s \log (1-x) dx$$
$$(s+1)J(s)=\frac{1}{s+1}+s J(s-1)$$
It's easy to check that $J(0)=1$.
Introducing a new function:
$$Y(s+1)=(s+1) J(s)$$
We see that:
$$Y(s+1)=\frac{1}{s+1}+Y(s) \\ Y(1)=1$$
But this is exactly the definition of harmonic numbers.
So we have:
$$I(p)= \sum_{n=1}^\infty \frac{1}{n} J(pn-1)=\sum_{n=1}^\infty \frac{1}{n} \frac{Y(pn)}{pn}=\frac{1}{p} \sum_{n=1}^\infty \frac{H_{pn}}{n^2}$$
