Closed forms of Nielsen polylogarithms $\int_0^1\frac{(\ln t)^{n-1}(\ln(1-z\,t))^p}{t}dt$? (This summarizes my posts on Nielsen polylogs.)
I. Question 1: How to complete the table below? Consider the special cases $z=-1$ and $z=\frac12$. Given the Nielsen generalized polylogarithm,
$$S_{n,p}(z) = C_1\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$
where,
$$C_1 = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}$$
then for what $n,p$ are there closed-forms in terms of ordinary polylogarithms $\rm{Li}_m(x)$?
Note: For $p=1$, then the Nielsen polylog just reduces to $\rm{Li}_m(x)$.

MSE has a lot of posts asking for close-forms (see this, this, etc), most of which do not name the integral as a Nielsen polylog. The table below summarizes known results (so far),
$$\begin{array}{|c|c|c|c|c|}
\hline
n+p&n&p&z=-1&z=\tfrac12\\
\hline
3&1 &2 &Y&Y\\
3&2 &1 &Y&Y\\
\hline
4&1 &3 &Y&Y\\
4&2 &2 &Y&Y\\
4&3 &1 &Y&Y\\
\hline
5&1 &4 &Y&Y\\
5&2 &3 &Y&Y\\
5&3 &2 &Y&Y\\
5&4 &1 &Y&Y\\
\hline
6&1 &5 &Y&Y\\
6&2 &4 &-&-\\
6&3 &3 &\color{red}N&\color{red}Y\\
6&4 &2 &-&-\\
6&5 &1 &Y&Y\\
\hline
7&1 &6 &Y&Y\\
7&2 &5 &-&-\\
7&3 &4 &-&-\\
7&4 &3 &-&-\\
7&5 &2 &\color{red}Y&\color{red}N\\
7&6 &1 &Y&Y\\
\hline
\end{array}$$
Surprisingly, $S_{3,3}\big(\tfrac12\big)$ is expressible, but $S_{3,3}(-1)$ is not. Let $a=\ln 2$, then,
$$2\,S_{3,3}\big(\tfrac12\big) =\tfrac{23}{16}\zeta(6)-2a\zeta(5)+\tfrac18a^2\zeta(4)-\tfrac1{16}a^3\zeta(3)+\tfrac1{72}a^6-\zeta^2(3)+a\big(S_{3,2}\big(\tfrac12\big)-S_{2,3}\big(\tfrac12\big)+\zeta(2)\zeta(3)\big)$$
Since the two Nielsen polylogs in RHS are expressible as ordinary polylogs (see here), then so is the LHS. Conversely, $S_{5,2}(-1)$ is expressible, 
$$128S_{5,2}(-1) = 64\zeta(2)\zeta(5)+112\zeta(3)\zeta(4)-251\zeta(7)$$
but $S_{5,2}\big(\tfrac12\big)$ apparently is not.

II. Question 2: What simple relations are there between the inexpressible(?) Nielsen polylogs in the table? Two are,
$$16S_{3,3}(-1)-24S_{4,2}(-1)=-4\zeta^2(3)+5\zeta(6)$$
$$128S_{3,4}(-1)-192S_{4,3}(-1)=-64\zeta(2)\zeta(5)-160\zeta(3)\zeta(4)+315\zeta(7)$$
which is mentioned in this and this post. (The two seem to belong to a general form.) Are there others?
 A: Consider the generalization of this integral as follows:
$$I(a,b,c)=\int_0^\infty\ln^a(x)\ln^b(1+x)\ln^c\left(1+\frac1x\right)~\frac{\mathrm dx}x$$
We can apply the substitution $x\mapsto1/x$ to get $I(a,b,c)=(-1)^aI(a,c,b)$. We can also apply the substitution $x\mapsto1/x$ on $(1,\infty)$ to get two integrals:
$$I_1(a,b,c)=\int_0^1\ln^a(x)\ln^b(1+x)\ln^c\left(1+\frac1x\right)~\frac{\mathrm dx}x$$
$$I_2(a,b,c)=(-1)^a\int_0^1\ln^a(x)\ln^c(1+x)\ln^b\left(1+\frac1x\right)~\frac{\mathrm dx}x$$
Since $\ln(1+1/x)=\ln(1+x)-\ln(x)$, it is then possible to rewrite our initial integral in terms of the Nielsen polylog at $z=-1$. Let $\displaystyle S_{n,p}=\int_0^1\ln^n(x)\ln^p(1+x)~\frac{\mathrm dx}x$. Then we have
$$I_1(a,b,c)=\sum_{j=0}^c\binom cj(-1)^jS_{a+j,b+c-j}$$
$$I_2(a,b,c)=(-1)^a\sum_{k=0}^b\binom bk(-1)^kS_{a+k,b+c-k}$$
$$I(a,b,c)=\sum_{j=0}^c\binom cj(-1)^jS_{a+j,b+c-j}+(-1)^a\sum_{k=0}^b\binom bk(-1)^kS_{a+k,b+c-k}$$

On the other hand, one can perform the substitution $x\mapsto\frac1x-1$ to get
\begin{align}I(a,b,c)&=\int_0^1\ln^a\left(\frac{1-x}x\right)\ln^b\left(\frac1x\right)\ln^c\left(\frac1{1-x}\right)~\frac{\mathrm dx}{x(1-x)}\\&=(-1)^{b+c}\int_0^1[\ln(1-x)-\ln(x)]^a\ln^b(x)\ln^c(1-x)\left[\frac1{1-x}+\frac1x\right]~\mathrm dx\end{align}
For brevity we define another integral:
$$J(a,b,c)=\int_0^1[\ln(1-x)-\ln(x)]^a\ln^b(x)\ln^c(1-x)~\frac{\mathrm dx}x$$
Splitting this into two integrals and applying $x\mapsto1-x$ for the part with $1/(1-x)$ gives us
$$(-1)^{b+c}I(a,b,c)=J(a,b,c)+(-1)^aJ(a,c,b)$$
Let $J(b,c)=J(0,b,c)$. It is possible to rewrite $J(a,b,c)$ in terms of the two argument version by binomially expanding it out:
$$J(a,b,c)=\sum_{k=0}^a\binom ak(-1)^kJ(b+k,c+a-k)$$
and it is easy to solve $J(b,c)$ as partial derivatives of the Beta function, which gives us solutions in terms of the Riemann zeta function at natural arguments:
$$J(b,c)=\lim_{(\mu,\nu)\to(0,0)}\frac{\partial^{b+c}}{\partial\mu^b\partial\nu^c}B(\mu,\nu+1)-\frac1\mu$$

Conclusion: we can generate identities for sums of the form $\displaystyle\sum_{n+p=a+b+c}a_{a,b,c,n,p}S_{n,p}$ in terms of the Riemann zeta function.
For example, with $(a,b,c)=(2,1,2)$ we get
$$2S_{2,3}-3S_{3,2}+S_{4,1}=\frac{\pi^6}{60}+6\zeta^2(3)$$
and with $(a,b,c)=(0,2,3)$ we get
$$2S_{0,5}-5S_{1,4}+4S_{2,3}-S_{3,2}=\frac{\pi^6}{36}-18\zeta^2(3)$$
and with $(a,b,c)=(0,1,4)$ we get
$$2S_{0,5}-5S_{1,4}+6S_{2,3}-4S_{3,2}+S_{4,1}=12\zeta^2(3)-\frac{2\pi^6}{45}$$
which sadly doesn't give any values.
