This post, after a complicated analysis, evaluates the integral
$$I=\int_0^1\frac{\ln^2(x)\,\ln^3(1+x)}xdx$$
simply as
$$I
=-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)\tag1$$
where,
$$\zeta(\bar{5},1)=\frac{1}{24}\int^1_0\frac{\ln^4{x}\ln(1+x)}{1+x}{\rm d}x$$
More succinctly,
$$I = -12\,S_{3,3}(-1)\tag2$$
with Nielsen generalized polylogarithm $S_{n,p}(z)$.
Question: How do we show that $\zeta(\bar{5},1)$ is also a Nielsen generalized polylogarithm in disguise? More generally, for $-1\leq z\leq1$, how to show
$$\begin{aligned}S_{n,p}(z)
&= C_1\int_0^1\frac{(\ln x)^{n-1}\big(\ln(1-z\,x)\big)^p}{x}dx\\
&\overset{?}= C_2\int_0^1\frac{(\ln x)^{n}\;\big(\ln(1-z\,x)\big)^{p-1}}{1-z\,x}dx\end{aligned}\tag3$$
where,
$$C_1 = \frac{(-1)^{n+p-1}}{(n-1)!\,p!},\qquad C_2 = \frac{(-1)^{n+p-1}}{n!\,(p-1)!}\color{red}z$$
If true, this implies,
$$\zeta(\bar{5},1) \overset{\color{red}?}= S_{4,2}(-1)\tag4$$
Edit: It turns out the notation $\zeta(\bar{5},1)$ is a multiple zeta function so,
$$\zeta(\bar{a},1)=\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^a}\,(-1)^{n+1} = S_{a-1,2}(-1)$$
with harmonic numbers $H_n$, hence $(4)$ indeed is true and is just the case $a=5$. However, $(3)$ still needs to be proved in general.
Best Answer
Without knowing much about the context for this problem, the relationship between the two integrals seems to be pretty direct from integration by parts. For $n, p \geq 1$, writing $u(x) = (\ln x)^n$ and $v(x) = (\ln(1 - zx))^p$, we have $$\frac{du}{dx} = \frac{n(\ln x)^{n-1}}{x} \qquad \text{and} \qquad \frac{dv}{dx} = -\frac{pz (\ln (1 - zx))^{p-1}}{1 - zx}$$ hence using integration by parts: \begin{align*} n\int_0^1 \frac{(\ln x)^{n-1} (\ln(1 - zx))^p}{x} \,dx &= \int_0^1 \frac{du}{dx} v \,dx \\ &= (uv)|_0^1 - \int_0^1 u \frac{dv}{dx} \,dx \\ &= pz\int_0^1 \frac{(\ln x)^n (\ln (1 - zx))^{p-1}}{1 - zx} \,dx \end{align*} where the last equation holds since $u(x)v(x) \to 0$ as $x \to 0$ or $x \to 1$.
This is $(3)$, up to the factor $(-1)^{n+p-1}/n! p!$.