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In this post, the OP asks about the integral,

$$I = \int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$$

I. User DavidH gave a beautiful (albeit long) answer in terms of the Nielsen generalized polylogarithm,

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

namely,

$$I = \tfrac32 S_{2,2}(-1)+\tfrac{11}{8} S_{1,3}(1)-S_{1,3}(-1) + \tfrac32 S_{3,1}(-1) \approx 0.223076$$

with the last addend tweaked by yours truly. A session with Mathematica shows that these explicitly are,

$$S_{3,1}(-1) = -\tfrac78\zeta(4) \\ S_{1,3}(1) = \zeta(4) \\ S_{2,2}(-1) = 2S_{1,3}(-1)-\tfrac18\zeta(4)$$

and,

$$S_{1,3}(-1) = \tfrac18\ln^3(2)\,\rm{Li}_1\big(\tfrac12\big)+\tfrac12\ln^2(2)\,\rm{Li}_2\big(\tfrac12\big)+\ln(2)\,\rm{Li}_3\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)-\zeta(4)$$

Since $S_{1,3}(-1)$ and $S_{2,2}(-1)$ have a linear relation, then the integral can be simplified as,

$$\color{blue}{I = 2S_{1,3}(-1)+\tfrac14\zeta(4)}$$

Note that $\rm{Li}_n\big(\tfrac12\big)$ for $n=1,2,3$ have closed-forms.

II. User nospoon gave an equal but alternative form as,

$$I=\tfrac52\ln(2)\zeta(3)-\tfrac{11}{576}\pi^4-\tfrac1{2}\ln^2(2)\zeta(2)+\tfrac1{16}\ln^4(2)+\tfrac32\rm{Li}_4\big(\tfrac12\big)-A+\tfrac12B\\ \approx 0.223076$$

where

$$A = \int_0^1\frac{\rm{Li}_3(x)}{1+x}dx$$ $$B= \int_0^1\frac{\ln(1-x^2)\,\rm{Li}_2\big(\tfrac{1-x}2\big)}{x}dx$$

III. Question

After guessing on various candidate variables, is it true that the closed-forms of $A$ and $B$ are,

$$A = -4S_{2,2}(-1)+6S_{1,3}(-1) +\ln(2)\zeta(3) = 0.339545\dots$$ $$B = -\tfrac12S_{2,2}(-1)-2S_{1,3}(-1)-\tfrac38\ln(2)\zeta(3)+\tfrac14\ln^2(2)\zeta(2) = -0.1112606\dots$$

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