On the closed-form of the triple integral $\int_0^\infty\int_0^\infty\int_0^\infty\frac1{xyz\left(x+y+z+1/x+1/y+1/z\right)^2}\rm{dx\,dy\,dz}$

Question

While doing research for my recent post on the Clausen function $\rm{Cl}_m(x)$, I came across in p. 19 of this paper (by one of the Borwein brothers) the remarkable integral,

$$I_3 =\frac4{3!}\int_0^\infty\int_0^\infty\int_0^\infty\frac1{xyz\left(x+y+z+1/x+1/y+1/z\right)^2}\rm{dx\,dy\,dz}=\frac4{3\sqrt3}\rm{Cl}_2\left(\frac\pi3\right)$$

where $\rm{Cl}_2\left(\frac\pi3\right)$ is Gieseking's constant. More generally,

$$I_n = \frac{2^n}{n!}\int_0^\infty x K_0^n(x)\,\rm{dx}$$

with modified Bessel function of the second kind $K_m(x)$. Hence,

$$\begin{aligned} I_2 &= \frac12\\ I_3 &= \frac4{3\sqrt3}\rm{Cl}_2\left(\frac\pi3\right)\\ I_4 &= \,\frac7{4}\;\rm{Cl}_3\left(\frac\pi3\right)=\frac{7\zeta(3)}{12}\\ I_5 &= \,?? \end{aligned}$$

where,

$$\rm{Cl}_n(x) =\sum_{m=1}^\infty \frac{\sin(mx)}{m^n},\quad \text{even}\;n$$ $$\rm{Cl}_n(x) =\sum_{m=1}^\infty \frac{\cos(mx)}{m^n},\quad \text{odd}\;n$$

Questions:

1. How do we show that triple integral $I_3$ is equal to its RHS?
2. And what is the closed form of $I_5$? (Naturally, I checked its ratio with $\rm{Cl}_4\left(\frac\pi3\right)$ but it doesn't seem to be algebraic.)