Theorem: Mean Distance between Two Random Points in Cuboid

Let $B$ be a cuboid in the Cartesian $3$-space $\R^3$ as:
:$\size x \le a$, $\size y \le b$, $\size z \le c$
Let $E$ denote the mean distance $D$ between $2$ points chosen at random from the interior of $B$.
Then:
{{begin-eqn}}
{{eqn | l = E
      | r = \dfrac {2 r} {15} - \dfrac 7 {45} \paren {\paren {r - r_1} \paren {\dfrac {r_1} a}^2 + \paren {r - r_2} \paren {\dfrac {r_2} b}^2 + \paren {r - r_3} \paren {\dfrac {r_3} c}^2}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 8 {315 a^2 b^2 c^2} \paren {a^7 + b^7 + c^7 - {r_1}^7 - {r_2}^7 - {r_3}^7 + r^7}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a b^2 c^2} \paren {b^6 \sinh^{-1} \dfrac a b + c^6 \sinh^{-1} \dfrac a c - {r_1}^2 \paren { {r_1}^4 - 8 b^2 c^2} \sinh^{-1} \dfrac a {r_1} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b c^2} \paren {c^6 \sinh^{-1} \dfrac b c + a^6 \sinh^{-1} \dfrac b a - {r_2}^2 \paren { {r_2}^4 - 8 c^2 a^2} \sinh^{-1} \dfrac b {r_2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b^2 c} \paren {a^6 \sinh^{-1} \dfrac c a + b^6 \sinh^{-1} \dfrac c b - {r_3}^2 \paren { {r_3}^4 - 8 a^2 b^2} \sinh^{-1} \dfrac c {r_3} }
      | c = 
}}
{{eqn | o =
      | ro= -
      | r = \dfrac 4 {15 a b c} \paren {a^4 \arcsin \dfrac {b c} {r_2 r_3} + b^4 \arcsin \dfrac {a c} {r_3 r_1} + c^4 \arcsin \dfrac {a b} {r_1 r_2} }
      | c = 
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = r
      | r = \sqrt {a^2 + b^2 + c^2}
}}
{{eqn | l = r_1
      | r = \sqrt {b^2 + c^2}
}}
{{eqn | l = r_2
      | r = \sqrt {a^2 + c^2}
}}
{{eqn | l = r_3
      | r = \sqrt {a^2 + b^2}
}}
{{end-eqn}}

Proof:
Let $X_1$, $X_2$; $Y_1$, $Y_2$; $Z_1$, $Z_2$ be pairs of independent random variables with a continuous uniform distribution in $\closedint {-a} a$, $\closedint {-b} b$ and $\closedint {-c} c$.
The random variables $U = \size {X_1 - X_2}$, $V = \size {Y_1 - Y_2}$, $W = \size {Z_1 - Z_2}$ are likewise independent.
Thus the probability density functions $\map f U$, $\map g V$, $\map h W$ can be shown to be:
{{begin-eqn}}
{{eqn | l = \map f u
      | r = \dfrac {2 a - u} {2 a^2}
}}
{{eqn | l = \map f v
      | r = \dfrac {2 a - v} {2 a^2}
}}
{{eqn | l = \map f w
      | r = \dfrac {2 a - w} {2 a^2}
}}
{{end-eqn}}
Hence the sought expectation is given by:
:$E = \ds \dfrac 1 {8 a^2 b^2 c^2} \int_0^{2 c} \int_0^{2 b} \int_0^{2 a} \sqrt {u^2 + v^2 + v^2} \paren {2 a - u} \paren {2 b - v} \paren {2 c - w} \rd u \rd v \rd w$
Let $\map P {a, b, c}$ be the pyramid defined by the planes $u = 2 a$, $v = 0$, $w = 0$, $a v = b u$ and $a w = c u$.
Set:
:$\map F {a, b, c} = \ds \dfrac 1 {8 a^2 b^2 c^2} \iiint_{\map P {a, b, c} } \sqrt {u^2 + v^2 + v^2} \paren {2 a - u} \paren {2 b - v} \paren {2 c - w} \rd u \rd v \rd w$
By symmetry:
:$\map F {a, b, c} = \map F {a, c, b}$
and so:
:$(1): \quad E = \dfrac 1 {8 a^2 b^2 c^2} \paren {\map F {a, b, c} + \map F {c, a, b} + \map F {b, c, a} }$
All that remains is to calculate $\map F {a, b, c}$.
Using spherical coordinates we obtain:
$\map F {a, b, c} = \ds \int_0^{\arctan \frac b a} \int_{\arccot \frac c a \cos \theta}^{\frac \pi 2} \int_0^{2 a \cosec \phi \sec \theta} \Phi \rd \rho \rd \phi \rd \theta$
where:
:$\Phi = \rho^3 \paren {2 a - \rho \sin \phi \cos \theta} \paren {2 b - \rho \sin \phi \sin \theta} \paren {2 c - \rho \cos \phi} \sin \phi$
After integration:
{{begin-eqn}}
{{eqn | l = \map F {a, b, c}
      | r = \dfrac {64} {315} a^7
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \dfrac 8 {315} a^2 \paren {8 a^4 - 19 a^2 b^2 - 6 b^4} r_3
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \dfrac 8 {315} a^2 \paren {8 a^4 - 19 a^2 c^2 - 6 c^4} r_2
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \dfrac 8 {315} a^2 \paren {8 a^4 - 6 b^4 - 6 c^4 + 19 a^2 b^2 + 30 b^2 c^2} r
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \dfrac 8 {15} a^6 b \sinh^{-1} \dfrac b a + \dfrac 8 {15} a^6 c \sinh^{-1} \dfrac c a
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \dfrac 8 {15} a^2 b \paren {a^4 - 4 a^2 c^2 - c^4} \sinh^{-1} \dfrac b {r_2}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \dfrac 8 {15} a^2 c \paren {a^4 - 4 a^2 b^2 - b^4} \sinh^{-1} \dfrac c {r_3}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \dfrac {32} {15} a^5 b c \arcsin \dfrac {b c} {r_2 r_3}
      | c = 
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = r
      | r = \sqrt {a^2 + b^2 + c^2}
}}
{{eqn | l = r_1
      | r = \sqrt {b^2 + c^2}
}}
{{eqn | l = r_2
      | r = \sqrt {a^2 + c^2}
}}
{{eqn | l = r_3
      | r = \sqrt {a^2 + b^2}
}}
{{end-eqn}}
Substituting in $(1)$ we obtain:
{{begin-eqn}}
{{eqn | l = E
      | r = \dfrac {2 r} {15} - \dfrac 7 {45} \paren {\paren {r - r_1} \paren {\dfrac {r_1} a}^2 + \paren {r - r_2} \paren {\dfrac {r_2} b}^2 + \paren {r - r_3} \paren {\dfrac {r_3} c}^2}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 8 {315 a^2 b^2 c^2} \paren {a^7 + b^7 + c^7 - {r_1}^7 - {r_2}^7 - {r_3}^7 + r^7}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a b^2 c^2} \paren {b^6 \sinh^{-1} \dfrac a b + c^6 \sinh^{-1} \dfrac a c - {r_1}^2 \paren { {r_1}^4 - 8 b^2 c^2} \sinh^{-1} \dfrac a {r_1} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b c^2} \paren {c^6 \sinh^{-1} \dfrac b c + a^6 \sinh^{-1} \dfrac b a - {r_2}^2 \paren { {r_2}^4 - 8 c^2 a^2} \sinh^{-1} \dfrac b {r_2} }
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \dfrac 1 {15 a^2 b^2 c} \paren {a^6 \sinh^{-1} \dfrac c a + b^6 \sinh^{-1} \dfrac c b - {r_3}^2 \paren { {r_3}^4 - 8 a^2 b^2} \sinh^{-1} \dfrac c {r_3} }
      | c = 
}}
{{eqn | o =
      | ro= -
      | r = \dfrac 4 {15 a b c} \paren {a^4 \arcsin \dfrac {b c} {r_2 r_3} + b^4 \arcsin \dfrac {a c} {r_3 r_1} + c^4 \arcsin \dfrac {a b} {r_1 r_2} }
      | c = 
}}
{{end-eqn}}
{{qed}}
{{Proofread|Bound to be numerous transcription errors}}