6-dimensional integral with strange result I am working through a physics paper where I encounter the following 6-dimensional integral:
$$ G(\mathbf{q}) = \frac{2}{(2\pi)^6 n^2} \int_{k \leq q_F} \int_{k' \leq q_F} \frac{\mathbf{q} \cdot (\mathbf{q} + \mathbf{k} -\mathbf{k}')}{|\mathbf{q} + \mathbf{k} -\mathbf{k}'|^2} \mathrm{d}^3 \mathbf{k} \mathrm{d}^3 \mathbf{k}'.$$
Here $\mathbf{q},\mathbf{k},\mathbf{k}'$ are vectors in $\mathbb{R}^3$ and $q,k,k'$ are their euclidean norms.
The authors present a solution to this intergal which is claimed to be exact, but sadly without any ansatz or idea to solve it. It reads:
\begin{align} 
G(\mathbf{q}) = \frac{9}{32} \left( \frac{q}{q_F} \right)^2 \left\lbrace \frac{2}{105} \left[ 24 \left( \frac{q_F}{q} \right)^2 + 44 + \left( \frac{q}{q_F} \right)^2 \right] \\
 - 2 \frac{q_F}{q} \left[ \frac{8}{35} \left( \frac{q_F}{q} \right)^2 - \frac{4}{15} + \frac{1}{6} \left( \frac{q}{q_F} \right)^2 \right] \ln \Bigg \vert \frac{q + 2q_F}{q - 2q_F} \Bigg \vert \\ 
 + \left( \frac{q}{q_F} \right)^2 \left[ \frac{1}{210} \left( \frac{q}{q_F} \right)^2 - \frac{2}{15} \right] \ln \Bigg \vert \frac{q^2 - 4q_F^2}{q^2} \Bigg \vert\right\rbrace  
\end{align}
I find this result rather strange because of all the numerical fractions and I have never seen any result similar to this, which is why I don´t have any idea how to arrive at this result.
So far the only idea I had was to substitute with a relative coordinate $\mathbf{t} = \mathbf{k} -\mathbf{k}'$, but I have trouble finding out the new integral boundaries. Further I would take $\mathbf{q}$ parallel to the $z$-axis, such that I can use $\mathbf{q} \cdot \mathbf{k} = qk \cos\theta$ and use spherical coordinates. Also the definitions $n = \frac{q_F^3}{3\pi^2}$ and $\int_{k \leq q_F} \mathrm{d}^3 \mathbf{k} = \frac{4\pi}{3}q_F^3$ could be helpful. Maybe someone has done something similar before and knows how to approach an integral like this.
Any help or ideas would be greatly appreciated, thanks in advance!

So for anyone still interested or coming across this later, here is the solution. A colleague and I finally managed to solve it. It is however quite long and tedious, so I will only present the main ideas and the end results.
We start with a transformation to $\mathbf{r} = \frac{\mathbf{k} + \mathbf{k}'}{2}$ and $\mathbf{t} = \mathbf{k} - \mathbf{k}'$ such that
\begin{align}
\mathbf{k} = \mathbf{r} + \frac{\mathbf{t}}{2} \text{ and } \mathbf{k}' = \mathbf{r} 
- \frac{\mathbf{t}}{2} \,.
\end{align}
One can then verify that the absolute value of the determinant of the Jacobian is equal to $1$, so for the differentials we now have $\mathrm{d} \mathbf{k} \mathrm{d} \mathbf{k}' = \mathrm{d} \mathbf{r} \mathrm{d} \mathbf{t}$. From the integration bounds the conditions
$$|\mathbf{k}| = |\mathbf{r} + \frac{\mathbf{t}}{2}| \leq q_F \text{ and } |\mathbf{k}'| = |\mathbf{r} - \frac{\mathbf{t}}{2}| \leq q_F$$
follow. This can be visualized by two spheres with radius $q_F$ overlapping. $\mathbf{t}$ is then the vector connection their centers. $\mathbf{r}$ has its origin at the midpoint of the line connecting their centers and all possible $\mathbf{r} \pm \frac{\mathbf{t}}{2}$ are confined to the overlapping region. For a similar picture see Fetter, Walecka - Quantum theory of many-particle systems,2003 pg. 28. We then express the integration over $\mathrm{d} \mathbf{r}$ using step functions as
\begin{align}
\int \Theta(q_F - |\mathbf{r} + \tfrac{\mathbf{t}}{2}|)\Theta(q_F - |\mathbf{r} - \tfrac{\mathbf{t}}{2}|) \mathrm{d} \mathbf{r} = 2V \Theta(1-x)
\end{align}
where we have set $x = \frac{t}{2q_F}$ and $V$ is the volume of the spherical cap which makes up the overlapping region and has to be multiplied by $2$ because of symmetry. $V$ can be calculated using the formulas form https://en.wikipedia.org/wiki/Spherical_cap with $h = q_F - \frac{t}{2}$ and $a^2 = q_F^2 - \frac{t^2}{4}$, so the result of the integral in terms of $x$ is
\begin{align}
\frac{4 \pi}{3} q_F^3 \left(1 - \tfrac{3}{2} x + \tfrac{1}{2} x^3 \right) \Theta(1-x) \,.
\end{align}
The step function is there to account for values of $\frac{t}{2}>q_F$ in which case there would be no overlap, so everything would be equal to $0$. We can also plug in $n = \frac{q_F^3}{3\pi^2}$ to rewrite the prefactor, leaving us with the integral
\begin{align}
G(\mathbf{q}) = \frac{3}{8\pi q_F^3} \int \left( 1 - \tfrac{3}{2} x(t) + \tfrac{1}{2} x^3(t) \right) \Theta(1-x(t)) \frac{\mathbf{q} \cdot (\mathbf{q}+\mathbf{t})}{|\mathbf{q} + \mathbf{t}|^2} \mathrm{d} \mathbf{t} \,.
\end{align}
If we set $\mathbf{q}$ parallel to the $z$-axis we can make use of spherical coordinates and we arrive at
\begin{align}
G(\mathbf{q}) = \frac{3}{4q_F^3} \int_0^{2q_F} \left( 1 - \tfrac{3}{4q_F} t + \tfrac{1}{16q_F^3} t^3 \right) \int_0^\pi \frac{q^2 t^2 \sin\theta + q t^3 \cos\theta \sin\theta}{q^2 + t^2 + 2qt\cos\theta} d\theta dt \,.
\end{align}
Evaluating the angular part gives
\begin{align}
A(q,t) \equiv q t \ln \left| \frac{q+t}{q-t} \right| + \frac{t(q^2 + t^2)}{2q} \ln \left| \frac{q-t}{q+t} \right| + t^2 \,.
\end{align}
We now have to perform the $t$-integration over the product of $A(q,t)$ with the polynomial. As already said at the beginning this is a very long calculation, but it is elementary and does not require any special tricks, you just have to be really careful because there are going to be a lot of terms. I will present the solution of each part of the polynomial:
\begin{align}
    \frac{3}{4q_F} \int_0^{2q_F} A(q,t) dt &= \left[ \frac{3}{4} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{3}{2} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| + \frac{3}{8} \left( \frac{q}{q_F} \right)^2 + \frac{3}{2} \\
    - \frac{9}{16 q_F^4} \int_0^{2q_F} t A(q,t) dt &= - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ - \frac{3}{4} \left( \frac{q}{q_F} \right) + \frac{9}{5} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \\
    & \quad - \frac{3}{20} \left( \frac{q}{q_F} \right)^2 - \frac{9}{5} \\
    \frac{3}{64q_F^6} \int_0^{2q_F} t^3 A(q,t) dt &= \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{7} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right|\\
    & \quad + \frac{3}{280} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{3}{7} \,.
\end{align}
If we add up all terms and simplify we end up with
\begin{align}
    G(q) &= \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{9}{70} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \notag \\
    & \quad + \left[ - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 + \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \right] \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| \notag \\
    & \quad + \frac{33}{140} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{9}{70}
\end{align}
which after factorizing and grouping is exactly the desired equation. You can also check by expanding the expression of $G(\mathbf{q})$ given in the question.
