What is the closed form of the following integral? I want to know the closed form of the following master integral in (any) $D$ dimension
\begin{equation}
\int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2(k+r)^2(k+p)^2}.
\end{equation}
The references that I can find have formulas for the integral with very general propagators $\frac{1}{p^{2\alpha}}$, but they involve some hideous looking hypergeometric functions and are half-page long, which makes the formulas almost impossible to use. I wonder if there is a simpler fomula that is specific to the simplest propagator $\frac{1}{p^2}$. Thanks in advance!
BTW, the real integral that I am interested in is the following
\begin{equation}
\int\frac{d^D k}{(2\pi)^D}\frac{k^\mu k^\nu}{k^2(k+r)^2(k+p)^2}.
\end{equation} 
I know that it is related to the master integral by the Passarino-Veltman reduction.
 A: The integral to evaluate is
\begin{equation}
I = \int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2(k+r)^2(k+p)^2}.
\end{equation}

The usual way to do this to use Feynman parameters. The formula is given by 
$$ {\displaystyle {\frac {1}{A_{1}\cdots A_{n}}}=\left(n-1\right)!\int _{0}^{1}d\alpha _{1}\cdots \int _{0}^{1}d\alpha _{n}{\frac {\delta \left(1-\alpha _{1}-\cdots -\alpha _{n}\right)}{[\alpha _{1}A_{1}+\cdots +\alpha _{n}A_{n}]^{n}}}.}
$$
Using this for the case of $n=3$ (your case) we have that 
$$
\frac{1}{k^2(k+r)^2(k+p)^2} = 2! \int dx_1 \int dx_2 \int dx_3 \frac{\delta(1 - x_1 - x_2 - x_3)}{[x_1 k^2 + x_2(k+r)^2 + x_3 (k+p)^2]^3}
 $$
Then swap the order of integration so that 
$$ I =  2!\iiint dx_1 dx_2 dx_3\delta(1 - x_1 - x_2 - x_3) \int\frac{d^D k}{(2\pi)^D}\frac{1}{[x_1 k^2 + x_2(k+r)^2 + x_3 (k+p)^2]^3} $$
Then expand the denominator into an order 2 polynomial in $k$ so it is of the form 
$$ I = 2!\iiint dx_1 dx_2 dx_3  \delta(1 - x_1 - x_2 - x_3) \int\frac{d^D k}{(2\pi)^D}\frac{1}{[f(x_i) k^2 + g(x_i)k + h(x_i)]^3} \tag{a} $$
where $f(x_i):= f(x_1, x_2, x_3)$. Now, the innermost integral in $(a)$ is given in closed form in the appendix of Peskin in terms of $\Gamma$ functions and other stuff. 

Now is when I leave you, as integrating over the feynman parameters is usually the ugliest part of the problem. Moreover, if all you want is a perturbative answer then this is certainly the way to go.
I could work through it, but it would be of little value to you (and me). So I will merely wish you good luck and a welcome you to the QFT Integral Club.
