Loop integral using Feynman's trick I am trying to show for the one-loop integral with three propagators with different internal masses $m_1$, $m_2$, $m_3$, and all off-shell external momenta $p_1$, $p_2$, $p_3$ the following formula appearing in 't Hooft(1979), Bardin (1999), Denner (2007):  (unfortunate metric $-,+,+,+$)
$$\int d^d q\frac{1}{(q^2+m_1^2)((q+p1)^2+m_2^2)((q+p_1+p_2)^2+m_3^2)}
$$
$$=i\pi^2\int_0^1dx\int_0^xdy\frac{1}{ax^2+by^2+cxy+dx+ey+f}$$
where $a$, $b$, $c$, ... are coefficients depending on the momenta in the following way:
$a=-p_2^2$,
$b=-p_1^2$,
$c=-2p_1.p_2$,
$d=m_2^2-m_3^2+p_2^2$,
$e=m_1^2-m_2^2+p_1^2+2(p_1.p_2)$,
$f=m_3^2-i\epsilon$.
I don't really care about factors in fromt like $i\pi^2$.  My simple problem is:  I am totally unable to reproduce coefficients $d$, $e$ and $f$.  The problem is, when I integrate over the third Feynman parameter, $m_3$ appears in all three coefficients $d$, $e$ and $f$.  How do I squeeze the denominators to reproduce this formula?
 A: Define the LHS of the equation above:
$$I=\int d^d q\frac{1}{(q^2+m_1^2)((q+p_1)^2+m_2^2)((q+p_1+p_2)^2+m_3^2)}$$
The first step is to squeeze the denominators using Feynman's trick:
$$I=\int_0^1 dx\,dy\,dz\,\delta(1-x-y-z)\int d^d q\frac{2}{[y(q^2+m_1^2)+z((q+p_1)^2+m_2^2)+x((q+p_1+p_2)^2+m_3^2)]^3}$$
The square in $q^2$ may be completed in the denominator by expanding:
$$[\text{denom}]=q^2+2q.(z p_1+x(p_1+p_2))+y m_1^2+z (p_1^2+m_2^2)+x(m_3^2+(p_1+p_2)^2)$$
$$=q^2+2q.Q+A^2\,$$
where $Q^\mu=z p_1^\mu+x(p_1+p_2)^\mu$ and $A^2=y m_1^2+z (p_1^2+m_2^2)+x(m_3^2+(p_1+p_2)^2)$, and by shifting the momentum, $q^\mu=(k-Q)^\mu$ as a change of integration variables.  Upon performing the $k$ integral, we are left with integrals over Feynman parameters (because this integral has three propagators, it is UV finite):
$$I=i\pi^2\int_0^1 dx\,dy\,dz\,\delta(1-x-y-z)\frac{1}{[-Q^2+A^2]}$$
Now integrate over $z$ with the help of the Dirac delta:
$$I=i\pi^2\int_0^1 dx\int_0^{1-x}dy \frac{1}{[-Q^2+A^2]_{z\rightarrow1-y-z}}$$
To arrive at the RHS of the OP's equation(which is the part I forgot to do), we make a final change of variables: $x=1-x'$:
So that the denominator reads $ax^2+by^2+cxy+dx+ey+f$, with the coefficients $a,b,c,\ldots$ exactly defined in the question of OP.  Note the change in the range of integration in $y$.
$$I=i\pi^2\int_0^1dx\int_0^xdy\frac{1}{ax^2+by^2+cxy+dx+ey+f}$$
