OK. It is going to be long and extremely boring, but, as promised, here goes. We'll get the precision $O(mL^2)$ where $m=\max(a,b,c), L=\max(\log\frac 1a,\log\frac 1b,\log\frac 1c)$, which should be enough for any reasonable interpretation of "$a,b,c$ going to $0$ at about the same rate".

We start with the approximation of an auxiliary integral:
$$
I(a,b)=\iint_{[a,+\infty)\times[b,+\infty)}\frac{1}{(1+x+y)^2}\frac{dx}{x}\frac{dy}{y}
$$
as $a,b\to 0+$.

First, we have
$$
\frac{1}{(1+x+y)^2}-\frac 1{(1+x)^2}\frac 1{(1+y)^2}=\frac{xy(2+2x+2y+xy)}{(1+x+y)^2(1+x)^2(1+y)^2}
\\
\le
\frac{2xy}{(1+x)^{2}(1+y)^{2}}\,,
$$
which means that $I(a,b)=J(a)J(b)+D_1+O(a+b)$ where
$$
J(a)=\int_{a}^\infty\frac{1}{(1+x)^2}\frac{dx}x\,,
$$
and
$$
D_1=\iint_{(0,\infty)^2}\frac{(2+2x+2y+xy)dx\,dy}{(1+x+y)^2(1+x)^2(1+y)^2}
$$
(below we will always denote by $D_k$ various numerical constants whose values are expressed by some convergent integrals. Some of them are easy to evaluate and some aren't, but that distinction won't bother us here).

Now
$$
J(a)=\int_a^\infty\left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}+\log\frac 1a\,.
$$
Since for $x<1$, we have $\frac 1{1+x}-1=-\frac x{1+x}$, we conclude that
$$
J(a)=\log\frac 1a+D_2+O(a)
$$
with
$$
D_2=\int_0^\infty \left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}
$$
Multiplying out, we get
$$
I(a,b)=\log\frac 1a\log\frac 1b+D_2(\log\frac 1a+\log\frac 1b)+D_3
\\
+O\left((a+b)(\log\frac 1b+\log\frac 1a)\right)\,.
$$
Now we can return to our quadruple integral. Let us rewrite the integrand in the form
$$
\frac{1}{(\lambda_1+\lambda_2+c)^2}\frac{1}{\left(1+\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_1}+
\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_2}\right)^2}
\,d\lambda_1\,d\lambda_2\,\frac{d\alpha_1}{\alpha_1}\,\frac{d\alpha_2}{\alpha_2}
$$
For fixed $\lambda_1,\lambda_2$, we can now happily integrate $\alpha_1,\alpha_2$ out using the computation above to reduce our integral to
$$
\iint_{[0,1]^2}
\frac 1{(\lambda_1+\lambda_2+c)^2}\left[\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}
\\+
D_2\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)+D_3\right]\,d\lambda_1\,d\lambda_2+E_1
$$
where the error term $E_1$ is controlled by
$$
\iint_{[0,1]^2}\frac{1}{(\lambda_1+\lambda_2+c)^2}\left[
\left(\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}+
\frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\right)
\\
\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)
\right]\,d\lambda_1\,d\lambda_2
$$

Now we do not need a special investigation of $E_1$ because the first factor in the brackets is bounded by $a+b$ and if we look at the rest, it is just one of the terms in the main asymptotic integral. Next, once we open the parentheses and expand logarithms of products, we see that we just need to approximate three double integrals:
$$
I_2=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_2^2}\,,
$$
$$
I_1=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\,,
$$
and
$$
I_0=\iint_{[0,1]^2}
\frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}
$$
IT is not hard to convince yourself that in all $3$ cases the integral over the domain $\lambda_1+\lambda_2>1$ tends to its value at $c=0$ and the deviation from that value is $O(c)$, so we can replace those parts by appropriate constants. In the other part it is convenient to make the change of variable $\lambda_1=v\lambda,\lambda_2=(1-v)\lambda$ ($0<v,\lambda<1$) and to integrate $v$ out (again splitting logarithms of products into sums). We see that we need to approximate
$$
I_2'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log^2\frac{\lambda+c}{\lambda^2}\,,
$$
$$
I_1'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log\frac{\lambda+c}{\lambda^2}
$$
$$
I_0'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}
$$

To be continued...

pathintegral. By "Feynman integral" in his edit, David S-D means an integral arising from the evaluation of a Feynman diagram (which are related, as they appear in perturbative calculations of path integrals). See en.wikipedia.org/wiki/Feynman_diagram and en.wikipedia.org/wiki/Feynman_parametrization $\endgroup$ – j.c. May 20 '18 at 15:453more comments