Finding a bound on the integral of function of exponentials I am currently trying to show that an function has a finite integral. I am trying to find the iterated integral of:
$$
\frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}}\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}}
$$
over $-\infty < \alpha, \beta <\infty$. 
I am able to see that $\frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}} < e^{|\alpha + 0.3\beta|}$, and that $\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}} < e^{|\alpha + 0.05\beta|}$.
Hence, I have that:
$$
\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \frac{e^{\alpha + 0.3\beta}}{1+e^{\alpha + 0.3\beta}}\frac{1}{1+e^{\alpha + 0.3\beta}}\frac{e^{\alpha + 0.05\beta}}{1+e^{\alpha + 0.05\beta}}\frac{1}{1+e^{\alpha + 0.05\beta}} d\alpha d\beta
$$
$$
< \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}e^{|\alpha + 0.3\beta|} e^{|\alpha + 0.05\beta|}d\alpha d\beta
$$
However, I am not sure where to proceed from here. One option seems to be doing the Jacobian and using a transformation. Is there any easy way to bound this integral? Thanks!

We transform coordinates with
$$x=\alpha +0.3\beta$$
$$y=\alpha+0.05\beta$$
Then, the Jacobian of transformation is $4$, and the integral of interest $I$ is
$$\begin{align}
I=&4\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{x}}{\left(1+e^x\right)^2}\frac{e^y}{\left(1+e^y\right)^2}\,dx\,dy\\\\
&=4\left(\int_0^\infty\frac{1}{(1+u)^2}\, du\right)^2\\\\
&=4
\end{align}$$
