Show that a definite integral vanishes for all values of  $a,b,c > 0$. Show $$\int_0^\infty \int_0^\infty \frac{(ax-by) {\rm e}^{-x} {\rm e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}} \,dx \,dy = 0$$ for any $a,b,c > 0$.
I came upon the above double integral when simplifying an expression for a probability density function. How can I demonstrate that the integral is zero for any positive constants, $a$, $b$ and $c$?
I've verified the result numerically. At the moment it seems rather fascinating to me that the integral should always be zero given the presence of three free variables, and I would expect there to be a relatively simple derivation.
I've tried all sorts of approaches (integral substitutions including polar coordinates, differentiating under the integral, splitting the domain of integration into pieces) without seeming to make progress.

To answer Ali's comment, below I give my crude Matlab code for evaluating the double integral (with the trapezoidal rule) and one output. Regardless of what positive values of $a$,$b$,$c$ I put in, I get something close enough to zero for my liking.
Output of program:
double_int = getDoubleInt(2,3,4)

double_int =
            -6.1549e-010

Program code:
function double_int = getDoubleInt(a,b,c)

x_max = 25;
y_max = 25;
NN_x = 1000;
NN_y = 1000;

x_vec = logspace(log10(x_max/10^10),log10(x_max),NN_x);

y_vec = logspace(log10(y_max/10^10),log10(y_max),NN_y);

XX = x_vec'*ones(1,NN_y);

YY = ones(NN_x,1)*y_vec;

ZZ = (a*XX-b*YY)./(a^2*XX+b^2*YY+c*XX.*YY).^(3/2).*exp(-XX).*exp(-YY);

int_1 = zeros(size(x_vec));

for i = 1:NN_x

int_1(i) = trapz(y_vec,ZZ(i,:));

end

double_int = trapz(x_vec,int_1);

 A: Using the substitution
$$
r\frac{1+t}{1-t}=\frac1r\frac{1+s}{1-s}
$$
maps $[-1,1]\mapsto[-1,1]$ and has proven useful in several situations similar to this. Some formulas pertinent to this substitution are
$$
\frac{\mathrm{d}t}{1-t^2}=\frac{\mathrm{d}s}{1-s^2}
$$
$$
\frac{r(1+t)+\frac1r(1-t)}{1-t^2}=\frac{\frac1r(1+s)+r(1-s)}{1-s^2}
$$
$$
\frac{r(1+t)-(1-t)}{\sqrt{1-t^2}}=\frac{(1+s)-r(1-s)}{\sqrt{1-s^2}}
$$
Change variables $u=x+y$ and $v=x-y$, then $v=ut$:
$$
\begin{align}
&\int_0^\infty\int_0^\infty\frac{(ax-by)\,\mathrm{e}^{-x}\,\mathrm{e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}}\,\mathrm{d}x\,\mathrm{d}y\\[6pt]
&=2\int_0^\infty\int_{-u}^u\frac{(a-b)u+(a+b)v}{\left(cu^2-cv^2+2(a^2+b^2)u+2(a^2-b^2)v\right)^{3/2}}\,\mathrm{e}^{-u}\,\mathrm{d}v\,\mathrm{d}u\\[6pt]
&=2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)t}{\left(cu^2(1-t^2)+2u(a^2(1+t)+b^2(1-t))\right)^{3/2}}\,\mathrm{d}t\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{1}
\end{align}
$$
Using the substitution $\frac{a}{b}\frac{1+t}{1-t}=\frac{b}{a}\frac{1+s}{1-s}$ with $(1)$ yields
$$
2\int_0^\infty\int_{-1}^1\frac{(b-a)+(a+b)s}{\left(cu^2(1-s^2)+2u(b^2(1+s)+a^2(1-s))\right)^{3/2}}\,\mathrm{d}s\,\mathrm{e}^{-u}u^2\,\mathrm{d}u
$$
Then substituting $w=-s$ yields
$$
-2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)w}{\left(cu^2(1-s^2)+2u(a^2(1+w)+b^2(1-w))\right)^{3/2}}\,\mathrm{d}w\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{2}
$$
Since $(1)$ and $(2)$ are equal, yet negatives, they are both $0$.
A: Take $a=c=1$ and $b$ very small.
A: The result is surprising because one would think that the combination with the exponentials would be too complicated to make the integral vanish for all $a,b,c$. However, the exponentials don't enter into it because the integral along every line with constant sum $x+y$ is zero, and the exponential factor is constant on these lines. Consider
$$
x=u+v\;,
\\
y=u-v\;,
$$
which, up to a constant factor from the Jacobian, transforms the integral into
$$\int_0^\infty\mathrm du\mathrm e^{-2u}\int_{-u}^u\mathrm dv\frac{a(u+v)-b(u-v)}{\left(a^2(u+v)+b^2(u-v)+c(u^2-v^2)\right)^{3/2}}\;.$$
The inner integral is readily performed; Wolfram|Alpha gives
$$\int_0^\infty\mathrm du\mathrm e^{-2u}\frac2{(a-b)^2+2cu}\left[\frac{a(u+v)+b(u-v)}{\sqrt{a^2(u+v)+b^2(u-v)+c(u^2-v^2)}}\right]_{v=-u}^{v=u}\;,$$
and the result is $0$ since the antiderivative takes the same value at $u$ and $-u$.
