A special improper integral I need to evaluate the following integral:
$$\int_{0}^\infty  dx \left(\frac{x^2}{2x^2+1}\right)^a\left(\frac{x^2}{2x^2+t_y^2}\right)^b\left(\frac{x^2}{2x^2+t_z^2}\right)^c\frac{1}{x^3}$$
where, $a$, $b$, $c$, $t_y$ and $t_z$ are real.
I have succeeded to perform this integral only for two special cases:
1. $t_y$ = $t_z$ =1 and 2. $t_z$ = $t_y$
I shall be very thankful if someone could suggest a way to perform the integral above for $t_z \neq t_y$. Looking forward to hearing.  
 A: $$\int_0^\infty\left(\dfrac{x^2}{2x^2+1}\right)^a\left(\dfrac{x^2}{2x^2+t_y^2}\right)^b\left(\dfrac{x^2}{2x^2+t_z^2}\right)^c\dfrac{1}{x^3}~dx$$
$$=-2^{-a-b-c}\int_0^\infty\left(\dfrac{1}{1+\dfrac{1}{2x^2}}\right)^a\left(\dfrac{1}{1+\dfrac{t_y^2}{2x^2}}\right)^b\left(\dfrac{1}{1+\dfrac{t_z^2}{2x^2}}\right)^c~d\left(\dfrac{1}{2x^2}\right)$$
$$=2^{-a-b-c}\int_0^\infty\left(1+x\right)^{-a}\left(1+t_y^2x\right)^{-b}\left(1+t_z^2x\right)^{-c}~dx$$
$$=2^{-a-b-c}\int_0^1\left(1+\dfrac{u}{1-u}\right)^{-a}\left(1+\dfrac{t_y^2u}{1-u}\right)^{-b}\left(1+\dfrac{t_z^2u}{1-u}\right)^{-c}~d\left(\dfrac{u}{1-u}\right)$$
$$=2^{-a-b-c}\int_0^1(1-u)^{a+b+c-2}(1+(t_y^2-1)u)^{-b}(1+(t_z^2-1)u)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}F_1(1,b,c,a+b+c;1-t_y^2,1-t_z^2)}{a+b+c-1}~\text{when}~0<t_y,t_z<\sqrt2$$
(according to http://en.wikipedia.org/wiki/Appell_series#Integral_representations)
$$=2^{-a-b-c}\int_0^1u^{a+b+c-2}((t_y^2-1)+(2-t_y^2)u)^{-b}((t_z^2-1)+(2-t_z^2)u)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}}{(t_y^2-1)^b(t_z^2-1)^c}\int_0^1u^{a+b+c-2}\left(1-\dfrac{(t_y^2-2)u}{t_y^2-1}\right)^{-b}\left(1-\dfrac{(t_z^2-2)u}{t_z^2-1}\right)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}}{(a+b+c-1)(t_y^2-1)^b(t_z^2-1)^c}F_1\left(a+b+c-1,b,c,a+b+c;\dfrac{t_y^2-2}{t_y^2-1},\dfrac{t_z^2-2}{t_z^2-1}\right)~\text{when}~t_y,t_z\geq\sqrt2$$
(according to http://en.wikipedia.org/wiki/Appell_series#Integral_representations)
