In view of the identity $$\frac{c_1a^2}{c_1a^2+c_2b^2}=1-\frac{c_2b^2}{c_1a^2+c_2b^2}$$ and the exchangeability of $a$ and $b$, we may assume that $c_2\le c_1$.
The expectation in question is $I/c_1$, where $$I:=E\frac{a^2}{a^2+cb^2},\quad c:=\frac{c_2}{c_1}\in(0,1].$$ Using the identity $1/A=\int_0^\infty du\, e^{-Au}$ for real $A>0$ and the Fubini--Tonelli theorem, we have $$I=\int_0^\infty du\, Ea^2 e^{-(a^2+cb^2)u} =\int_0^\infty du\, Ea^2 e^{-u a^2}\, Ee^{-cu b^2}.$$ Next, for real $u>0$, $$Ee^{-u b^2}=(1+2u)^{-1/2}$$ and hence $$Ea^2 e^{-u a^2}=-((1+2u)^{-1/2})'=(1+2u)^{-3/2}.$$ So, using the substitution $u=(x^2-1)/2$, we get $$I=\frac1{\sqrt c}\int_1^\infty \frac{dx}{x^2\,\sqrt{x^2+r^2}},$$ where $r:=\sqrt{\dfrac{1-c}c}$. So, $I$ is easily found if $c=1$. If $c\in(0,1)$, use the standard substitution $x=r\tan t$ to find $I$. Collecting all the pieces, we get get that the expectation in question is $$\frac1{c_1+\sqrt{c_1c_2}}.$$