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, 
$$I=\int_0^\infty du\, (1+2u)^{-3/2} (1+2cu)^{-1/2}.$$
Using now 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 that the expectation in question is
$$\frac1{c_1+\sqrt{c_1c_2}}.$$

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Similarly and more generally, the calculation of 
$$E\frac{Z_1^2}{c_1^2Z_1^2+\cdots+c_k^2Z_k^2}$$
for any natural $k\ge2$ reduces to the calculation of 
$$I_k:=\int_1^\infty \frac{dx}{x^2\,\sqrt{x^2+R_2}\cdots\sqrt{x^2+R_k}}$$
for real $R_2,\dots,R_k>-1$. 

For $k=3$, the integral $I_k$ can be expressed in terms of elliptic functions, as seen from the following image of a Mathematica notebook: 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/lroti.png