I know that the probability density function of the $r$th order statistics is given by $$f_r(x)=C_{r:n}[F(x;\theta)^{r-1}[1-F(x;\theta)]^{n-r}f(x;\theta)],$$ where $C_{r:n}=\frac{n!}{(r-1)!(n-r)!}.$
So I am wondering is there a similar expression of the density of the order statistics of $X_1$, $X_2$,...,$X_n$ if $X_i$ is distributed as $F(x;\theta_i)$ (so the same form but different parameters)?
In general, you can use a same principle to derive the pdf but the you cannot simplify the sum as nice as this one, unless in some special case where the parameters are in a particular sequences which allow you to simplify. I just write out the case explicitly for $n = 3$ as example: $$ \begin{align} f_{X(1)}(x) &=& f_{X_1}(x)[1 - F_{X_2}(x)][1 - F_{X_3}(x)] \\ &+& f_{X_2}(x)[1 - F_{X_1}(x)][1 - F_{X_3}(x)] \\ &+& f_{X_3}(x)[1 - F_{X_1}(x)][1 - F_{X_2}(x)] \end{align}$$
$$ \begin{align} f_{X(2)}(x) &=& f_{X_1}(x)F_{X_2}(x)[1 - F_{X_3}(x)] \\ &+& f_{X_1}(x)F_{X_3}(x)[1 - F_{X_2}(x)] \\ &+& f_{X_2}(x)F_{X_1}(x)[1 - F_{X_3}(x)] \\ &+& f_{X_2}(x)F_{X_3}(x)[1 - F_{X_1}(x)] \\ &+& f_{X_3}(x)F_{X_1}(x)[1 - F_{X_2}(x)] \\ &+& f_{X_3}(x)F_{X_2}(x)[1 - F_{X_1}(x)] \\ \end{align}$$
$$ \begin{align} f_{X(3)}(x) &=& f_{X_1}(x)F_{X_2}(x)F_{X_3}(x) \\ &+& f_{X_2}(x)F_{X_1}(x)F_{X_3}(x) \\ &+& f_{X_3}(x)F_{X_1}(x)F_{X_2}(x) \end{align}$$
For the general $n$ it is similar - you need to sum over the possible combinations