This is following on from Douglas Zare's answer. While not an answer in its own right,it was getting too long for a comment. Briefly, we can hit the question with brute force and look for a generating function for the desired expected values.

So, put
$$f_j(y) = \sum_{k=0}^{j-1} {n \choose k} y^k (1-y)^{n-k}$$
so that using the notation of Douglas' answer, $g_j(x)=f_j(F(x))$, and the expectation of the $j$th order statistic is
$$ E_j := \int_0^\infty f_j(F(x)) \ dx $$

Put $G(y,z) = \sum_{j=1}^n f_j(y) z^j$ where $z$ is a formal variable. By linearity we have
$$ \sum_{j=1}^n E_j z^j = \int_0^\infty G(F(x),z) \ dx $$

We can try to write $G$ as a rational function in $y$ and $z$.
Expanding out and interchanging the order of summation gives
$$ \eqalign{
G(y,z) & = \sum_{j=1}^n\sum_{k=0}^{j-1} {n\choose k} y^k (1-y)^{n-k} z^j \\\\
& = \sum_{k=0}^{n-1} \sum_{j=k+1}^n z^j {n\choose k} y^k (1-y)^{n-k} \\\\
& = \sum_{k=0}^{n-1} \left( \sum_{j=1}^{n-k} z^j \right) \cdot z^k{n\choose k} y^k (1-y)^{n-k} \\\\
}
$$

Now
$$ \eqalign{
\sum_{j=1}^{M-1}jz^j
= z \frac{d}{dz}\sum_{j=0}^{M-1} z^j
& = z \frac{d}{dz}\left[\frac{1-z^M}{1-z}\right] \\\\
& = \frac{z-z^{M+1}}{(1-z)^2} - \frac{Mz^M}{1-z}
& = \frac{z-z^M}{(1-z)^2} - \frac{(M-1)z^M}{1-z}
} $$
so substituting this back in we get
$$ \eqalign{
G(y,z)
& = \sum_{k=0}^{n-1} \left[ \frac{z-z^{n-k+1}}{(1-z)^2} - \frac{(n-k)z^{n-k+1}}{1-z} \right] \cdot z^k{n\choose k} y^k (1-y)^{n-k} \\\\
& = \sum_{k=0}^{n-1} \frac{z}{(1-z)^2}{n\choose k} (yz)^k (1-y)^{n-k} \\\\
& \ \ - \sum_{k=0}^{n-1} \frac{z^{n+1}}{(1-z)^2} {n\choose k} y^k (1-y)^{n-k} \\\\
& \ \ - \sum_{k=0}^{n-1} \frac{nz^{n+1}}{1-z} { {n-1} \choose k } y^k(1-y)^{n-k} \\\\
& = \frac{z}{(1-z)^2} \left[ (1-y+yz)^n - (yz)^n \right]
- \frac{z^{n+1}}{(1-z)^2} (1-y^n)
- \frac{nz^{n+1}}{1-z}
} $$

I guess that in theory one could plug this back in to obtain a "formula" for the generating function $\sum_{j=1}^n E_j z^j$, but I can't see how that formula might then simplify to something calculable, unless $F$ has a rather special form.

`$1-(1-a)=a$`

– not that it helps much, but it had me scratching my head for a moment, wondering if something had got lost in all the parentheses. $\endgroup$1more comment