Is there a closed form for the average energy of a QAM constellation?
From $\bar{E}=\frac{(M-1)d^2}{6}$ as the average energy in $M$-QAM, $M$ a perfect square and a square constellation, we have that the total energy is
$$E_{\operatorname{tot}} = M\frac{(M-1)d^2}{6}$$
Now, since the $M$-QAM signal is the sum of two orthogonal $\sqrt{M}$-PAM
signals of equal energy, then, writing $N$ instead of $\sqrt{M}$,
we have each of the orthogonal signals carrying a total energy of 
$$\displaystyle \frac{1}{2}\frac{M(M-1)d^2}{6}
= \frac{N^2(N^2-1)d^2}{12}.$$
Since there are $M = N^2$ constellation points, the average energy of
a $N$-PAM signal with spacing $d$ is
$\displaystyle \frac{(N^2-1)d^2}{12}$.

Now, if $b$ is odd, a $2^b$-QAM signal constellation is
a rectangular grid with $2^{(b-1)/2}$ points in one direction and
$2^{(b+1)/2}$ points in the other.  Thus, the $2^b$ points contribute an
average of
$$\frac{(2^{b-1}-1)d^2}{12} +  \frac{(2^{b+1}-1)d^2}{12}
= \frac{(2.5M-2)d^2}{12}$$

Note that it is possible to use constellations other than rectangular when
$b$ is odd. For example, the $32$-CROSS constellation is a (rotated) version
of a $36$-AM square constellation with the corner points deleted. Rotation
does not change the energy and so we can get the answer for $32$-CROSS by
finding the total energy for a $36$-QAM square constellation, subtracting
off the energy of the $4$ corner points, and dividing what is left by $32$.