Suppose $X_t$ is a Semi-martingale and $H_t$ is $X_t$-predictable.
I know that if $X_t=W_t$ is a Wiener process then
$$
\mathbb{E}[H\cdot W_T^2] = \mathbb{E}\bigg[\int_0^TH_t^2dt\bigg],
$$
where $H\cdot W_T$ denotes the stochastic integral of $H_t$ against $W_t$ up to time $T$.
My question is if $W_t$ is not a Weiner process then what is $$ \mathbb{E}[H\cdot W_T^2] $$ equal to?
If we assume that $X_t$ is $L^2$ and that $H_t$ is also $L^2$, in their respective senses then we may proceed as follows...
The result is again called the Ito isometry and given your setting is as follows:
Itô Isometry
$$ \mathbb{E}\left[\left( \int_0^T H_t dX_t\right)^2 \right] = \mathbb{E}\left[ \int_0^T H_t^2 d[X]_t \right], $$ where $[X]_t$ denotes the quadratic variation of $X$. Theorem 5 in this blog shows the details of the result.
In particular if $X_t$ is an Ito process, that is $X_t$ satisfies the SDE $$ dX_t= \mu_tdt +\Sigma_tdW_t, $$ then $[X_t]=\Sigma^{\star}\Sigma_t$. In this case the Ito isomtery simplifies to
$$ \mathbb{E}\left[\left( \int_0^T H_t dX_t\right)^2 \right] = \mathbb{E}\left[ \int_0^T H_t^2 \Sigma^{\star}\Sigma dt \right]. $$ Hope this helped :)