Is there a way to calculate easily $\mathrm{Cov}(X,XY)$? Supposing to have two random dependent variables $X$ and $Y$ how can one calculate the covariance between the two variables $X$ and $X \cdot Y$? That is
$$\mathrm{Cov}(X,XY)$$
Trying using the definition does not lead to a simple result, so are there some properties about this particular covariance?
(For istance if $Y$was a constant then it would be 
$$\mathrm{Cov}(X,XY)=Y \mathrm{Cov}(X,X)=Y\mathrm{Var}(X)$$
But this is not the case)

Nevertheless the  following reasoning makes me think that it should be $\mathrm{Cov}(X,XY)=Y\mathrm{Var}(X)$ also in this case.
Infact take two indpendent variables $a$ and $b$ and say I want to calculate the error $\sigma_C$ on the quantity $$C=ab-a=a(b-1)$$
Since I can see $C$ in two ways (as highlighted above) I can calculate $\sigma_C$ in two ways: 
$C=ab-a \to \mathrm{ab \,\, and \, \, a \,\, are \,\,\, dependent} \to \sigma_C^2=\sigma_a^2 b^2 +\sigma_b^2 a^2 +\sigma_a^2-2 \mathrm{Cov}(a,ab)$
$C=a(b-1) \to \mathrm{a \,\, and \, \, (b-1) \,\, are \,\, not \,\, dependent} \to   \sigma_C^2=\sigma_b^2 a^2 +\sigma_a^2 (b-1)^2 $
These two must be equal, so
$$\sigma_b^2 a +\sigma_a^2 (b-1)^2=\sigma_a^2 b^2 +\sigma_b^2 a^2 +\sigma_a^2-2 \mathrm{Cov}(a,ab) \implies \mathrm{Cov}(a,ab)=b^2 \sigma_a^2$$
 A: This is a nice problem for testing the development code in the next version of mathStatica. 
Note that $\text{Cov}(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment), which is why I am requesting the {1,1} CentralMoment of {X, X Y} ... when the variables are {X,Y}:

where $\mu_{2,1}  = E\left[(X-E[X])^2 \;(Y-E[Y])^1\right]$
If $X$ and $Y$ are independent (information not stated in the problem), then the answer simplifies further:

A: My first reaction is that you won't be able to find this value without knowing the dependence structure between $X$ and $Y$. This is further verified by using law of total covariance as shown below,
\begin{align*}
Cov(X,XY)&  = E[Cov(X,XY\mid X)] + Cov(E[X\mid X],E[XY \mid X])\\
& = E[X^2Cov(1,Y \mid X)] + Cov(X, X\, E[Y \mid X])\\
& = 0 + Cov(X, X\, E[Y \mid X])\\
& = Cov(X, X\, E[Y \mid X]).
\end{align*}
$Y$ goes away due to the expectation, so your claim cannot hold. If you know the expectation of $Y \mid X$ then you can find this. If not, then you won't be able to solve this.
