Section 2.5 - Sum Expected Values

What is the expected value of the sum $X + Y$ given $X$ and $Y$ are (not necessarily independent) random variables?

From the definition of expected value, $$E(X + Y) = \sum_x\sum_y (x + y)\,p(X = x \wedge Y = y)$$ Evaluating this sum as given requires work proportional to $n^2$; can it be evaluated more efficiently? Because $X$ and $Y$ aren’t necessarily independent, the probability $p(X = x \wedge Y = y)$ can’t be simplified in the same way it was when computing the expected value of random-value products.^?That is, without independence $p(X = x \wedge Y = y)$ is not necessarily equal to $p_X(X=x)p_Y(Y=y)$. Are there some other tricks that can simplify evaluation?