[Not an answer to the exact question above, which was answered in comments, but an additional optimization.]
If division by $2$ is considered "easy," you can do it in two multiplications.
Specifically:
$(a+b)(c+d)+(a-b)(c-d) = 2(ac+bd)$ and $(a+b)(c+d)-(a-b)(c-d)=2(ad+bc)$
So you only need to compute the multiplications $(a+b)(c+d)$ and $(a-b)(c-d)$ (if division by 2 is "easy.")
Otherwise, use the method in the comments above.
[I got this answer by considering your output as the output of a matrix multiplication of the $2\times 2$ matrix $A=[a,b;b,a]$ with the column vector $(c,d)^T$. I then found the eigenvectors of $A$ to be $v_1=(1,1)^T$ and $v_2=(1,-1)^T$ with corresponding eigenvlues $a+b$ and $a-b$. Then I wrote $(c,d)^T=\frac{c+d}{2}v_1 + \frac{c-d}{2}v_2.$ In retrospect, the answer is obvious, but often faster computations are found this way - representation as a linear algebra computation, then using eigenvalues and eigenvectors to "simplify."]