Given your Eq-3
$$ (x * h)[n]=\sum_{k=\max (a, n-d)}^{\min (b, n-c)} x[k] h[n-k] . \tag{1} $$
where $x[n]$ and $h[n]$ are as in Eq-1, then you will have
$$ \sum_{k=\max (a, n-d)}^{\min (b, n-c)} \alpha \beta = \alpha \beta (\min (b, n-c)- \max (a, n-d) + 1). \tag{2} $$
Where you should evaluate the closed form expression Eq.2 for the non-zero range of output given by the range of $n$ : $a+c \leq n \leq b+d $