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

$$
\sum_{k=\max (a, n-d)}^{\min (b, n-c)} \alpha \beta = \alpha \beta (\min (b, n-c)- \max (a, n-d)). \tag{2}
$$

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 < n < b+d $