expArg=Expand[((((L-x))^2)/(s1^2))-2 r (((L-x))/s1)(x/s2)+((x^2)/(s2^2))];
Collect[expArg,x]

(* Since we ultimatly want to complete the square, collect the leading coefficient of x and call it a *)
a=((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)));

Simplify[ (((L^2)/(s1^2))/a)]

Simplify[(((-((2 L)/(s1^2))-((2 L r)/(s1 s2))) )/a)]

We now have our exponent factored in the following form :

((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)))( (x^2)-(((2 L s2 (r s1+s2))/((s1^2)+2 r s1 s2+(s2^2))))x)+((L^2)/(s1^2))

Lets complete the square of the term with respect to x :

c=(1/2)(((2 L s2 (r s1+s2))/((s1^2)+2 r s1 s2+(s2^2))))

((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)))( (x^2)-(((2 L s2 (r s1+s2))/((s1^2)+2 r s1 s2+(s2^2))))x+(c^2))+((L^2)/(s1^2))-((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2))) (c^2)

So we need to simplify the following expression

Simplify[((L^2)/(s1^2))-((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2))) (c^2)]

to get

((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)))((( x-(((L s2 (r s1+s2))/((s1^2)+2 r s1 s2+(s2^2))))))^2)-(((L^2) (-1+(r^2)))/((s1^2)+2 r s1 s2+(s2^2)))

Next we multiply by

(1/((1-(r^2))))

to get

(1/(1-(r^2)))((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)))((( x-(((L s2 (r s1+s2))/((s1^2)+2 r s1 s2+(s2^2))))))^2)+(((L^2) )/((s1^2)+2 r s1 s2+(s2^2)))

as part of the argument of the exponential

Simplify[(1/(s1^2))(1/(s2^2))(1/(1-(r^2)))(((1-(r^2)))/(((1/(s1^2))+(1/(s2^2))+((2 r)/(s1 s2)))))]