Controls: Getting the feedback transfer function of a Plant and Controller
It depends how the calculation is done. Take as an example \$\small G= \large \frac{1}{s+a}\$, and \$\small K= 1\$. We may write the CLTF as:

$$\small\frac{G}{1+GK}=G.\frac{1}{1+GK}=\frac{1}{s+a}.\frac{1}{\left(1+\frac{1}{s+a}\right)}=\frac{1}{s+a}.\frac{s+a}{(s+a+1)}$$

This expands to:

\$\large\frac{G}{1+GK}=\frac{s+a}{s^2+(2a+1)s+(a^2+a)}\$

which is correct but inconvenient, and which reduces to:

\$ \large \frac{G}{1+GK}=\frac{1}{s+(a+1)}\$ when the \$(s+a)\$ factor is cancelled.

To force the reduce form, you could try writing \$\small G\$ and \$\small K\$ in numerator/denominator form: \$\small G=\large \frac{N_g}{D_g};\:\:\:\:\small K=\large\frac{N_k}{D_k}\$, then write the CLTF as:

$$\small\frac{G}{1+GK}=\frac{N_g\:D_k}{D_g\:D_k+N_g\:N_k}$$

Note that Mathworks recommends using the FEEDBACK command to derive closed-loop TFs, as this gives the reduced form.