Transfer Function Derivation
There are many ways a transfer function can be written. The best and only way is the low-entropy form, a term forged by Dr. Middlebrook when working on Design-Oriented Analysis. The transfer function must be expressed in such a way you "see" where the poles and zeroes are located and how they affect the response.

In your expression, factor the numerator \$\frac{\omega_0s}{Q}\$ while in the denominator you factor \$\frac{\omega_0s}{Q}\$ also. Re-arrange the whole thing and you should get \$H(s)=-H_0\frac{1}{1+Q(\frac{\omega_0}{s}+\frac{s}{\omega_0})}\$ which is the correct form for expressing the transfer function of a second-order band-pass filter.

The below Mathcad sheet shows that responses are identical. In your circuit, determine the denominator \$D(s)\$ using the fast analytical circuit techniques (FACTs) from which you obtain \$Q\$ and \$\omega_0\$. Factor \$N(s)\$ to match the formula I gave and your leading term represents the peaking gain \$H_0\$ (20 dB in the given example).