Step response of third-order continuous-time transfer function
It does matter that $a=b=2$, because this gives a relatively nice looking solution. As mentioned in a comment, you should split up the given transfer function $H(s)$. But first, let's introduce a normalized variable $p$:

$$p=\frac{s}{\omega_n}\tag{1}$$

Now we can write the given transfer function as

$$\hat{H}(p)=\frac{2p^2+2p+1}{p^3+2p^2+2p+1}\tag{2}$$

The denominator can be factored as

$$p^3+2p^2+2p+1=(p+1)(p^2+p+1)\tag{3}$$

Partial fraction expansion of $\hat{H}(p)$ gives

$$\begin{align}\hat{H}(p)&amp;=\frac{1}{p+1}+\frac{p}{p^2+p+1}\\&amp;=\frac{1}{p+1}+\frac{p}{(p+\frac12)^2+\frac34}\tag{4}\end{align}$$

The Laplace transform of the step response equals the transfer function divided by $p$:

$$\begin{align}\hat{A}(p)=\frac{\hat{H}}{p}&amp;=\frac{1}{p(p+1)}+\frac{1}{(p+\frac12)^2+\frac34}\\&amp;=\frac{1}{p}-\frac{1}{p+1}+\frac{1}{(p+\frac12)^2+\frac34}\tag{5}\end{align}$$

The inverse Laplace transform of $(5)$ is (table)

$$\begin{align}\hat{a}(t)&amp;=u(t)-e^{-t}u(t)+\frac{2}{\sqrt{3}}e^{-t/2}\sin(\sqrt{3}t/2)u(t)\\&amp;=\left[1-e^{-t}+\frac{2}{\sqrt{3}}e^{-t/2}\sin(\sqrt{3}t/2)\right]u(t)\tag{6}\end{align}$$

Finally, denormalization according to $(1)$ taking into account that $\hat{A}(p)=\omega_nA(s)$ (because of division by $p$) gives the desired step response:

$$\begin{align}a(t)&amp;=\hat{a}(\omega_nt)\\&amp;= \left[1-e^{-\omega_nt}+\frac{2}{\sqrt{3}}e^{-\omega_nt/2}\sin(\sqrt{3}\omega_nt/2)\right]u(t)\tag{7}\end{align}$$