You made a mistake there somewhere: the value of $X(0)$ makes no sense. Here's what I get for the LT:
$$X(s) = \frac{\dot{x}_0 + (s+\delta) x_0}{s^2+\delta s+\omega_0^2} + \frac{\gamma s}{(s^2+\delta s+\omega_0^2)(s^2+\omega^2)} $$
where $x_0 = x(0)$ and $\dot{x}_0 = \dot{x}(0)$. As you point out, the ILT is given by
$$\frac{1}{i 2 \pi} \lim_{R \to \infty} \int_{c-i R}^{c+i R} ds \, e^{s t} \, X(s)$$
I think you are under the impression that you actually have to evaluate this integral directly. I'm sure it could be done, but that's not the best way to do it. The way to evaluate this integral is to use the Residue theorem. If you are not familiar with it, then I cannot give a entire course on it here, but recommend that you read up in some introductory text in Complex Analysis such as, e.g., Churchill & Brown, so you can follow what I do below.
For $t>0$, close a contour $C$ to the left of $\Re{s}=c$ in the complex-$s$ plane with a semicircle of radius $R$; the integral about the circular arc approaches zero as $R \to \infty$, so the integral about the closed loop $C$ is equal to $i 2 \pi$ times the ILT. The integral about this closed loop $C$, on the other hand, is $i 2 \pi$ times the sum of the residues of the poles inside $C$; therefore, the ILT is the sum of these residues.
The poles of course are given by the zeroes of the denominator:
$$s^2+\omega^2=0 \implies s=\pm i \omega$$
$$s^2+ \delta s+\omega_0^2 = 0 \implies s_{\pm} = \frac12 \left ( -\delta \pm \sqrt{\delta^2-4 \omega_0^2}\right)$$
Now, for simple poles like these, the residue of a pole $s_0$ of a function of the form $p(s)/q(s)$ is $p(s_0)/q'(s_0)$. So finding the ILT is just a matter of handling the algebra behind evaluating the residues of the LT times $e^{s t}$ at the four poles. The sum of the residues due to the first two poles at $s=\pm i \omega$ above is straightforward:
$$x_1(t) = \frac{\gamma}{2} \frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+4 \delta^2 \omega^2} \cos{\omega t} + \frac{2 \gamma \delta \omega}{(\omega_0^2-\omega^2)^2+4 \delta^2 \omega^2} \sin{\omega t} $$
This first part corresponds to the oscillations induced by the forcing term. The other contribution to the full solution determined by the other poles are more tricky, because you have to examine 3 cases: 1) $\delta > 2 \omega_0$, 2) $\delta < 2 \omega_0$, and 3) $\delta = 2 \omega_0$ exactly. Let's examine the first case. The algebra involved in evaluating these residues is obviously tedious, but there are clear ways to simplify a little. For example, $2 s_{\pm}+\delta = \pm \sqrt{\delta^2-4 \omega^2}$. Otherwise, there is no good way to illustrate the algebra other than to just show the result:
$$x_2(t)=\frac{e^{-\delta t/2}}{\sqrt{\delta^2-4 \omega_0^2}} \left \{ \left [ -\dot{x}_0 + \frac12 x_0 (\delta+\sqrt{\delta^2-4 \omega_0^2})\right] e^{\sqrt{\delta^2-4 \omega_0^2} t/2} + \left [ \dot{x}_0 +\frac12 (\delta-\sqrt{\delta^2-4 \omega_0^2}) \right] e^{-\sqrt{\delta^2-4 \omega_0^2} t/2} \right \}$$
$$x_3(t) =\frac{\gamma}{2} \frac{e^{-\delta t/2}}{\sqrt{\delta^2-4 \omega_0^2}} \left \{ \frac{-\delta + \sqrt{\delta^2-4 \omega_0^2}}{\omega^2-\omega_0^2+\frac12 \delta^2 -\frac12 \delta \sqrt{\delta^2-4 \omega_0^2}} e^{\sqrt{\delta^2-4 \omega_0^2} t/2} + \frac{\delta + \sqrt{\delta^2-4 \omega_0^2}}{\omega^2-\omega_0^2+\frac12 \delta^2 +\frac12 \delta \sqrt{\delta^2-4 \omega_0^2}} e^{-\sqrt{\delta^2-4 \omega_0^2} t/2}\right \}$$
Then the solution to the original equation is
$$x(t) = x_1(t) + x_2(t) + x_3(t)$$
This corresponds to the case of pure damping even though the $x_1$ solution is oscillatory. Note that this solution is valid for the case $\delta<2 \omega_0$, although you will have some algebra involving complex numbers to do. There, the solution has both a damping and oscillatory component.
I leave the case $\delta = 2 \omega_0$ to the reader.
For $t<0$, close the contour to the right of $\Re{s}=c$; by design, there are no poles here, so the integral is zero as you'd expect.
