I found the following integrals in "Tables of Integrals, Series and Products Vol. 8" p. 494-495
$$\int_0^{\infty}e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{\beta \, \gamma}{A} \, \mathrm{K}_1(\gamma A)$$
$$\int_0^{\infty} \sqrt{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{\beta^2 \, \gamma^2}{A^2} \, \mathrm{K}_0(\gamma A)+\left(\frac{2\beta^2 \, \gamma}{A^3}-\frac{\gamma}{A}\right) \mathrm{K}_1(\gamma A)$$
$$\int_0^{\infty} (\gamma^2+x^2) e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \left(\frac{-3\beta \, \gamma^2}{A^2}+\frac{4\beta^3 \gamma^2}{A^4}\right) \, \mathrm{K}_0(\gamma A)+ \\ +\left(\frac{-6 \beta \, \gamma}{A^3}+\frac{8\beta^3\gamma}{A^5}+\frac{\beta^3\gamma^3}{A^3}\right)\mathrm{K}_1(\gamma A)$$
$$\int_0^{\infty}\frac{1}{\sqrt{\gamma^2+x^2}}e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \mathrm{K}_0(\gamma A)$$
with $A=\sqrt{\beta^2+b^2}$, Re$[\beta]>0$, Re$[\gamma]>0$, $b>0$.
It seems to me that there is some kind of system behind it. Can anybody give me a hint to proof this formulars? There are similar ones for $\mathrm{sin}(b \, x)$.
I would like to solve the integral
$$\int_0^{\infty} \frac{1}{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x)\, \mathrm{d}x$$
There is another one in the tables
$$\int_0^{\infty} \left(\frac{1}{\beta(\gamma^2+x^2)^{3/2}}+\frac{1}{\gamma^2+x^2}\right) e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{1}{\beta \gamma} A \, \mathrm{K}_1(\gamma A)$$
Let $F(b,\beta,\gamma)$ be given by the integral
$$\begin{align} F(b,\beta,\gamma)&=\int_0^\infty \frac{e^{-\beta \sqrt{x^2+\gamma^2}}}{\sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx\tag 1 \end{align}$$
In a similar approach to the one that I took in developing THIS ANSWER, we enforce the substitution $x=\gamma \sinh(t)$ in $(1)$ to obtain
$$\begin{align} F(b,\beta,\gamma)&=\int_0^\infty e^{-\beta \gamma \cosh(t)}\,\cos(b\gamma \sinh(t))\,dt \\\\ &=\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dt+\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)-ib\gamma \sinh(t)}\,dt\\\\ &=\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dx+\frac12\int_{-\infty}^0 e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dx\\\\ &= \frac12 \int_{-\infty}^\infty e^{-\gamma \sqrt{\beta^2+b^2} \cosh(t)}\,dt\\\\ &=\int_0^\infty e^{-\gamma \sqrt{\beta^2+b^2} \cosh(t)}\,dt \tag 2\\\\ &=K_0(\gamma \sqrt{\beta^2+b^2}) \tag 3 \end{align}$$
where in going from $(2)$ to $(3)$ we used the well-known integral representation for the modified Bessel function of the second kind as found, for example, HERE. Therefore, we find the alternative integral representation for $K_0(\gamma \sqrt{\beta^2+b^2})$ as
We can use the integral representation given in $(4)$ to derive several relationships by differentiating under the integral sign. Proceeding, we have
$$\begin{align} -\frac{\partial}{\partial \beta}\left(K_0(\gamma \sqrt{\beta^2+b^2})\right)&=\int_0^\infty e^{-\beta \sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx\\\\ \end{align}$$
Then, using $K_0'(x)=-K_1(x)$, we can write
$$\bbox[5px,border:2px solid #C0A000]{\frac{\gamma \beta}{\sqrt{\beta^2+b^2}}K_1(\gamma \sqrt{\beta^2+b^2})=\int_0^\infty e^{-\beta \sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx}\tag 5$$
Note that $(5)$ is the first relationship presented in the OP. And continuing to differentiate under the integral and exploiting the identities for $K_0'(x)=-K_1(x)$ and $K_1'(x)=-K_0(x)-\frac1x K_1(x)$ we can arrive at the second and third identities rather straightforwardly.
Finally, the integral of interest $I(b,\beta,\gamma)=\int_0^\infty \frac{e^{-\beta \sqrt{x^2+\gamma^2}}}{x^2+\gamma^2}\,\cos(bx)\,dx$ satisfies the relationship
$$-\frac{\partial I(b,\beta,\gamma)}{\partial \beta}=K_0(\gamma \sqrt{\beta^2+b^2})$$
Further development is left to the reader