This is an integral in <A HREF="https://books.google.nl/books?id=X7HtCAAAQBAJ">Oberhettinger's Table of Fourier Transforms</A> (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does indicate why a fully closed-form expression will not be forthcoming (you would need the _indefinite_ integral of a Bessel function of argument $\sqrt{1-x^2}$).

<IMG SRC="https://ilorentz.org/beenakker/MO/cosineintegral_Oberhettinger.png"/>

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$
$$\qquad\qquad-\frac{\pi}{2}  \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$
$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

Notes:    
<sub>• As a further check, note that the derivative with respect to $b$ gives formula 3.876.2 of Gradshteyn.</sub>    
<sub>• The corresponding formulas in Erdelyi contain several typo's. These have been corrected in Gradshteyn, the formulas in question have a label $^6$.</sub>    
<sub>• The integral also implies the identity
$$\int_0^1 K_0\left(\sqrt{1-t^2}\right)\,dt=\tfrac{1}{2e} \text{Ei}(1)-\tfrac{e}{2} \text{Ei}(-1).$$</sub>