Note that for $r>0$ one has integral representation
$$J_0(r)=\frac{1}{2\pi}\int_0^{2\pi}e^{ir\cos\phi}d\phi$$
Hence
$$I=\int_0^{\infty}J_0\left(\alpha\sqrt{x^2+z^2}\right)\cos \beta x\,dx=
\frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha\sqrt{x^2+z^2}\cos\phi}\cos\beta x \, d\phi \, dx.\tag{1}$$
On the other hand,
$$\sqrt{x^2+z^2}\cos\phi=z\cos(\phi-\phi_0)+x\sin(\phi-\phi_0),$$
where $\tan\phi_0=-\frac{x}{z}$. Exchanging the order of integration in (1) and shifting $\phi$ by $\phi_0$, we arrive at
$$I=\frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha(z\cos\phi+x\sin\phi)}\cos\beta x \, d\phi \, dx.$$
Finally, using that $\displaystyle\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}dx=\delta(\omega)$ we obtain
$$I=\frac{1}{4}\int_0^{2\pi}e^{i\alpha z\cos\phi}\Bigl[\delta\left(\alpha\sin\phi+\beta\right)+\delta\left(\alpha\sin\phi-\beta\right)\Bigr]d\phi$$
It remains to use $\delta(f(x))=\sum\limits_{\text{zeros of }f}\frac{1}{|f'(x_k)|}\delta(x-x_k)$ and compute the two contributions coming from each of the two delta-functions.
We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$
(See this question for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases}
\frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\
\frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0
\end{cases} $$
Best Answer
By exploiting the integral representation for the $J_0$ function: $$ J_0(z) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\cos t}\,dt \tag{1}$$ we have: $$ \int_{0}^{+\infty}\sin(ax)\, J_0(b\sqrt{1+x^2})\,dx =\frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)e^{ib\sqrt{x^2+1}\cos t}\,dt\,dx \tag{2}$$ where: $$ \sqrt{x^2+1}\cos t = \cos(t-t_0)-x\sin(t-t_0),\qquad x=\tan t_0 \tag{3}$$ so: $$\begin{eqnarray*} I &=& \frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)\,e^{ib\cos t-ibx\sin t}\,dt\,dx\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}e^{ib\cos t}\,dt\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}\,\cos(b\cos t)\,dt\tag{4} \end{eqnarray*}$$ but since: $$ \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt = \frac{ (2n-1)!!}{b^n}\, J_n(b) \tag{5}$$ we have: $$\begin{eqnarray*} I &=& \frac{1}{a}\sum_{n\geq 0}\left(\frac{b}{a}\right)^{2n} \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt \\&=&\frac{1}{a}\sum_{n\geq 0}\frac{(2n)!}{n!}\left(\frac{b}{2a^2}\right)^n J_n(b)\\&=&\frac{1}{a}\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!}\left(\frac{b}{2}\right)^m\sum_{n\geq 0}\left(\frac{b}{2a}\right)^{2n}\frac{(2n)!}{n!(n+m)!}\\&=&\frac{1}{\sqrt{a^2-b^2}}+\frac{1}{a}\int_{0}^{1}\sum_{m\geq 1}\frac{(-1)^m}{m!(m-1)!}\left(\frac{b}{2}\right)^m(1-u)^{m-1}\sum_{n\geq 0}\binom{2n}{n}\left(\frac{b^2}{4a^2}\right)^{n}u^n\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}+\int_{0}^{1}\sum_{m\geq 1}\left(\frac{b}{2}\right)^m \frac{(-1)^m}{m!(m-1)!}\frac{(1-u)^{m-1}}{\sqrt{a^2-b^2 u}}\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}-\sqrt{\frac{b}{2}}\int_{0}^{1}\frac{J_1(\sqrt{2b(1-u)})}{\sqrt{(a^2-b^2 u)(1-u)}}\,du\\&=&\frac{1}{\sqrt{a^2-b^2}}-\sqrt{2b}\int_{0}^{1}\frac{J_1(t\sqrt{2b})}{\sqrt{(a^2-b^2)+b^2 t^2}}\,dt.\tag{6}\end{eqnarray*}$$