Let $I_{a,b,c} = \int_0^\infty K_\nu(ax) K_\nu(bx) \sin(cx) dx$. This answer will prove the following identity:
$$\boxed{I_{a,b,c} = \frac{c}{4ab} \pi \nu \csc(\pi \nu) \int_0^1 (1-x)^{-1/2} \,_{2}F_1 \left(1+\nu,1-\nu;\, 3/2;\, u(x)\right) dx}$$
where $u(x) = (-(a-b)^2-c^2x)(4ab)^{-1}$.
In particular, for $a=b$ we can apply Euler's integral transform for generalized hypergeometric functions to obtain the following closed form:
$$I_{a,a,c} = \frac{c}{2a^2} \pi \nu \csc(\pi\nu) \,_{3}F_2 (1+\nu,1-\nu,1;\,3/2,3/2; -c^2(4a^2)^{-1})$$
We also have $_{2}F_1(1+\nu,1-\nu;\, 3/2;\,-z) = \sinh(2\nu \operatorname{arcsinh}(\sqrt{z}))(z(1+z))^{-1/2}(2 \nu)^{-1}$ which allows to rewrite the integral using only elementary functions. I will also explain why the integral with $\sin(cx)$ is much harder than the one with $\cos(cx)$.
For $c \ll |a-b|$ we can also obtain:
$$I_{a,b,c} \approx \frac{c}{2ab} \pi \nu \csc(\pi\nu)\,_{2}F_1(1+\nu,1-\nu;\, 3/2;\, -(a-b)^2(4ab)^{-1})$$
Proof. Consider $\tilde{I}_{abc} = \int_0^\infty K_\nu(ax) K_\nu(bx) e^{icx} dx$. Using [1] and [2] we obtain
$$\tilde{I}_{abc} = \frac12 \int_0^\infty e^{icx} \int_0^\infty \exp\left(-\frac{t}{2} -\frac{a^2+b^2}{2t} x^2 \right) \frac{1}{t} \int_0^\infty \exp \left( -\frac{abx^2}{t} \cosh r\right) \cosh(\nu r) dr dt dx = $$
$$=\frac12 \int_0^\infty \cosh (\nu r) \int_0^\infty \frac{1}{t} \exp \left( -\frac{t}{2} \right) \int_0^\infty \exp \left( -x^2 \left( \frac{a^2+b^2+2ab\cosh r}{2t}\right)\right) e^{icx} dx dt dr$$
Using [3]:
$$\tilde{I}_{abc} = \frac{\sqrt{\pi}}{2\sqrt{2}} \int_0^\infty \frac{\cosh (\nu r)}{\sqrt{a^2+b^2+2ab \cosh r}} \int_0^\infty \frac{1}{\sqrt{t}} \exp \left( -\frac{t}{2} \left( 1 + \frac{c^2}{a^2+b^2+2ab \cosh r}\right)\right) \left(1+i\operatorname{erfi} \left(\frac{c\sqrt{t}}{\sqrt{2}\sqrt{a^2+b^2+2ab \cosh r}}\right)\right) dt dr$$
Using [4] and [5]:
$$\tilde{I}_{abc} = \frac{\pi}{2}\int_0^\infty \frac{\cosh (\nu r)}{\sqrt{a^2+b^2+c^2+2ab \cosh r}} \left(1+\frac{2i}{\pi}\operatorname{arctanh} \left(\frac{c}{\sqrt{a^2+b^2+c^2+2ab \cosh r}}\right)\right) dr$$
Here we can use [6] to have
$$\operatorname{Re} \tilde{I}_{a,b,c} = \frac{\pi}{2} \int_0^\infty \frac{\cosh (\nu r) dr}{\sqrt{a^2+b^2+c^2+2ab \cosh r}} = $$
$$=\frac{\pi}{2 \sqrt{2ab}} \frac{\pi P_{\nu-1/2}((a^2+b^2+c^2)(2ab)^{-1})}{\sqrt{2}} \sec(\pi \nu) = $$
$$=\frac{\pi^2}{4\sqrt{ab}} \sec(\pi \nu) P_{\nu-1/2}((a^2+b^2+c^2)(2ab)^{-1})$$
which gives an expression for the integral with $\cos(cx)$. But for $\sin(cx)$ we have a harder integral:
$$I_{a,b,c} = \operatorname{Im} \tilde{I}_{a,b,c} = \int_0^\infty \frac{\cosh (\nu r) dr}{\sqrt{a^2+b^2+c^2+2ab \cosh r}} \operatorname{arctanh} \left(\frac{c}{\sqrt{a^2+b^2+c^2+2ab \cosh r}}\right)$$
Here we are going to use Feynman’s technique to simplify this:
$$I_{a,b,c}(z) = \int_0^\infty \frac{\cosh (\nu r) dr}{\sqrt{a^2+b^2+c^2+2ab \cosh r}} \operatorname{arctanh} \left(\frac{zc}{\sqrt{a^2+b^2+c^2+2ab \cosh r}}\right)$$
$$\frac{dI_{a,b,c}(z)}{dz} = \int_0^\infty \frac{\cosh (\nu r) dr}{a^2+b^2+c^2+2ab \cosh r} \frac{c}{1 - \frac{z^2c^2}{a^2+b^2+c^2+2ab\cosh r}}$$
$$=\frac{c}{2ab}\int_0^\infty \frac{\cosh (\nu r)dr}{(a^2+b^2+c^2(1-z^2))(2ab)^{-1}+ \cosh r}$$
Using [6] we have
$$\frac{dI_{a,b,c}(z)}{dz} = \frac{c}{2ab} \sqrt{\frac{\pi}{2}} \frac{\Gamma(1+\nu)\Gamma(1-\nu)}{(\omega(z)^2-1)^{1/4}} P_{\nu-1/2}^{-1/2}(\omega(z)),$$
where $\omega(z) = (a^2+b^2+c^2(1-z^2))(2ab)^{-1}$.
Since $I_{a,b,c}(0) = 0$ we can also write
$$I_{a,b,c} = \frac{c}{2ab} \sqrt{\frac{\pi}{2}} \Gamma(1+\nu)\Gamma(1-\nu) \int_0^1 \frac{P_{\nu-1/2}^{-1/2}(\omega(z))}{(\omega(z)^2-1)^{1/4}} dz$$
Since $\Gamma(1+\nu)\Gamma(1-\nu) = \nu\Gamma(\nu)\Gamma(1-\nu) = \pi \nu \csc (\pi \nu)$:
$$I_{a,b,c} = \frac{c}{2ab} \sqrt{\frac{\pi}{2}} \pi \nu \csc (\pi \nu) \int_0^1 \frac{P_{\nu-1/2}^{-1/2}(\omega(z))}{(\omega(z)^2-1)^{1/4}} dz$$
Using [7] we can rewrite:
$$I_{a,b,c} = \frac{c}{2ab} \pi \nu \csc (\pi \nu) \int_0^1 \left(\frac{1+\omega(z)}{2}\right)^{-1/2} \,_{2}F_1 \left(-\nu+1/2,\nu + 1/2;\, 3/2;\, {\frac {1-\omega(z)}{2}}\right) dz = $$
$$= \frac{c}{2ab} \pi \nu \csc (\pi \nu) \int_0^1 \,_{2}F_1 \left(1+\nu,1-\nu;\, 3/2;\,{\frac {1-\omega(z)}{2}}\right) dz$$
where in the last line we have used Euler's transformation.
Using substitution $x = 1-z^2$ we obtain
$$I_{a,b,c} = \frac{c}{4ab} \pi \nu \csc(\pi \nu) \int_0^1 (1-x)^{-1/2} \,_{2}F_1 \left(1+\nu,1-\nu;\, 3/2;\, u(x)\right) dx$$
where $u(x) = (-(a-b)^2-c^2x)(4ab)^{-1}$.
[1]
$K_\nu (ax) K_\nu(bx) = \frac12 \int_0^\infty \exp\left(-\frac{t}{2} -\frac{a^2+b^2}{2t} x^2 \right) K_\nu \left( \frac{abx^2}{t}\right) \frac{dt}{t}$
[2] $K_\nu(z) = \int_0^\infty e^{-z \cosh t} \cosh (\nu t) dt$
[3] $\int_0^\infty \exp(-px^2) e^{icx} dx = \frac{\sqrt{\pi}}{2\sqrt{p}} \exp\left(-\frac{c^2}{4p}\right) \left(1 + i\operatorname{erfi} \left( \frac{c}{2\sqrt{p}}\right)\right)$
[4] $\int_0^\infty \frac{1}{\sqrt{t}} \exp(-pt) dt = \frac{\sqrt{\pi}}{\sqrt{p}}$
[5] $\int_0^\infty \frac{1}{\sqrt{t}} \exp(-pt) \operatorname{erfi} (k\sqrt{t}) dt = \frac{2}{\sqrt{\pi p}} \operatorname{arctanh} \left( \frac{k}{\sqrt{p}}\right)$
[6] $\int_0^\infty \frac{\cosh (\nu r) dr}{(z + \cosh r)^{\mu+1/2}} = \sqrt{\frac{\pi}{2}} \frac{\Gamma(\nu+\mu+1/2)\Gamma(\mu-\nu+1/2)}{\Gamma(\mu+1/2)(z^2-1)^{\mu/2}} P^{-\mu}_{\nu-1/2}(z)$
[7] $P_{\nu}^{\mu}(z) = \frac{1}{\Gamma(1-\mu)} \left(\frac{z+1}{z-1}\right)^{\mu/2}\,_{2}F_1 \left(-\nu ,\nu +1;1-\mu ;{\frac {1-z}{2}}\right)$
