3.471.9
I will start with 3.471.9, since it is much simpler. The key ingredient
Slater theorem, saying that
$$
\int_0^\infty t^{\alpha-1} f(t) g\left(\frac{x}{t}\right) \mathrm{d} t = \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} \mathcal{M}_f(s+\alpha) \mathcal{M}_g(s) x^{-s} \mathrm{d} s = \mathcal{M}^{-1} \left\{ \mathcal{M}_f(s+\alpha) \mathcal{M}_g(s) \right\}(x)
$$
where $\mathcal{M}_f(s) = \int_0^\infty t^{s-1} f(t) \mathrm{d} t$ is the Mellin transform of $f$, valid within some strip $\min < \Re(s) < \max $, and $\gamma$ is a real constant, such that $\gamma$ is within the strip of validity of the Mellin transform of $g$, and $\gamma + \Re(\alpha)$ in within the strip of the Mellin transform of $f$.
The proof is easy:
$$ \begin{eqnarray}
\int_0^\infty t^{\alpha-1} f(t) g\left(\frac{x}{t}\right) \mathrm{d} t &=& \int_0^\infty t^{\alpha-1} f(t) \left( \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma+\infty} \mathcal{M}_g(s) x^{-s} t^s \mathrm{d} s \right) \mathrm{d} t \\ &=& \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma+\infty} \left( \int_0^\infty t^{s+\alpha-1} f(t) \mathrm{d} t \right) \mathcal{M}_g(s) x^{-s} \mathrm{d} s \\
&=& \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma+\infty} \mathcal{M}_f(s+\alpha) \mathcal{M}_g(s) x^{-s} \mathrm{d} s
\end{eqnarray}
$$
We now apply this theorem to 3.471.9, using Cahen-Mellin integral $\mathcal{M}_{\exp(-\bullet)}(s) = \Gamma(s) $:
$$\begin{eqnarray}
\int_0^\infty t^{\alpha-1} \exp\left(-y t\right) \exp\left( - \frac{x}{t}\right) \mathrm{d} t &=& \frac{1}{2 \pi i} \int_{\gamma - i \gamma}^{\gamma + i \infty} \Gamma(s+\alpha) \Gamma(s) y^{-\alpha -s} x^{-s} \mathrm{d}s \\ &=& y^{-\alpha} \left( \frac{1}{2 \pi i} \int_{\gamma - i \gamma}^{\gamma + i \infty} \Gamma(s+\alpha) \Gamma(s) \left( x y \right)^{-s} \mathrm{d}s \right) \\ &=& y^{-\alpha} \left( 2 (x y)^{\alpha/2} K_{\alpha}(2 \sqrt{x y} ) \right) \\
&=& 2 \left( \frac{x}{y} \right)^{\alpha/2} K_\alpha\left(2 \sqrt{x y}\right)
\end{eqnarray}
$$
where DMLF 10.43.19 was used.
6.635.3
A formula of kin is worked out in Bateman, Erdelyi et al, "Higher Transcendental Functions", chapter 7, section 7.7.6 on Macdonald's and Nochilson's formulas, formula (37). It says:
$$
\int_0^\infty \exp\left(-\frac{t}{2} - \frac{x^2+X^2}{2 t} \right) I_\nu \left( \frac{x X}{t} \right) \frac{\mathrm{d} t}{t} = \cases{2 I_\nu\left( x \right) K_\nu\left( X \right) & x < X \\ 2 I_\nu\left( X \right) K_\nu\left( x \right) & x > X }
$$
By doing an analytic continuation $x \to \mathrm{e}^{i \pi/2} x$ of the (the only possible) first branch, we get
$$
\int_0^\infty \exp\left(-\frac{t}{2} - \frac{X^2-x^2}{2 t} \right) J_\nu \left( \frac{x X}{t} \right) \frac{\mathrm{d} t}{t} = 2 J_\nu(x) K_\nu(X)
$$
Now, performing reparameterization $X = \sqrt{2 \alpha \left( \sqrt{\beta^2 + \gamma^2}+ \beta \right)}$, $x = \sqrt{2 \alpha \left( \sqrt{\beta^2 + \gamma^2} - \beta \right)}$, accompanied by a change of variables $t \to \left(2 \alpha t\right)^{-1}$ we arrive at 6.635.3.
Derivation of the point of departure formula relies on two auxiliary results:
$$
\int_0^\infty J_{\nu}(x u) J_{\nu}(X u) \exp\left(-\frac{t u^2}{2}\right) u \mathrm{d} u = \frac{1}{t} \exp\left(-\frac{x^2+X^2}{2t}\right) I_\nu\left( \frac{x X}{t}\right)
$$
which is obtained by expanding the exponential in series and integrating term-wise. Then, using the above integral representation in the formula under consideration, and carrying out simple integration with respect to $t$, we get
$$
\int_0^\infty \frac{\mathrm{e}^{-t/2}}{t} \exp\left(-\frac{x^2+X^2}{2t}\right) I_\nu\left( \frac{x X}{t}\right) \mathrm{d} t =
\int_0^\infty J_\nu(x u) J_\nu(X u) \frac{2 u \mathrm{d} u}{1+u^2}
$$
The resulting integral is of Sonine-Gegenabuer-type, and equals to $2 I_\nu( \min(x,X)) K_\nu(\max(x,X))$.
Assuming $0<c<1$
$$I=\int\limits_{0}^{ \infty } {\frac{{\exp \left( { - ax - \frac{b}{x}} \right)}}{{1 + cx}}dx}=\sum_{n=0}^\infty (-1)^n c^n \int\limits_{0}^{ \infty } x^n\exp \left( { - ax - \frac{b}{x}} \right)\,dx$$
$$I=2\sum_{n=0}^\infty (-1)^n c^n a^{-\frac{n+1}{2} } b^{\frac{n+1}{2}} K_{-(n+1)}\left(2 \sqrt{a} \sqrt{b}\right) \qquad \Re(b)>0\land \Re(a)>0$$ that you could make nicer.
Best Answer
The definition of the modified Bessel function $I_0$ is $$ I_0(z)=\frac{1}{\pi}\int_0^\pi e^{z\cos x}dx=\frac{1}{2\,\pi}\int_{-\pi}^\pi e^{z\cos x}dx $$ Now $$ a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x+\phi) $$ for some angle $\phi$. Since the integral is over a full period, you get the result.