Let $$I_{n}=\int_{0}^{\infty}x^{n}e^{-(ax+b/x)}$$
Then, differentiating under the integral sign, $$\frac{dI_{0}}{da}=\int_{0}^{\infty}\frac{\partial}{\partial a}\left(e^{-(ax+b/x)}\right)dx=\int_{0}^{\infty}-xe^{-(ax+b/x)}dx=-I_{1}$$
So, similarly, $$I_{n}=(-1)^{n}\frac{d^{n}I_{0}}{da^{n}}$$
Following on from Claude Lerbovici's answer, we can prove that closed form by induction. The base case $n=0$ is covered here. Now, assume that for $n=\nu$, $$I_{\nu}=2 \left(\frac{b}{a}\right)^{\frac{\nu+1}{2}} K_{\nu+1}\left(2 \sqrt{ab}\right)$$
Then, by the product rule,
$$\frac{dI_{\nu}}{da}=-\left(\frac{\nu+1}{2}\right)2\frac{b^{(\nu+1)/2}}{a\cdot a^{(\nu+1)/2}}K_{\nu+1}\left(2 \sqrt{ab}\right)+2 \left(\frac{b}{a}\right)^{\frac{\nu+1}{2}} \frac{1}{2}\sqrt\frac{b}{a}K_{\nu+1}'\left(2 \sqrt{ab}\right)$$
Now, there exists an identity which states that $$K'_{\nu+1}(z)=\frac{\nu+1}{z}K_{\nu+1}(z)-K_{\nu+2}(z)$$
Which simplifies the above to $$-2\left(\frac{b}{a}\right)^{(\nu+2)/2}\left(\frac{\nu+1}{2\sqrt{ab}}K_{\nu+1}(\sqrt{2ab})-\frac{\nu+1}{2\sqrt{ab}}K_{\nu+1}(\sqrt{2ab})+K_{\nu+2}(\sqrt{2ab})\right)$$
Thus, $$I_{\nu+1}=-\frac{dI_{\nu}}{da}=-2\left(\frac{b}{a}\right)^{(\nu+2)/2}K_{\nu+2}(\sqrt{2ab})$$
as required.