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This is sort of follow up to my previous question (less difficult integral) here. How do I find $$\large\int_{0}^{\infty}x^ne^{-\left(ax+\frac{b}{x}\right)}dx$$ where $a$ and $b$ are positive reals, $n$ is a positive integer. A variation of this question without the term $x^n$ is answered here.
EDIT: This is not a homework question. This comes as part of my work in signal processing.
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.
Starting from Daniel Littlewood's answer and using a CAS (just as for my answer to the specific case where $n=0$ your post refers to), I arrived to $$I_{n}=\int_{0}^{\infty}x^{n}e^{-(ax+b/x)}=2 \left(\frac{b}{a}\right)^{\frac{n+1}{2}} K_{n+1}\left(2 \sqrt{a} \sqrt{b}\right)$$ provided that $\Re(b)>0$ and $\Re(a)>0$.
