Integral $\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx$ $$I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx=\frac{(1+\alpha)\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.$$
This one looks very nice.  It has stumped me.
Differentiation with respect to parameter does not seem to work either if I try $I(\alpha)$ and $I'(\alpha)$. at x=0 there seems to be a problem with the integrand also however I am not sure how to go about using this. Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty  x^n /n!$, however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being. 

Assuming the $3\alpha x$ term is in fact $3\alpha x^2$ (otherwise the numerical results do not match).
$$\begin{align*}
I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx\\
&=\int_0^\infty \ln x\, d\left(-\alpha x^{-3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\
&=-\alpha\left(\left.\frac{\ln x}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\
&=\alpha\int_0^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\
&=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\
&=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -x^2\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\
&=\alpha\int_0^1 (x^2+x^{-4})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\
&=\alpha\int_0^1 (x^2-1+x^{-2})\exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_0^1 (1+(x-1/x)^2)\exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_{-\infty}^0 (1+y^2)\exp\left(-\frac{y^2}{2\alpha }\right)dy\\
&=\alpha e^{-1/\alpha}(\alpha+1)\sqrt{\frac{\alpha\pi}{2}}.
\end{align*}$$
