Nice integral involving omega constant, Lambert's function and logarithms 

$$\int_1^e \Big(\operatorname{W}(x)^2 \ln(x)0.25-\operatorname{W}(x) 0.5 + \operatorname{W}(x) \ln(x) \Big) \, dx=\operatorname{W}(1)0.5 + \operatorname{W}(1)^2 0.25$$


I can express the antiderivative wich involves exponential integral and Lambert's function and calculate it using the fundamental theorem of calculus but I would like to know if there are smarter methods (maybe complex analysis)
Thanks a lot for your time and patience.
 A: I unfortunately do not have a smart way to evaluate this interesting integral. My approach is more just "plug and chug". 
Let $u = \operatorname{W}_0 (x)$, then $x = ue^u, dx = e^u(1 + u) \, du$ while the limits of integration become: $u = \operatorname{W}_0 (1) = \Omega$ and $u = \operatorname{W}_0 (e) = 1$. Here $\Omega$ is the Omega constant. Denoting the integral by $I$, we have
\begin{align}
I &= \int_\Omega^1 \left [\frac{u^2}{4} (\ln u + u) - \frac{u}{2} + u(\ln u + u) \right ] e^u (1 + u) \, du\\
&= \int_\Omega^1 \left [e^u \ln u \left (u + \frac{5}{4} u^2 + \frac{1}{4} u^3 \right ) + e^u \left (\frac{1}{4} u^4 + \frac{5}{4} u^3 + \frac{1}{2} u^2 - \frac{1}{2} u \right ) \right ] \, du.
\end{align}
Observing that
$$\int e^x \left (\frac{x^4}{4} + \frac{5}{4} x^3 + \frac{x^2}{2} - \frac{x}{2} \right ) \, dx = \frac{1}{4} e^x x^2 (x^2 + x - 1) + C,$$
and
$$\int e^x \ln x \left (x + \frac{5}{4} x^2 + \frac{x^3}{4} \right ) \, dx = \frac{1}{4} e^x x^2 [(x + 2) \ln x - 1] + C,$$
we find
$$I = \frac{1}{2} \Omega + \frac{1}{4} \Omega^2,$$
as required.
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align} 
I&=
\int_1^\e (\W(x)^2\ln(x)-2\,\W(x)+4\,\W(x)\,\ln(x))\, dx
\tag{1}\label{1}
\\
&=
\int_1^\e ((\W(x)^2+4\,\W(x))\,\ln(x)-2\,\W(x))\, dx
\tag{2}\label{2}
.
\end{align} 
The antiderivative of the integrand in \eqref{1}
has a compact form
\begin{align} 
&x\,\W(x)\,(2+\W(x))(\ln(x)-1)+\mathrm{C}
,\\
\text{so }\quad
I&=(x\,\W(x)\,(2+\W(x))(\ln(x)-1))\mid_{x=1}^{\e}
\\
&=\W(1)(2+\W(1))=\Omega\,(2+\Omega)
\approx 1.455938
\end{align}
$=$A246823
$\endgroup$
