Integrals invariant to the choice of the real branch of the Lambert W function used in the integrand $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Consider for example, this integral:
\begin{align}
\int_0^1 \W(-\tfrac t\e) \, dt
\tag{1}\label{1}
.
\end{align}
Its value depends on the choice of the real branch used:
\begin{align}
\int_0^1 \Wp(-\tfrac t\e) \, dt
&=\e-3
\tag{2}\label{2}
,\\
\int_0^1 \Wm(-\tfrac t\e) \, dt
&=-3
\tag{3}\label{3}
.
\end{align}
On the other hand,
the value of the integral 
\begin{align}
\int_0^1 \frac{-\W(-\tfrac t\e)\,(2+\W(-\tfrac t\e))}{1+\W(-\tfrac t\e)}\, dt
\tag{4}\label{4}
\end{align}
is invariant to the choice of the real branch used,
and despite that the integrands represent
completely different curves,
in both cases the integral value is the same:
\begin{align}
\int_0^1 \frac{-\Wp(-\tfrac t\e)\,(2+\Wp(-\tfrac t\e))}{1+\Wp(-\tfrac t\e)}\, dt
&=1
\tag{5}\label{5}
,\\
\int_0^1 \frac{-\Wm(-\tfrac t\e)\,(2+\Wm(-\tfrac t\e))}{1+\Wm(-\tfrac t\e)}\, dt
&=1
\tag{6}\label{6}
.
\end{align}
Question: are there more known interesting examples with this kind of invariance?
$\endgroup$
 A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Two more examples:

\begin{align} 
\int_0^1 \sin(-\W(-\tfrac t\e))\, dt
&=
\int_0^1 \sin(-\Wp(-\tfrac t\e))\, dt
\\
&=
\int_0^1 \sin(-\Wm(-\tfrac t\e))\, dt
\\
&=\tfrac12\,\cos(1)
\approx .270151152934
.
\end{align} 
\begin{align} 
\int_0^1 \frac{1+\ln(-\W(-\tfrac t\e))(1+\W(-\tfrac t\e))}
{1+\W(-\tfrac t\e)}\, dt 
&=
\int_0^1 \frac{1+\ln(-\Wp(-\tfrac t\e))(1+\Wp(-\tfrac t\e))}
{1+\Wp(-\tfrac t\e)}\, dt 
\\
&=
\int_0^1 \frac{1+\ln(-\Wm(-\tfrac t\e))(1+\Wm(-\tfrac t\e))}
{1+\Wm(-\tfrac t\e)}\, dt 
\\
&=0
.
\end{align} 

Update:
This one showed up in 
a-little-game-around-lamberts-function-and-simple-and-beautiful-integral
\begin{align} 
\int_0^1 \left(2\,\sqrt{-\W(-\tfrac t\e)}+\frac 1{\sqrt{-\W(-\tfrac t\e)}} \right)\, dt
&=\int_0^1 \left(2\,\sqrt{-\Wp(-\tfrac t\e)}+\frac 1{\sqrt{-\Wp(-\tfrac t\e)}} \right)\, dt
\\
&=\int_0^1 \left(2\,\sqrt{-\Wm(-\tfrac t\e)}+\frac 1{\sqrt{-\Wm(-\tfrac t\e)}} \right)\, dt
\\
&=4
.
\end{align}
$\endgroup$
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
One more non-trivial real-branch-invariant function:
\begin{align} 
f_{\W}(x)&=25\,\cos(2\,\W(-\tfrac x\e))-16\,\cos(\W(-\tfrac x\e))
.
\end{align}
Again, the graphs of $f_{\Wp}$ and $f_{\Wm}$ 
are essentially different,

but the integral is invariant to the choice of the branch used:
\begin{align} 
\int_0^1 25\,\cos(2\,\W(-\tfrac x\e))&-16\,\cos(\W(-\tfrac x\e))
\, dx
\\
&=
\int_0^1 25\,\cos(2\,\Wp(-\tfrac x\e))-16\,\cos(\Wp(-\tfrac x\e))
\, dx
\\
&=
\int_0^1 25\,\cos(2\,\Wm(-\tfrac x\e))-16\,\cos(\Wm(-\tfrac x\e))
\, dx
\\
&=
11+\sin(1)\,(4+3\,\sin(1)-4\,\cos(1))
\\
&\approx 14.6715
.
\end{align} 
$\endgroup$
