= Solution
For a uniform sphere of radius $R$, write $M(r)=M(r/R)^3$ and $dM=3Mr^2dr/R^3$. The self-gravitational <potential energy> is assembled by adding shells:
$$
U_G=-\int_0^R\frac{GM(r)}r\,dM(r)
=-\frac{3GM^2}{R^6}\int_0^Rr^4\,dr
=-\frac{3GM^2}{5R}.
$$
The <cosmological constant> produces the external potential per unit mass $\Phi_\Lambda=-\Lambda r^2/6$, since $-\nabla\Phi_\Lambda=\Lambda\mathbf r/3$ in units $c=1$. Its energy is
$$
U_\Lambda=\int_0^R\Phi_\Lambda\,dM=-\frac{\Lambda}{6}\frac{3M}{R^3}\int_0^Rr^4\,dr=-\frac{\Lambda MR^2}{10}.
$$
There is no additional factor $1/2$ for this fixed external potential, unlike the pair-counting factor for gravitational self-energy. At the maximum radius $R=r_m$, the kinetic energy vanishes, so
$$
\boxed{U_m=-\frac35\frac{GM^2}{r_m}-\frac1{10}\Lambda Mr_m^2.}
$$
This is the turnaround energy in <uniform-sphere collapse with a cosmological constant>. The original PDF's shell kinetic term is $\dot r^2/2$, not the $r^2/2$ produced by the TeX transcription.
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