Transformed distribution involving logistic equation solution

Suppose I solve the logistic equation (with $y(0) = 1$ at $t=10$),

$$y(10) = \frac{\kappa e^{10 r}}{\kappa +e^{10 r}-1},$$

where $\kappa$ is the carrying capacity and $r$ is the initial growth rate.

Further suppose that I have the following uncertainty about the inputs: $r\sim U(0,1)$ and $\kappa \sim U(10,20)$. I can simulate what the distribution of $y(10)$ looks like,

fSampleRK[] := {RandomReal[{0, 1}], RandomReal[{10, 20}]}
fLogistic[r_, \[Kappa]_, t_, a_] := (a E^(r t) \[Kappa])/(-a + a E^(r t) + \[Kappa])
fSampleQ[t_, a_] := Block[{lRK = fSampleRK[]}, fLogistic[lRK[], lRK[], t, a]]
Histogram[Table[fSampleQ[10, 1], {100000}], 100]


Yielding the monstrosity below, I rather optimistically tried the following,

aDist = TransformedDistribution[(E^(10 r) \[Kappa])/(-1 + E^(10 r) + \[Kappa]),
{r \[Distributed] UniformDistribution[{0, 1}],
\[Kappa] \[Distributed] UniformDistribution[{0, 10}]}]


But Mathematica falls over and returns the bottom line unevaluated.

Can anyone think of a way to determine the analytic PDF here?

(Of course I can sample and fit a kernel density estimator to the samples, but I am looking for an analytic result.)

\$Version

(* "11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)" *)

(E^(10 r) κ)/(-1 + E^(10 r) + κ),
{r \[Distributed] UniformDistribution[{0, 1}],
κ \[Distributed] UniformDistribution[{0, 10}]}];


Although Mathematica cannot determine a closed-form representation for aDist, it can still do some calculations with it.

mean = Mean[aDist] // FullSimplify

(* (1/200)*(-10 + 2*E^10*(5 - (-2 + E^10)*ArcTanh[5/(4 + E^10)]) - 100*Log +
Log[-1 + E^10] + 99*Log[9 + E^10]) *)

mean // N

(* 4.09884 *)

(* 1/(200*Sqrt[-(3/(-35700 + 406000*Log +
60000*Log*Log - 30000*Log^2 -
460*Log[-1 + E^10] - 600*Log*Log[-1 + E^10] +
3*Log[-1 + E^10]^2 + 200*Log*
(2030 + 300*Log - 3*Log[-1 + E^10] -
297*Log[9 + E^10]) + 3*E^40*
(Log[-1 + E^10] - Log[9 + E^10])^2 -
405540*Log[9 + E^10] - 59400*Log*
Log[9 + E^10] + 594*Log[-1 + E^10]*
Log[9 + E^10] + 29403*Log[9 + E^10]^2 +
2*E^20*(-3850 + 9*Log[-1 + E^10]^2 -
6*Log[-1 + E^10]*(-85 + 50*Log +
50*Log - 47*Log[9 + E^10]) +
30*(-17 + 10*Log + Log)*
Log[9 + E^10] - 291*Log[9 + E^10]^2) +
4*E^10*(10850 + 45*Log[-1 + E^10] -
3*Log[-1 + E^10]^2 + 300*Log*
(-5 + Log[-1 + E^10] - Log[9 + E^10]) +
300*Log*(-5 + Log[-1 + E^10] -
Log[9 + E^10]) + 1455*Log[9 + E^10] -
294*Log[-1 + E^10]*Log[9 + E^10] +
297*Log[9 + E^10]^2) -
4*E^30*(3*Log[-1 + E^10]^2 + Log[-1 + E^10]*
(185 - 6*Log[9 + E^10]) + Log[9 + E^10]*
(-185 + 3*Log[9 + E^10]))))]) *)

stdDev // N

(* 2.67158 *)


While working with the PDF is a problem, the CDF can be numerically evaluated

CDF[aDist, 10]

(* 1 *)

median = x /. FindRoot[CDF[aDist, x] == 1/2, {x, 3}]

(* 3.63448 *)

Epilog -> {Red, Dashed, Line[{{median, 0}, {median, 1/2}, {0, 1/2}}]}] data = RandomVariate[aDist, 10000];

bDist = FindDistribution[data, RandomSeeding -> 1]

(* MixtureDistribution[{0.317651, 0.406901,
0.275448}, {NormalDistribution[1.73085, 0.85307],
UniformDistribution[{0.00174172, 9.99195}],
UniformDistribution[{0.00174172, 9.99195}]}] *)

Show[
Histogram[data, Automatic, "PDF"],
Plot[PDF[bDist, x], {x, 0, 10}]] Show[Histogram[data, Automatic, "CDF"],
Plot[{CDF[aDist, x], CDF[bDist, x]}, {x, 0, 10},
PlotStyle -> {Red, Blue},
BaseStyle -> Thick,
PlotLegends -> Placed["Expressions", {.25, .75}]]] 