Can this previously solved three-dimensional constrained integration also be solved with certain added products in the integrand?

Question

In Solved3DConstrainedIntegration the constrained three-dimensional (Hilbert-Schmidt-metric-based HSmetric) integration problem for the absolute separability probability of the two-qubit (quantum bit) states AbsSepProb

Integrate[
 9081072000 (x - y)^2 (x - z)^2 (y - z)^2 (-1 + 2 x + y + z)^2 (-1 + 
     x + 2 y + z)^2 (-1 + x + y + 2 z)^2 Boole[
   x > y && y > z && z > 1 - y - z], {z, 0, 1}, {y, 0, 1}, {x, 0, 
  1}]

was solved by user JimB with the answer

29902415923/497664 - 50274109/(512 Sqrt[2]) - (3072529845 π)/(32768 Sqrt[2]) + (1024176615 ArcCos[1/3])/(4096 Sqrt[2])

I now would like to know if one multiplies the integrand above by various product terms (associated with the important class of monotone metrics MonotoneMetrics--including the prominent Bures/maximal BuresMetric one)--whether the so-extended problems can also be solved.

All the product terms of interest by which the integrand is multiplied contain two factors, a common one for all the monotone metrics,

((x - y)^2 (x - z)^2 (y - z)^2 (-1 + 2 x + y + z)^2 (-1 + x + 2 y + z)^2 (-1 + x + y + 2 z)^2)/(Sqrt[x] Sqrt[y] Sqrt[1 - x - y - z] Sqrt[z])

and a monotone-metric-specific one of the form

c[x, y] c[x, 1 - x - y - z] c[x, z] c[y, 1 - x - y - z] c[y, z] c[z, 1 - x - y - z]c[x, y] c[x, 1 - x - y - z] c[x, z] c[y, 1 - x - y - z] c[y, z] c[z, 1 - x - y - z]

where

c[u_, v_] := 1/(v f[v/u)]

$f$ is some choice of a "Morozova-Chentsov" function MC. Prominent examples of $f(t)$ are $\sqrt{t}$ (probably the simplest cf. LovasAndaiInvariance--if any to solve), $\frac{1+t}{2}$ (giving rise to the Bures metric, $f(t)=\frac{2 t}{t+1}$ and the Kubo-Mori $f(t)=\frac{t-1}{\ln{t}}$.

So, for the choice $f(t)=\sqrt{t}$, the general class of problems I am posing takes the specific form

Integrate[(9081072000 (x - y)^4 (x - z)^4 (y - z)^4 (-1 + 2 x + y + z)^4 (-1 + x + 2 y + z)^4 (-1 + x + y + 2 z)^4)/(x^2 y^2 z^2 (-1 + x + y + z)^2)Boole[x > y && y > z && z > 1 - y - z], {z, 0, 1}, {y, 0, 1}, {x, 0, 1}]

It would seem that much of the same approach of user JimB in his answer to the earlier question could be pursued but with further development apparently needed.

Answer 1

The answer to the specific problem posed at the very end based on the function $f(t)=\sqrt{t}$ appears to be

-(59126849405441003905/58549611741984) - (20093337600 Log[2])/323 + (148387043175 Log[3])/2432

which evaluates to 2.290169135502096*10^7, as numerical integration strongly supports.

But I am puzzled as this result--not lying in [0,1]--is clearly not a probability, which was the aim of the exercise. So, perhaps there is some normalization required.

The statement

 Boole[x > y && y > z && z > 1 - y - z]

imposes the the ordering requirement on the four eigenvalues ( x, y, z and 1 - y - z, summing to 1) of the (trace unity) $4 \times 4$ "density matrix" that must be satisfied for the property of "absolute separability" to hold. So, the normalization required would be when this eigenvalue ordering requirement is lifted.

Oh, well! For absolute separability, I should have also included the crucial property that $x - z < 2 \sqrt{y (1 - x - y - z)}$, which can be seen in the problem stated in Solved3DintegrationProblem.

So, without the imposition of this requirement, I am compelled to conclude that my calculation is simply that of the normalization constant.

Back to the drawing board!

If I use the integrand

integrand = 9081072000 (x - y)^2 (x - z)^2 (y - z)^2 (-1 + 2 x + y + z)^2 (-1 + x + 2 y + z)^2 (-1 + x + y + 2 z)^2

of specific interest here in the user JimB's code

(* Initial integrations ) a1 = Integrate[integrand, {y, 1 - z - 2 Sqrt[z - 2 z^2], 1/2 (1 - 2 z)}, {x, 1 - y - 2 z, -2 y + z + 2 Sqrt[y - 2 y z]}, Assumptions -> {1/8 (2 - Sqrt[2]) < z < 1/6}];( Around 6 minutes ) a2 = Integrate[integrand, {y, 1/2 (1 - 2 z), (2 - z)/9 + 2/9 Sqrt[1 - z - 2 z^2]}, {x, y, -2 y + z + 2 Sqrt[y - 2 y z]}, Assumptions -> {1/8 (2 - Sqrt[2]) < z < 1/6}]; ( Around 4 minutes ) a3 = Integrate[integrand, {y, z, 1/2 (1 - 2 z)}, {x, 1 - y - 2 z, -2 y + z + 2 Sqrt[y - 2 y z]}, Assumptions -> {1/6 < z < 1/4}]; a4 = Integrate[integrand, {y, 1/2 (1 - 2 z), (2 - z)/9 + 2/9 Sqrt[1 - z - 2 z^2]}, {x, y, -2 y + z + 2 Sqrt[y - 2 y z]}, Assumptions -> {1/6 < z < 1/4}]; ( Around 3 minutes ) a5 = Integrate[integrand, {y, z, (2 - z)/9 + 2/9 Sqrt[1 - z - 2 z^2]}, {x, y, -2 y + z + 2 Sqrt[y - 2 y z]}, Assumptions -> {1/4 < z < 1/3}]; ( Around 5 minutes *)

in the previous (Hilbert-Schmidt) solved problem, the indicated computations do not (appear) to come to completion.