I would like to know if it is possible to calculate in closed-form, or well what work can be done about it, the definite integral $$\int_0^1\int_0^1\int_0^1\frac{3dxdydz}{3-z(x+\sqrt{xy}+y)},\tag{1}$$
where I was inspired in a well-known integral representation for the Apéry constant involving the volume $x\cdot y\cdot z$ in the denominator, and in the formula for the volume of a square frustum of basis $\sqrt{x}$ and $\sqrt{y}$ and height $z$, as reference for all I add the Wikipedia *Heronian mean.*

It is easy to check the integration of the logartihm $$\int_0^1 \frac{1}{3-z(x+\sqrt{xy}+y)}dz=\frac{\log 3-\log(3-x-\sqrt{xy}-y)}{x+\sqrt{xy}+y}$$ but the computations using a CAS (and standard time of computation with my computer), that I tried after this step, seems to me very tedious to evaluate. I suspect that now an important key to evaluate it should be to exploit symmetry or a suitable change of variable.

Question.I would like to know if it is possible to evaluate in closed-form (in terms of well-known constants and/or particular values of special functions) previous definite integral $(1)$. If isn't feasible to get the closed-form that I evoke explain why or add what work can be done.Many thanks.

If this integral is in the literature feel free to refer the literature in your answer or comment and I can to read the result from the literature.

**Edit: (see comments, please).** I don't know if from this step one can to get the integral in closed-form. Feel free to do more feedback, many thanks.

Using previous hints in comments I can to write that it is possible to reduce the integral over $xy$ to one-dimensional integral by using polar coordinates in $xy$-plane

$$\int_0^1\int_0^1\frac{dxdy}{1-\frac{z}{3}(x+\sqrt{xy}+y)}=2\int_0^{\frac{\pi}{4}}\int_0^{\sec \theta}\frac{drd\theta}{\frac{3}{r}-z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)},$$ where the inner integral is equals to $$\int_0^{\sec\theta}\frac{dr}{\frac{3}{r}-z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)}=\frac{-A\sec\theta-3\log(3-A\sec\theta)+3\log 3}{A^2},$$ being $A=z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)$.

Since the integral seems very difficult I am going to accept an answer showing what work can be done (see the **Question**) as soon as expires the bounty.

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