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.*](https://en.wikipedia.org/wiki/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.



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> **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.**
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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 $A=z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)$.




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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.



On the closed-form of $\int_0^1\int_0^1\int_0^1\frac{dxdydz}{1-\frac{z}{3}(x+\sqrt{xy}+y)}$
Let me expand the integrand in powers of $z$ and integrate over $z$,
$$I=\int\_0^1 dx\int\_0^1 dy\int\_0^1 dz\;\frac{3}{3-z(x+\sqrt{xy}+y)}$$
$$\qquad\qquad=\sum\_{n=0}^\infty\,\frac{3^{-n}}{n+1} \int\_0^1 dx\int\_0^1 dy\;(x+\sqrt{xy}+y)^n.$$
The integral over $x$ and $y$ is an element $c\_{n}\in\mathbb{Q}$,
$$I=\sum\_{n=0}^\infty\,\frac{3^{-n}}{n+1}c\_n,\;\;c\_n=\int\_0^1 dx\int\_0^1 dy\;(x+\sqrt{xy}+y)^n,$$
$$\{c\_0,c\_1,c\_2,\ldots\}=\{1,\tfrac{13}{9},\tfrac{149}{60},\tfrac{1667}{350},\tfrac{18623}{1890},\tfrac{69667}{3234},\ldots\}.$$
I have not yet succeeded in finding a closed-form expression for $c\_n$ valid for all integer $n$, but that seems doable. At least we would then have the desired integral as a sum over rational coefficients.




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 With FusRoDah's input for $c\_n$, and carrying out the sum over $n$, I find
$$I=\sum\_{a,b=0}^\infty\frac{12 \Gamma (a+b+1) \left(3^{-a-\frac{b}{2}} B\_{\frac{1}{3}}\left(\frac{b}{2}+1,-a-b\right)-B\_{\frac{1}{3}}(a+b+1,-a-b)\right)}{(2 a+b) (2 a+b+2) \Gamma (a+1) \Gamma (b+1)},$$
(with $B$ the incomplete beta function).


