Quadruple Integral $\int_0^1\int_0^1\int_0^1\int_0^1\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\kern 2pt}{\rm\,d}x{\rm\,d}y{\rm\,d}z{\rm\,d}w$

Question

\begin{align*} \int_0^1\int_0^1\int_0^1\int_0^1\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\rm\,d}x\!{\rm\,d}y\!{\rm\,d}z\!{\rm\,d}w &=\,\iiiint\limits_{\substack{0\,\leqslant\,x\,\leqslant\,1 \\ 0\,\leqslant\,y\,\leqslant\,1\\ 0\,\leqslant\,z\,\leqslant\,1\\0\,\leqslant\,w\,\leqslant\,1}}\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\rm\,d}x\!{\rm\,d}y\!{\rm\,d}z\!{\rm\,d}w\\ &=\,{\color{red}{4}}\iiiint\limits_{V_1:\substack{0\,\leqslant\,y\,\leqslant\,x\,\leqslant\,1 \\ 0\,\leqslant\,z\,\leqslant\,x\,\leqslant\,1\\ \\ 0\,\leqslant\,w\,\leqslant\,x\,\leqslant\,1}}\sqrt{x^{\overset{\,}{2}}+y^2+z^2+w^2}{\rm\,d}x\!{\rm\,d}y\!{\rm\,d}z\!{\rm\,d}w\\ \,\\ \overset{\begin{cases} x=r\sin\psi\sin\theta\cos\varphi\\ y=r\sin\psi\sin\theta\sin\varphi\\ z=r\sin\psi\cos\theta\\ w=r\cos\psi \end{cases}}{\overline{\overline{\hspace{4cm}}}}\quad&\,4\iiiint\limits_{V_1}r^4\sin^2\psi\sin\theta\,{\rm\,d}r\!{\rm\,d}\psi\!{\rm\,d}\theta\!{\rm\,d}\varphi\\ \end{align*} \begin{align*} &=4\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\int_{\arctan\left(\csc\theta\sec\varphi\right)}^{\frac{\pi}{2}}\int_{0}^{\csc\psi\csc\theta\sec\varphi}r^4\sin^2\psi\sin\theta\,{\rm\,d}r\!{\rm\,d}\psi\!{\rm\,d}\theta\!{\rm\,d}\varphi\\ &=\frac{4}{5}\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\int_{\arctan\left(\csc\theta\sec\varphi\right)}^{\frac{\pi}{2}}\frac{1}{\sin^3\psi\sin^4\theta\cos^5\varphi}\,{\rm\,d}\psi\!{\rm\,d}\theta\!{\rm\,d}\varphi\\ &=\frac{4}{5}\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\dfrac{\displaystyle\int_0^{\sin\theta\cos\varphi}\sqrt{1+u^2}{\rm\,d}u}{\sin^4\theta\,\cos^5\varphi}\,{\rm\,d}\theta\!{\rm\,d}\varphi\\ &=\frac{\color{blue}{2}}{5}\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\left(\dfrac{\sqrt{1+\sin^2\theta\,\cos^2\varphi}}{\sin^3\theta\,\cos^4\varphi}+\dfrac{\ln\left(\sin\theta\,\cos\varphi+\sqrt{1+\sin^2\theta\,\cos^2\varphi}\right)}{\sin^4\theta\,\cos^5\varphi}\right)\,{\rm\,d}\theta\!{\rm\,d}\varphi\\ \end{align*} I have got the following result, but I don't know whether this result can be simplified. \begin{align*} \frac{8}{15}+\frac{14\ln3}{15}+\frac{8\sqrt{2}}{15}\arctan\left(\frac{\sqrt{2}}{5}\right)-\frac{8\sqrt{2}}{15}\arctan\left(\frac{1}{\sqrt{2}}\right)\\ +\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{2}{\sqrt{1+u^2}}+\sqrt{1+\frac{2}{1+u^2}}\right){\rm\,d}u\\ -\frac{8}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}+\sqrt{1+\frac{1}{1+u^2}}\right){\rm\,d}u\\ -\frac{4}{15}\int_0^1\left(\sqrt{1+u^2}\right)^3\arctan\left(\frac{1}{\sqrt{1+u^2}}\right){\rm\,d}u \end{align*}

Various closed forms

https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf

Answer 1

Although not a complete answer, I thought it might be of some use to explain how the integrals in my earlier comments were obtained.

The $\arctan$ expressions of the $u$-integral can be reduced with the help of the identities

$$\arctan a - \arctan b = \arctan\frac{a-b}{1+ab} \\ \arctan c = 2 \arctan \frac c{1+\sqrt{1+c^2}}$$

to yield the somewhat simpler expression,

$$\mathcal I = \frac1{15} \left(8 + 14\log 3 - 8 \sqrt2 \arctan\frac1{2\sqrt2} + 8 \mathcal J\right)$$

where

$$\mathcal J = \int_0^1 \left(1+u^2\right)^{3/2} \arctan\left(\sqrt{1+u^2}\sqrt{3+u^2} - \left(2+u^2\right)\right) \, du$$

Integration by parts gives

$$\mathcal J = -\frac{\left(7\sqrt2+3\operatorname{arsinh}1\right)\arctan\frac1{2\sqrt2}}{16} - \frac18 \int_0^1 \frac{u^2\left(5+2u^2\right)}{\left(2+u^2\right)\sqrt{3+u^2}} \, du - \frac38 \mathcal K$$

where

$$\mathcal K = \int_0^1 \frac{u\operatorname{arsinh}u}{\left(2+u^2\right) \sqrt{1+u^2} \sqrt{3+u^2}} \, du$$

The first of these integrals can be simplified with an Euler substitution of $u=\dfrac{3-v^2}{2v}$ and partial fraction expansion, and one of the subsequent integrals can be further reduced by completing the square, reducing powers, and simple substitutions:

$$\begin{align*} \int_0^1 \frac{u^2\left(5+2u^2\right)}{\left(2+u^2\right)\sqrt{3+u^2}} \, du &= \int_1^{\sqrt3} \left(\frac v2 - \frac2v + \frac9{2v^3} - \frac{8v}{9+2v^2+v^4}\right) \, dv \\ &= 2 - \log3 - \sqrt2 \int_\tfrac1{\sqrt2}^{\sqrt2} \frac{dv}{1+v^2} \\ &= 2 - \log3 - \sqrt2 \arctan\frac1{2\sqrt2} \end{align*}$$

Substituting $u\to\sinh u$ in $\mathcal K$, integrating by parts, then substituting back $u\to\operatorname{arcosh}u$, we can eliminate the mixed trig-hyperbolic constant and we're left with

$$\mathcal I = \frac25 + \log3 - \frac{7\sqrt2}{10} \arctan\frac1{2\sqrt2} - \frac15 \int_0^1 \frac{\arctan\left(2+u^2-\sqrt{1+u^2}\sqrt{3+u^2}\right)}{\sqrt{1+u^2}} \, du \tag{$*$}$$

Next, substituting $v=2+u^2-\sqrt{1+u^2}\sqrt{3+u^2}$, then enforcing $v\to\dfrac{1-v}{1+v}$ and applying the identity $\arctan\dfrac{1-a}{1+a}=\dfrac\pi4-\arctan a$, followed by $v\to\dfrac1v$ and $\arctan\dfrac1a=\dfrac\pi2-\arctan a$,

$$\begin{align*} \mathcal I &= \frac25 + \log3 - \frac{7\sqrt2}{10} \arctan\frac1{2\sqrt2} - \frac15 \int_{3-2\sqrt2}^{2-\sqrt3} \frac{(1+v)\arctan v}{2v\sqrt{1-4v+v^2}} \, dv \\ &= \frac25 + \log3 - \frac{7\sqrt2}{10}\arctan\frac1{2\sqrt2} - \frac\pi{20} \operatorname{artanh}\frac1{\sqrt2} - \frac{\sqrt2}5 \int_\tfrac1{\sqrt3}^\tfrac1{\sqrt2} \frac{\arctan v}{\left(v^2-1\right)\sqrt{3v^2-1}} \, dv \\ &= \frac25 + \log3 - \frac{7\sqrt2}{10}\arctan\frac1{2\sqrt2} + \frac\pi{20} \operatorname{arsinh}1 + \frac{\sqrt2}5 \int_{\sqrt2}^{\sqrt3} \frac{v\arctan v}{\left(1-v^2\right)\sqrt{3-v^2}} \, dv \end{align*}$$

The remaining integral bears a striking resemblance to the family of integrals mentioned here by @pisco except that we don't have the condition $a>b$.

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On the other hand, the $\arctan$ integral $(*)$ has a more condensed form as follows, though it's not clear to me whether this is of any practical use. Recall the identities $\arctan\sqrt{\dfrac{1-x}{1+x}}=\dfrac12\arcsin\sqrt{1-x^2}$ and $\arcsin x=\arctan\dfrac{\sqrt{1-x^2}}x$. (Alternatively, integrate $\mathcal K$ by parts in a different way.)

$$\begin{align*} (*) &= \frac12 \int_0^1 \frac{\arcsin\frac1{2+u^2}}{\sqrt{1+u^2}} \, du \\ &= \frac12 \int_0^1 \frac{\arctan\frac1{\sqrt{1+u^2}\sqrt{3+u^2}}}{\sqrt{1+u^2}} \, du \\ &= \frac12 \int_0^1 \int_0^\tfrac1{\sqrt{3+u^2}} \frac{dv\,du}{1+u^2+v^2} \end{align*}$$

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Per @MariuszIwaniuk's comment above, the original integral appears to have a closed form of

$$\mathcal I = \frac25 + \log3 - \frac{7\sqrt2}{10}\arctan\frac1{2\sqrt2} - \frac C{10} + \frac3{10} \Im \operatorname{Li}_2\left(\left(3-2\sqrt2\right)i\right)$$

($C$ is Catalan's constant.) I don't know how it was obtained, but based on what seems to be a closely related box integral (see @JackD'Aurizio's result here),

$$\frac13 \iiint\limits_{[0,1]^3} \frac{dx\,dy\,dz}{x^2+y^2+z^2} = \frac\pi2\operatorname{arsinh}1 - C + \Im \operatorname{Li}_2\left(\left(3-2\sqrt2\right)i\right),$$

I expect $(*)$ can be unraveled after converting it to an integral that more resembles the dilogarithm, likely using the same approach as @Zacky has shown here.