\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