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Let us define a following function: \begin{eqnarray} {\mathcal J}^{(d)}(\vec{A}) := \int\limits_0^\infty e^{-u^2} \prod\limits_{\xi=1}^d erf(A_\xi u) du \end{eqnarray} for $\vec{A}:=\left(A_\xi\right)_{\xi=1}^d \in {\mathbb R}_+^d$. This function is analyzed in An integral involving error functions and a Gaussian .

Now let $A_5 \ge 0$ be real and let us consider a a following definite integral: \begin{equation} I(A_5):= \int\limits_0^\infty e^{-u^2} [erf(u)]^4 erf[A_5 u] du = {\mathcal J}^{(5)}\left(1,1,1,1,A_5\right) \end{equation}

Now, by differentiating with respect to the parameter $A_5$ and then by integrating the result back again with respect to that parameter we obtained the following results: \begin{eqnarray} &&I(A_5)= \frac{8}{\pi}\arctan(\frac{A_5}{\sqrt{2+A_5^2}}) {\mathcal J}^{(3)}\left( \frac{1}{\sqrt{2+A_5^2}}, \frac{1}{\sqrt{2+A_5^2}}, \frac{1}{\sqrt{2+A_5^2}}\right) -\\ && \frac{96}{\pi^{5/2}} \int\limits_0^{A_5} \frac{t \sqrt{\frac{t^2+4}{t^2+2}} }{t^4+7 t^2+12} \arctan\left(\frac{t}{\sqrt{t^2+2}}\right) \arctan\left(\frac{1}{\sqrt{t^2+4}}\right) dt \end{eqnarray}

A5 = RandomReal[{0, 5}, WorkingPrecision -> 50];
NIntegrate[Exp[-u^2] Erf[u]^4 Erf[A5 u], {u, 0, Infinity}]
8/Pi NIntegrate[
  1/(1 + t^2) 1/Sqrt[2 + t^2] J30[1/Sqrt[2 + t^2], 1/Sqrt[2 + t^2], 
    1/Sqrt[2 + t^2]], {t, 0, A5}]
8/Pi (ArcTan[A5/Sqrt[2 + A5^2]] J30[1/Sqrt[2 + A5^2], 
     1/Sqrt[2 + A5^2], 1/Sqrt[2 + A5^2]] + 
   12/Pi^(3/2) NIntegrate[( 
      t Sqrt[(4 + t^2)/(2 + t^2)])/ (12 + 7 t^2 + t^4) ArcTan[t/Sqrt[
       2 + t^2]] ArcTan[1/Sqrt[4 + t^2] ], {t, 0, A5}])

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The message that I want to pass on is that we were not able to express the quantity $I(A_5)$ through elementary functions and di-logarithms only but instead we end up with an additional term that involves integrating a product of arc tangent function and a function containing square roots. My question is how do we deal with that last term? Is it possible to express the quantity in question through poly-logarithms ?

Note that integrals involving arc tangents and square roots were already analyzed in How do I evaluate this integral :$\int \frac{1}{ \sqrt{1-x^2} } \arctan( \frac{\sqrt{1-x^2}}{2})dx$? and in How find this integral $\int\frac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$ . However my example does not seem to be possible to be tackled by methods being used in those other questions.

Update: It is likely that the integral in question can be tackled by using integration by parts. Indeed the following identities hold true: \begin{eqnarray} &&\int\limits_0^{A_5} \frac{t \sqrt{\frac{t^2+4}{t^2+2}} }{t^4+7 t^2+12} \arctan\left(\frac{t}{\sqrt{t^2+2}}\right)dt=\arctan\left(\frac{A_5}{\sqrt{A_5^2+2}}\right) \arctan\left(\sqrt{\frac{A_5^2+2}{A_5^2+4}}\right)+\\ && \frac{\sqrt{2}}{8} \sum\limits_{\xi\in\{-1,1\}} \sum\limits_{\xi_1\in\{-1,1\}} \sum\limits_{\eta\in\{-1,1\}} \sum\limits_{\eta_1\in\{-1,1\}} \xi \eta_1 (-\sqrt{3}+\xi_1) \sqrt{2+\xi_1 \sqrt{3}}\cdot {\mathfrak F}^{(1,\frac{-\imath A_5 + \sqrt{4+A_5^2}}{\sqrt{2}\sqrt{2+A_5^2}})}_{\frac{\imath}{\sqrt{2}(\xi+\eta \sqrt{3})},-\xi_1 \eta_1 \sqrt{2-\xi_1 \sqrt{3}}}\\ &&\int\limits_0^{A_5} \frac{t \sqrt{\frac{t^2+4}{t^2+2}} }{t^4+7 t^2+12} \arctan\left(\frac{1}{\sqrt{t^2+4}}\right)dt=\left.\arctan\left(\frac{1}{\sqrt{t^2+4}}\right) \arctan\left(\sqrt{\frac{t^2+2}{t^2+4}}\right)\right|_{t=0}^{t=A_5}+\\ && \frac{1}{4} \sum\limits_{\xi\in\{-1,1\}} \sum\limits_{\xi_1\in\{-1,1\}} \sum\limits_{\eta\in\{-1,1\}} \sum\limits_{\eta_1\in\{-1,1\}} \xi \xi_1 \eta_1 {\mathfrak F}^{(\sqrt{2}-1,\frac{-\sqrt{2} + \sqrt{4+A_5^2}}{\sqrt{2+A_5^2}})}_{\imath(\xi+\eta \sqrt{2}),\frac{\sqrt{3}}{3} \imath \xi_1 \sqrt{1+\xi_1 \eta_1 2 \imath \sqrt{2}}} \end{eqnarray} where the function ${\mathfrak F}^{(A,B)}_{a,b}$ is related to di-logarithms and is defined in An integral involving error functions and a Gaussian .

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