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Is it possible to get a closed form of the following integral $$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$

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For real $a,b\neq 0$ Mathematica gives $$ \Phi(a,b) = \frac{1}{a\sqrt{b^2-a^2}}\cdot\arctan\left(\frac{\sqrt{b^2-a^2}}{a\sqrt{1+b^2}}\right) $$

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\begin{align} \int_{0}^{1}\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2+b^2}}&=\int_{0}^{\large k_1}\frac{b\sec^2t}{(b^2\tan^2t+a^2)\sqrt{b^2\tan^2t+b^2}}\mathrm dt\tag1\\ &=\int_{0}^{\large k_1}\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\mathrm dt\tag2\\ &=\int_{0}^{\large k_1}\frac{\cos t}{\left(b^2-a^2\right)\sin^2t+a^2}\mathrm dt\tag3\\ &=\frac{1}{a\sqrt{b^2-a^2}}\int_{0}^{\large k_2}\frac{\mathrm dy}{y^2+1}\tag4\\ \Phi(a,b)&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{a\sqrt{b^2-a^2}}\arctan\left(\frac{\sqrt{b^2-a^2}}{a\sqrt{1-b^2}}\right)}} \end{align}


Explanation :

$(1)\;$ Use substitution $\;\displaystyle x=b\tan t\;$ and $\;\displaystyle k_1=\arctan \left(\frac{1}{b}\right)\quad\Rightarrow\quad\tan k_1=\frac{1}{b}$

$(2)\;$ Use identities $\;\displaystyle \sec^2t=1+\tan^2t\;$ and $\;\displaystyle \tan t=\frac{\sin t}{\cos t}$

$(3)\;$ Use identity $\;\displaystyle \cos^2t=1-\sin^2t$

$(4)\;$ Use substitution $\;\displaystyle \sin t=\frac{ay}{\sqrt{b^2-a^2}}\,$ and $\;\displaystyle k_2=\frac{\sqrt{b^2-a^2}}{a\sqrt{1-b^2}}\,$ since $\;\;\;\quad\displaystyle\tan k_1=\frac{1}{b}\quad\Rightarrow\quad\sin k_2=\frac{1}{\sqrt{1-b^2}}$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\Phi\pars{a,b}\equiv \int_{0}^{1}{\dd x \over \pars{x^{2} + a^{2}}\root{x^{2} + b^{2}}}:\ {\large ?}}$.

With Euler-sub $\ds{\root{x^{2} + b^{2}} \equiv x + t}$ we found $\ds{x = {b^{2} - t^{2} \over 2t}}$ such that:

\begin{align} \Phi\pars{a,b}&=-4\int_{\verts{b}}^{\root{1 + b^{2}}\ -\ 1}{t\,\dd t\over t^{4} + 2\pars{2a^{2} - b^{2}}t^{2} + b^{4}} \end{align}

With $\ds{t\ \mapsto\ t^{1/2}}$:

\begin{align} \Phi\pars{a,b}& =-2\int_{b^{2}}^{2 + b^{2} - 2\root{1 + b^{2}}} {\dd t \over t^{2} + 2\pars{2a^{2} - b^{2}}t + b^{4}} \\[5mm]&=-2\int_{b^{2}}^{2 + b^{2} - 2\root{1 + b^{2}}} {\dd t \over \pars{t + 2a^{2} - b^{2}}^{2} - \pars{2a^{2} - b^{2}}^{2}+ b^{4}} \\[5mm]&=-2\int_{2a^{2}}^{2a^{2} + 2 - 2\root{1 + b^{2}}} {\dd t \over t^{2} + 4a^{2}b^{2} - 4a^{4}} \\[5mm]&=-\,{1 \over \verts{a}\root{b^{2} - a^{2}}} \int_{\verts{a}/\root{b^{2} - a^{2}}} ^{\pars{a^{2} + 1 - \root{1 + b^{2}}}/\pars{\verts{a}\root{b^{2} - a^{2}}}} {\dd t \over t^{2} + 1} \\[5mm]&={1 \over \verts{a}\root{b^{2} - a^{2}}}\bracks{% \arctan\pars{\verts{a} \over \root{b^{2} - a^{2}}}- \arctan\pars{a^{2} + 1 - \root{1 + b^{2}} \over \verts{a}\root{b^{2} - a^{2}}}} \end{align}

With the identity $\ds{\quad\arctan\pars{x} - \arctan\pars{y} =\arctan\pars{x - y \over 1 + xy}}$ we get

$$ \color{#66f}{\large\Phi\pars{a,b}} =\color{#66f}{\large{1 \over a\root{b^{2} - a^{2}}}\, \arctan\pars{\root{b^{2} - a^{2}} \over a\root{1 + b^{2}}}} $$

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