TITLE: Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$
QUESTION [5 upvotes]: Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$

REPLY [6 votes]: \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 $\;\;\;\quad\displaystyle\tan k_1=\frac{1}{b}\quad\Rightarrow\quad\sin k_2=\frac{1}{\sqrt{1-b^2}}$