I want to solve the following integral
\begin{align}
\int_{-\beta}^{\beta}\sqrt{\beta^2-x^2}\frac{\cos\left(\phi-\frac{3}{2}\arctan\left(\frac{\gamma+x}{\alpha}\right)\right)}{\left(\alpha^2+(\gamma+x)^2\right)^{3/4}}\mathrm{d}x
\end{align}
with $\alpha$, $\beta$, $\gamma$, $\phi\in\mathbb{R}_{>0}$
Best Answer
Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\times\mathbb{R}\rightarrow\mathbb{R}$ by the definite integral
$$\mathcal{I}{\left(\alpha,\beta,\gamma,\phi\right)}:=\int_{-\beta}^{\beta}\mathrm{d}x\,\frac{\sqrt{\beta^{2}-x^{2}}\cos{\left(\phi-\frac32\arctan{\left(\frac{\gamma+x}{\alpha}\right)}\right)}}{\left[\alpha^{2}+\left(\gamma+x\right)^{2}\right]^{3/4}}.$$
At first I was hopeful this integral could be reduced to elliptic integrals, but alas it appears to be hyper-elliptic. Here's what I found....
Suppose $\alpha>0\land\beta>0\land\gamma>0$. Without loss of generality, we may assume $0\le\phi<2\pi$.
Set $a:=\frac{\gamma-\beta}{\alpha}\land b:=\frac{\gamma+\beta}{\alpha}$, and note that $b>0\land b>a$. We can clean up $\mathcal{I}$ by rewriting it as
$$\begin{align} \mathcal{I}{\left(\alpha,\beta,\gamma,\phi\right)} &=\int_{-\beta}^{\beta}\mathrm{d}x\,\frac{\sqrt{\beta^{2}-x^{2}}\cos{\left(\phi-\frac32\arctan{\left(\frac{\gamma+x}{\alpha}\right)}\right)}}{\left[\alpha^{2}+\left(\gamma+x\right)^{2}\right]^{3/4}}\\ &=\int_{\frac{\gamma-\beta}{\alpha}}^{\frac{\gamma+\beta}{\alpha}}\mathrm{d}y\,\frac{\alpha\sqrt{\left(\beta+\gamma-y\alpha\right)\left(\beta-\gamma+y\alpha\right)}\cos{\left(\phi-\frac32\arctan{\left(y\right)}\right)}}{\left(\alpha^{2}+\alpha^{2}y^{2}\right)^{3/4}};~~~\small{\left[\frac{\gamma+x}{\alpha}=y\right]}\\ &=\int_{\frac{\gamma-\beta}{\alpha}}^{\frac{\gamma+\beta}{\alpha}}\mathrm{d}y\,\frac{\sqrt{\alpha}\sqrt{\left(\frac{\gamma+\beta}{\alpha}-y\right)\left(y-\frac{\gamma-\beta}{\alpha}\right)}\cos{\left(\phi-\frac32\arctan{\left(y\right)}\right)}}{\left(1+y^{2}\right)^{3/4}}\\ &=\sqrt{\alpha}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cos{\left(\phi-\frac32\arctan{\left(y\right)}\right)}.\\ \end{align}$$
Making use of trigonometric identities such as the angle difference formula
$$\cos{\left(z-w\right)}=\cos{\left(z\right)}\cos{\left(w\right)}+\sin{\left(z\right)}\sin{\left(w\right)},$$
the multiple-angle formulas
$$\sin{\left(2z\right)}=2\sin{\left(z\right)}\cos{\left(z\right)},$$
$$\cos{\left(2z\right)}=\cos^{2}{\left(z\right)}-\sin^{2}{\left(z\right)},$$
$$\sin{\left(3z\right)}=3\sin{\left(z\right)}-4\sin^{3}{\left(z\right)},$$
$$\cos{\left(3z\right)}=4\cos^{3}{\left(z\right)}-3\cos{\left(z\right)},$$
the power-reduction formulas
$$\sin^{2}{\left(z\right)}=\frac{1-\cos{\left(2z\right)}}{2},$$
$$\cos^{2}{\left(z\right)}=\frac{1+\cos{\left(2z\right)}}{2},$$
the half-angle formulas
$$\cos{\left(\frac{\tau}{2}\right)}=\sqrt{\frac{1+\cos{\left(\tau\right)}}{2}};~~~\small{-\pi<\tau<\pi},$$
$$\sin{\left(\frac{\tau}{2}\right)}=\frac{\sin{\left(\tau\right)}}{2\sqrt{\frac{1+\cos{\left(\tau\right)}}{2}}};~~~\small{-\pi<\tau<\pi},$$
and the composition formulas
$$\sin{\left(\arctan{\left(y\right)}\right)}=\frac{y}{\sqrt{1+y^{2}}},$$
$$\cos{\left(\arctan{\left(y\right)}\right)}=\frac{1}{\sqrt{1+y^{2}}},$$
we can rewrite the integrand of $\mathcal{I}$ in algebraic form as follows:
$$\begin{align} \frac{\mathcal{I}{\left(\alpha,\beta,\gamma,\phi\right)}}{\sqrt{\alpha}} &=\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cos{\left(\phi-\frac32\arctan{\left(y\right)}\right)}\\ &=\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cos{\left(\phi-\frac32\tau\right)};~~~\small{\tau:=\arctan{\left(y\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)}\\ &=\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[\cos{\left(\phi\right)}\cos{\left(\frac32\tau\right)}+\sin{\left(\phi\right)}\sin{\left(\frac32\tau\right)}\right]\\ &=\cos{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cos{\left(\frac32\tau\right)}\\ &~~~~~+\sin{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\sin{\left(\frac32\tau\right)}\\ &=\cos{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[4\cos^{3}{\left(\frac{\tau}{2}\right)}-3\cos{\left(\frac{\tau}{2}\right)}\right]\\ &~~~~~+\sin{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[3\sin{\left(\frac{\tau}{2}\right)}-4\sin^{3}{\left(\frac{\tau}{2}\right)}\right]\\ &=\cos{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[4\cos^{2}{\left(\frac{\tau}{2}\right)}-3\right]\cos{\left(\frac{\tau}{2}\right)}\\ &~~~~~+\sin{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[3-4\sin^{2}{\left(\frac{\tau}{2}\right)}\right]\sin{\left(\frac{\tau}{2}\right)}\\ &=\cos{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[2\cos{\left(\tau\right)}-1\right]\sqrt{\frac{1+\cos{\left(\tau\right)}}{2}}\\ &~~~~~+\sin{\left(\phi\right)}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[2\cos{\left(\tau\right)}+1\right]\frac{\sin{\left(\tau\right)}}{2\sqrt{\frac{1+\cos{\left(\tau\right)}}{2}}}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[2\cos{\left(\tau\right)}-1\right]\sqrt{1+\cos{\left(\tau\right)}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[2\cos{\left(\tau\right)}+1\right]\frac{\sin{\left(\tau\right)}}{\sqrt{1+\cos{\left(\tau\right)}}}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\left[2\cos{\left(\tau\right)}-1\right]\sqrt{1+\cos{\left(\tau\right)}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cdot\frac{\left[2\cos{\left(\tau\right)}+1\right]\left[1-\cos{\left(\tau\right)}\right]\sqrt{1+\cos{\left(\tau\right)}}}{\tan{\left(\tau\right)}\cos{\left(\tau\right)}}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cdot\frac{2-\sec{\left(\tau\right)}}{\sec{\left(\tau\right)}}\sqrt{1+\cos{\left(\tau\right)}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cdot\frac{\tan^{2}{\left(\tau\right)}-1+\sec{\left(\tau\right)}}{\tan{\left(\tau\right)}\sec{\left(\tau\right)}}\sqrt{1+\cos{\left(\tau\right)}}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cdot\frac{2-\sqrt{1+y^{2}}}{\sqrt{1+y^{2}}}\sqrt{1+\frac{1}{\sqrt{1+y^{2}}}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{\sqrt{\left(b-y\right)\left(y-a\right)}}{\left(1+y^{2}\right)^{3/4}}\cdot\frac{y^{2}-1+\sqrt{1+y^{2}}}{y\sqrt{1+y^{2}}}\sqrt{1+\frac{1}{\sqrt{1+y^{2}}}}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{2-\sqrt{1+y^{2}}}{\left(1+y^{2}\right)\sqrt{1+y^{2}}}\sqrt{1+\sqrt{1+y^{2}}}\sqrt{\left(b-y\right)\left(y-a\right)}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{y^{2}-1+\sqrt{1+y^{2}}}{y\left(1+y^{2}\right)\sqrt{1+y^{2}}}\sqrt{1+\sqrt{1+y^{2}}}\sqrt{\left(b-y\right)\left(y-a\right)}.\\ \end{align}$$
Transforming the integrals using the Euler substitution
$$\sqrt{1+y^{2}}=y+t\implies y=\frac{1-t^{2}}{2t},$$
we find
$$\begin{align} \frac{\mathcal{I}{\left(\alpha,\beta,\gamma,\phi\right)}}{\sqrt{\alpha}} &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{2-\sqrt{1+y^{2}}}{\left(1+y^{2}\right)\sqrt{1+y^{2}}}\sqrt{1+\sqrt{1+y^{2}}}\sqrt{\left(b-y\right)\left(y-a\right)}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{a}^{b}\mathrm{d}y\,\frac{y^{2}-1+\sqrt{1+y^{2}}}{y\left(1+y^{2}\right)\sqrt{1+y^{2}}}\sqrt{1+\sqrt{1+y^{2}}}\sqrt{\left(b-y\right)\left(y-a\right)}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{\sqrt{1+a^{2}}-a}^{\sqrt{1+b^{2}}-b}\mathrm{d}t\,\frac{(-1)(1+t^{2})}{2t^{2}}\cdot\frac{2-\left(\frac{1+t^{2}}{2t}\right)}{\left(\frac{1+t^{2}}{2t}\right)^{2}\left(\frac{1+t^{2}}{2t}\right)}\\ &~~~~~\times\sqrt{1+\frac{1+t^{2}}{2t}}\sqrt{\left(b-\frac{1-t^{2}}{2t}\right)\left(\frac{1-t^{2}}{2t}-a\right)};~~~\small{\left[y=\frac{1-t^{2}}{2t}\right]}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{\sqrt{1+a^{2}}-a}^{\sqrt{1+b^{2}}-b}\mathrm{d}t\,\frac{(-1)(1+t^{2})}{2t^{2}}\cdot\frac{\left(\frac{1-t^{2}}{2t}\right)^{2}-1+\left(\frac{1+t^{2}}{2t}\right)}{\left(\frac{1-t^{2}}{2t}\right)\left(\frac{1+t^{2}}{2t}\right)^{2}\left(\frac{1+t^{2}}{2t}\right)}\\ &~~~~~\times\sqrt{1+\frac{1+t^{2}}{2t}}\sqrt{\left(b-\frac{1-t^{2}}{2t}\right)\left(\frac{1-t^{2}}{2t}-a\right)};~~~\small{\left[y=\frac{1-t^{2}}{2t}\right]}\\ &=\frac{\cos{\left(\phi\right)}}{\sqrt{2}}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{1}{t}\cdot\frac{2(2t)^{2}-\left(\frac{1+t^{2}}{2t}\right)(2t)^{2}}{\left(\frac{1+t^{2}}{2t}\right)^{2}(2t)^{2}}\\ &~~~~~\times\sqrt{1+\frac{1+t^{2}}{2t}}\sqrt{\left(b-\frac{1-t^{2}}{2t}\right)\left(\frac{1-t^{2}}{2t}-a\right)}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{\sqrt{2}}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{1}{t}\cdot\frac{\left(\frac{1-t^{2}}{2t}\right)^{2}(2t)^{2}-(2t)^{2}+\left(\frac{1+t^{2}}{2t}\right)(2t)^{2}}{\left(\frac{1-t^{2}}{2t}\right)\left(\frac{1+t^{2}}{2t}\right)^{2}(2t)^{2}}\\ &~~~~~\times\sqrt{1+\frac{1+t^{2}}{2t}}\sqrt{\left(b-\frac{1-t^{2}}{2t}\right)\left(\frac{1-t^{2}}{2t}-a\right)}\\ &=\frac{\cos{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1+t\right)\left[3-\left(2-t\right)^{2}\right]}{t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(t^{2}+2bt-1\right)\left(1-2at-t^{2}\right)}{t}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1-t\right)\left[\left(2+t\right)^{2}-3\right]}{t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(t^{2}+2bt-1\right)\left(1-2at-t^{2}\right)}{t}}.\\ \end{align}$$
Setting $A:=\sqrt{1+a^{2}}-a\land B:=\sqrt{1+b^{2}}-b$, we have $0<B<1\land B<A$, and then
$$\begin{align} \frac{\mathcal{I}{\left(\alpha,\beta,\gamma,\phi\right)}}{\sqrt{\alpha}} &=\frac{\cos{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1+t\right)\left[3-\left(2-t\right)^{2}\right]}{t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(t^{2}+2bt-1\right)\left(1-2at-t^{2}\right)}{t}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1-t\right)\left[\left(2+t\right)^{2}-3\right]}{t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(t^{2}+2bt-1\right)\left(1-2at-t^{2}\right)}{t}}\\ &=\frac{\cos{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1+t\right)\left[3-\left(2-t\right)^{2}\right]}{t\left(1+t^{2}\right)^{2}}\\ &~~~~~\times\sqrt{\frac{\left(\sqrt{1+a^{2}}-a-t\right)\left(a+\sqrt{1+a^{2}}+t\right)\left(t-\sqrt{1+b^{2}}+b\right)\left(t+b+\sqrt{1+b^{2}}\right)}{t}}\\ &~~~~~+\frac{\sin{\left(\phi\right)}}{2}\int_{\sqrt{1+b^{2}}-b}^{\sqrt{1+a^{2}}-a}\mathrm{d}t\,\frac{\left(1-t\right)\left[\left(2+t\right)^{2}-3\right]}{t\left(1+t^{2}\right)^{2}}\\ &~~~~~\times\sqrt{\frac{\left(\sqrt{1+a^{2}}-a-t\right)\left(a+\sqrt{1+a^{2}}+t\right)\left(t-\sqrt{1+b^{2}}+b\right)\left(t+b+\sqrt{1+b^{2}}\right)}{t}}\\ &=\cos{\left(\phi\right)}\int_{B}^{A}\mathrm{d}t\,\frac{\left(1+t\right)\left[3-\left(2-t\right)^{2}\right]}{2t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(A-t\right)\left(t-B\right)\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}{t}}\\ &~~~~~+\sin{\left(\phi\right)}\int_{B}^{A}\mathrm{d}t\,\frac{\left(1-t\right)\left[\left(2+t\right)^{2}-3\right]}{2t\left(1+t^{2}\right)^{2}}\sqrt{\frac{\left(A-t\right)\left(t-B\right)\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}{t}}\\ &=\cos{\left(\phi\right)}\int_{B}^{A}\mathrm{d}t\,\frac{\left(1+t\right)\left[3-\left(2-t\right)^{2}\right]\left(A-t\right)\left(t-B\right)\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}{2t\left(1+t^{2}\right)^{2}\sqrt{\left(A-t\right)\left(t-B\right)t\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}}\\ &~~~~~+\sin{\left(\phi\right)}\int_{B}^{A}\mathrm{d}t\,\frac{\left(1-t\right)\left[\left(2+t\right)^{2}-3\right]\left(A-t\right)\left(t-B\right)\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}{2t\left(1+t^{2}\right)^{2}\sqrt{\left(A-t\right)\left(t-B\right)t\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)}}.\\ \end{align}$$
Since $P{(t)}=\left(A-t\right)\left(t-B\right)t\left(t+\frac{1}{A}\right)\left(t+\frac{1}{B}\right)$ is a polynomial of degree $5$ without multiple roots, the final two integrals above are hyper-elliptic. As such, I don't believe it's possible to find a closed form expression in terms of well-known special functions.