Suppose we have a finite continuous solenoid of radius $\rho$ and axial length $L$ where $I$ is the current carried by the wire and $n$ is the number of turns of wire per unit length. For convenience, we place the center of the solenoid at the origin and orient its axis to align with the $z$-axis.
The magnetic field $\vec{B}$ at the point $\vec{r}_{0}$ due to the solenoid is given by the iterated integral
$$\vec{B}{\left(\vec{r}_{0}\right)}=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}},$$
where $\vec{r}$ is the position vector for a point with cylindrical coordinates $\left(\rho,\phi,z\right)$ on the surface of the solenoid:
$$\vec{r}=\rho\cos{\left(\phi\right)}\,\hat{x}+\rho\sin{\left(\phi\right)}\,\hat{y}+z\hat{z}.$$
Similarly, the position vector $\vec{r}_{0}$ of a point with cylindrical coordinates $\left(\rho_{0},\phi_{0},z_{0}\right)$ can be written as
$$\vec{r}_{0}=\rho_{0}\cos{\left(\phi_{0}\right)}\,\hat{x}+\rho_{0}\sin{\left(\phi_{0}\right)}\,\hat{y}+z_{0}\hat{z}.$$
A closed-form expression for $\vec{B}{\left(\vec{r}_{0}\right)}$ in terms of complete elliptic integrals is derived below.
For reference, the complete elliptic integrals are defined here as follows:
$$K{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)},$$
$$E{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}};~~~\small{\kappa\in\left[-1,1\right]},$$
$$\Pi{\left(\nu,\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)\land\nu\in\left(-\infty,1\right)}.$$
As you correctly stated above, it can be shown that the distance between the two points in cylindrical coordinates is given by
$$\|\vec{r}_{0}-\vec{r}\|=\sqrt{\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}}.$$
The expression you gave for the vector quantity $\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)$, however, is wrong, so let's go over its calculation in detail.
Recall the cylindrical unit vectors $\hat{\rho}$ and $\hat{\phi}$ may be given in terms of Cartesian unit vectors by
$$\hat{\rho}=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y},$$
$$\hat{\phi}=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}.$$
Then, $\vec{r}=\rho\hat{\rho}+z\hat{z}$ and $\frac{\partial\vec{r}}{\partial\phi}=\rho\hat{\phi}$.
Similarly, $\vec{r}_{0}=\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}$, where the unit vectors $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$ can be expressed as
$$\hat{\rho}_{0}=\cos{\left(\phi_{0}\right)}\,\hat{x}+\sin{\left(\phi_{0}\right)}\,\hat{y},$$
$$\hat{\phi}_{0}=-\sin{\left(\phi_{0}\right)}\,\hat{x}+\cos{\left(\phi_{0}\right)}\,\hat{y}.$$
The above pair of equations can be inverted to give $\hat{x}$ and $\hat{y}$ in terms of $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$:
$$\hat{x}=\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0},$$
$$\hat{y}=\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}.$$
We then find
$$\begin{align}
\hat{\rho}
&=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y}\\
&=\cos{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\sin{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\
&=\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0},\\
\end{align}$$
$$\begin{align}
\hat{\phi}
&=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}\\
&=-\sin{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\cos{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\
&=\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}.\\
\end{align}$$
and thus,
$$\begin{align}
\hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right)
&=\hat{\phi}\times\left(\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}-\rho\hat{\rho}-z\hat{z}\right)\\
&=\hat{\phi}\times\left[\left(z_{0}-z\right)\hat{z}-\rho\hat{\rho}+\rho_{0}\hat{\rho}_{0}\right]\\
&=\left(z_{0}-z\right)\hat{\phi}\times\hat{z}-\rho\hat{\phi}\times\hat{\rho}+\rho_{0}\hat{\phi}\times\hat{\rho}_{0}\\
&=\left(z_{0}-z\right)\hat{\rho}+\rho\hat{z}-\rho_{0}\hat{\rho}_{0}\times\hat{\phi}\\
&=\left(z_{0}-z\right)\left[\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]+\rho\hat{z}\\
&~~~~~-\rho_{0}\hat{\rho}_{0}\times\left[\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]\\
&=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\
&~~~~~-\rho_{0}\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\rho}_{0}-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\phi}_{0}\\
&=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\
&~~~~~-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{z}\\
&=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\\
&~~~~~+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}.\\
\end{align}$$
We turn now to the calculation of $\vec{B}{\left(\vec{r}_{0}\right)}$ for the solenoid, and we start by simplifying the angular integral.
Given $\vec{r}_{0}\in\mathbb{R}^{3}$ such that $\neg\left[\rho_{0}=\rho\land\left(|z_{0}|\le\frac{L}{2}\right)\right]$, we have
$$\begin{align}
\vec{B}{\left(\vec{r}_{0}\right)}
&=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho\hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]},\\
\end{align}$$
where in the second-to-last line above we used the fact that the $\hat{\phi}_{0}$ component of the angular integral is identically zero because of its $2\pi$-periodic antiderivative, and in the last line above we've used the following lemma: for any $a\in\mathbb{R}$ and any continuous periodic function $f:\mathbb{R}\rightarrow\mathbb{R}$ with period $p\in\mathbb{R}_{>0}$, it can be shown that
$$\int_{0}^{p}\mathrm{d}x\,f{\left(x+a\right)}=\int_{0}^{p}\mathrm{d}x\,f{\left(x\right)}.$$
Then,
$$\begin{align}
\vec{B}{\left(\vec{r}_{0}\right)}
&=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{\left[z\mapsto z_{0}-z\right]}\\
&=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{symmetry}\\
&=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}},\\
\end{align}$$
where in the last line above we've set $B_{0}:=\frac{\mu_{0}nI}{4\pi}$ and $z_{\pm}:=z_{0}\pm\frac{L}{2}$.
Now, we could proceed to evaluate the angular integral in terms of elliptic integrals as others have already pointed out, but that would leave us with a very difficult non-elementary axial integral. It turns out, however, that changing the order of integration makes the integral much easier.
Consider the following derivative:
$$\frac{d}{dz}\left[\frac{z\vec{c}-a\vec{b}}{a\sqrt{z^{2}+a}}\right]=\frac{z\vec{b}+\vec{c}}{\left(z^{2}+a\right)^{3/2}};~~~\small{a\in\mathbb{R}_{>0}\land\vec{b}\in\mathbb{R}^{3}\land\vec{c}\in\mathbb{R}^{3}}.$$
Assuming $0<\rho_{0}\neq\rho$, we have $0<\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}$. Also note that $0<\frac{4\rho_{0}\rho}{z^{2}+\left(\rho_{0}+\rho\right)^{2}}<1$ for all real $z$.
Setting $\nu:=\frac{4\rho_{0}\rho}{\left(\rho_{0}+\rho\right)^{2}}\land k_{\pm}:=\sqrt{\frac{4\rho_{0}\rho}{z_{\pm}^{2}+\left(\rho_{0}+\rho\right)^{2}}}$, we then have
$$\begin{align}
\vec{B}{\left(\vec{r}_{0}\right)}
&=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\
&=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{d}{dz}\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\
&~~~~~+\frac{2z\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\
&=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\,\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}+\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\
&~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\
&~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\
&=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}-\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\
&~~~~~+\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\
&~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\
&~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\bigg{]};~~~\small{\left[\phi=2\varphi\right]}\\
&=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\
&~~~~~+\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{+}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{-}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\bigg{]}\\
&=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\\
&~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\bigg{]},\\
\end{align}$$
and finally,
$$\begin{align}
\vec{B}{\left(\vec{r}_{0}\right)}
&=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\bigg{]};~~~\small{\left[\varphi\mapsto\frac{\pi}{2}-\varphi\right]}\\
&=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~+\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&=\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{+}^{2}}{2}\right)}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~-\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{-}^{2}}{2}\right)}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~+\hat{z}\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&~~~~~-\hat{z}\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\
&=\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{+}\right)}-\left(1-\frac{k_{+}^{2}}{2}\right)K{\left(k_{+}\right)}\bigg{]}\hat{\rho}_{0}\\
&~~~~~-\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{-}\right)}-\left(1-\frac{k_{-}^{2}}{2}\right)K{\left(k_{-}\right)}\bigg{]}\hat{\rho}_{0}\\
&~~~~~+\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{+}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{+}\right)}\bigg{]}\hat{z}\\
&~~~~~-\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{-}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{-}\right)}\bigg{]}\hat{z}.\blacksquare\\
\end{align}$$
Best Answer
HINT
One stupid way $$ \frac{1}{2\pi} \int_0^{2\pi} f(\phi)d\phi = \frac{1}{2\pi} \int_0^1 d\rho \int_0^{2\pi} f(\phi)d\phi $$