Update: I don't use a trig-substitution because I don't see which one could work. To integrate $p^{2}/\sqrt{p^{2}+3}$ I applied a technique derived from this more general one.
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$
(Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.)
EDIT: simplified the exposition. The integral is separable, as already noted by J. M. in a comment. Assuming the limits of integration are as follows, we have
$\begin{eqnarray*} I &:&=\int_{\theta =0}^{2\pi }\int_{\phi =0}^{\pi }\int_{p=0}^{2}\frac{p^{2}\sin (\phi )}{\sqrt{p^{2}+3}}\;\mathrm{d}p\;% \mathrm{d}\phi \;\mathrm{d}\theta \\&=&\left( \int_{0}^{2\pi }\mathrm{d}\theta \right) \left( \int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{% p^{2}+3}}\;\mathrm{d}p\right) \\ &=&2\pi \left( \int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{% p^{2}+3}}\;\mathrm{d}p\right). \end{eqnarray*}$
The antiderivative of $\sin (\phi )$ is $-\cos \left( \phi \right) $; hence
$I =4\pi \int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p.$
For the evaluation of this last integral we are going to apply the following property:
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$\int \frac{P(x)}{\sqrt{x^{2}+c}}\;\mathrm{d}x=Q(x)\sqrt{x^{2}+c}+\int \frac{C}{\sqrt{% x^{2}+c}}\;\mathrm{d}x.$
In the present case, we have $P(x)=x^{2}$ and $Q(x)$ is of the form $Q(x)=Ax+B$
$\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\left( Ax+B\right) \sqrt{% x^{2}+c}+\int \frac{C}{\sqrt{x^{2}+c}}\;\mathrm{d}x.$
Differentiating both sides, we get
$\begin{eqnarray*} \frac{x^{2}}{\sqrt{x^{2}+c}} &=&A\sqrt{x^{2}+c}+\frac{x\left( Ax+B\right) }{% \sqrt{x^{2}+c}}+\frac{C}{\sqrt{x^{2}+c}} \\ &=&\frac{1}{\sqrt{x^{2}+c}}\left[ A\left( x^{2}+c\right) +x\left( Ax+B\right) +C\right], \end{eqnarray*}$
and so
$x^{2} =A\left( x^{2}+c\right) +x\left( Ax+B\right) +C=2Ax^{2}+Bx+Ac+C.$
Comparing coefficients, we get $A=1/2,\;B=0,\;C=-c/2$. Hence
$\begin{eqnarray*} \int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x &=&\frac{1}{2}x\sqrt{x^{2}+c}-% \frac{c}{2}\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x \\ &=&\frac{1}{2}x\sqrt{x^{2}+c}-\frac{c}{2}\ln \left( x+\sqrt{x^{2}+c}\right) \end{eqnarray*}$
because
$\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\ln \left( x+\sqrt{x^{2}+c}% \right)+\text{ constant}.$ Thus
$\int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p=\sqrt{7}-\frac{3}{2}% \ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right).$
And finally
$I=4\pi \left( \sqrt{7}-\frac{3}{2}\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right) \right) .$