= Solution
Group the two terms containing $p$ to obtain the <mixture distribution>
$$
f(x)=p\frac1{\pi(1+x^2)}+(1-p)\frac{|x|}{(1+x^2)^2}.
$$
The first component is a <Standard Cauchy distribution>. The second is normalized because
$$
2\int_0^\infty\frac{r}{(1+r^2)^2}\,dr=1.
$$
For this symmetric component, let $R=|X|$. Its density is $2r/(1+r^2)^2$, and hence its <cumulative distribution function> and <quantile function> are
$$
P(R\leq r)=1-\frac1{1+r^2},\qquad R=\sqrt{\frac{V}{1-V}}.
$$
The independent sign is positive or negative with equal probability. For one sample, use the next three independent uniforms $U,V,H$ in the given sequence and return
$$
\boxed{X=\begin{cases}\tan\bigl(\pi(V-\tfrac12)\bigr),&U\leq p,\\\sqrt{V/(1-V)},&U>p,\ H\leq\tfrac12,\\-\sqrt{V/(1-V)},&U>p,\ H>\tfrac12.\end{cases}}
$$
The Cauchy branch follows from its <quantile function>, while the other branch follows from the magnitude law and fair sign just derived. Independence of the branch selection and component draws therefore gives exactly the displayed <mixture distribution>. Discard the zero-probability endpoint draw $V=1$ to keep the computation finite, and use a fresh block of three uniforms for the next sample. This is the <symmetric rational mixture sampler>.
One can alternatively use the root solution's <rejection sampling> with a <Standard Cauchy distribution> proposal. The density ratio is $p+\pi(1-p)|x|/(1+x^2)$, bounded by $M=p+\pi(1-p)/2$ because $2|x|\leq1+x^2$. Accept the Cauchy draw $Y$ with probability $[p+\pi(1-p)|Y|/(1+Y^2)]/M$ using a fresh independent uniform.
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