<s>[INST] Derivation of the pdf of RV $X (\left|X \right| + 1)$ How to derive the probability density function $p_Y(y)$ in case, when $Y = X (\left|X \right| + 1)$, where $X  \sim \mathrm{Uniform}(-1, 1)$.
I know that if $X < 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(1- \sqrt{1 - 4y} \right)$ and also if $X \geq 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(\sqrt{1 + 4y}  - 1 \right)$. But I do not know how to find CDF in this case. How correctly to combine these conditions into one?
Thank you very much!
 [/INST] Your results so far are correct and you are almost there. 
Note that the density of $X$ is symmetric with respect to zero as in $\Pr\{ X < 0 \} = \Pr\{ X > 0 \}  = \frac12$. Meanwhile, $X < 0$ actually means $-1 < X < 0$ by definition, and similarly $X > 0$ implicitly means $0 < X < 1$.
Also note that $Y$ has the same sign as $X$. That is, $Y < 0$ whenever $X < 0$ and when $X$ is positive then so is $Y$.
Last thing to note is that the density of $Y$ is also symmetric with respect to zero. This is due to $|X|+1$ having the range of $[1,2]$ and the product with the symmetric $X$ just produces a copy on the negative side. Thus we already have $\Pr\{ Y < 0 \} = \Pr\{ Y > 0 \}  = \frac12$ before doing any calculation.

On the negative $X$ side, we have
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X < 0 \bigr\} &= \Pr\Bigl\{ -1 < X < 0 ~~\& ~~X < \frac{1}{2} \left(1- \sqrt{1 - 4y} \right)\Bigr \} \\
& = \frac{1}{2} \left(1- \sqrt{1 - 4y} \right) - (-1)
\end{align}
where you already knew that $r_m \equiv \frac12 \left(1- \sqrt{1 - 4y} \right)$ is smaller than zero, hence we do the $r_m -(-1)$ to take the length from the left end point (negative one) to $r_m$ as the probability for the uniform distribution.
Putting $\Pr\{ X < 0 \}$ back, unconditionally we have on the negative $Y$ part (since $Y$ follows the sign of $X$)
$$F_Y(y) = \frac12 \cdot \frac12 \left( 3 - \sqrt{1 - 4y} \right) \qquad \text{for}~~ y < 0$$
On the positive side, the conditional CDF (ignoring the negative side) is
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} &= \Pr\Bigl\{ 0 < X < 1 ~~\& ~~X < \frac12 \left(-1 + \sqrt{1 + 4y} \right) \Bigr \} \\
& = \frac12 \left(-1 + \sqrt{1 + 4y} \right) - 0
\end{align}
where we take the length between the left end point (now zero) and $r_p \equiv \frac12 \left(-1 + \sqrt{1 + 4y} \right)$ which you already knew is between $0$ and $1$.
Now, unconditionally we are no longer ignoring the negative side and have to put things on top of it. Namely, we already have accumulated the $1/2$ probability mass from $y<0$.
\begin{align} 
F_Y(y) &= \Pr\bigl\{ Y < \frac12 \bigr\} + \Pr\bigl\{ X > 0 \bigr\} \cdot \Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} \\
&= \frac12 + \frac12 \cdot \frac12 \left(-1 + \sqrt{1 + 4y} \right) \\
& = \frac14 \left( 1 + \sqrt{1 + 4y} \right)  \qquad \text{for}~~ y > 0
\end{align}

As mentioned before, the density of $Y$ is symmetric (even) while the cumulative function is rotationally symmetric (odd) with respect to $(0,\frac12)$. This can be emphasized by using the sign function $Sgn(\cdot)$, which is odd, and the absolute value that is even.
\begin{align}
f_Y(y) &\equiv \frac{d F_Y(y) }{ dy } = \frac1{2 \sqrt{ 1 + 4 |y| } } & &\text{for} ~~ -2 < y < 2  \\
F_Y(y) &= \frac12 - \frac{ Sgn(y) }4 \Bigl( 1 - \sqrt{1 + 4 |y|} \Bigr)  & &\text{for} ~~ -2 < y < 2
\end{align}
The density written in this form is obviously an even function. As for the CDF, note that the sign fucntion (odd) multiplying the big parenthesis (even) is odd.
</s>