5. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $E$ is the midpoint of $D C$, and $F$ is the midpoint of $B B_{1}$. Then the volume of the tetrahedron $A D_{1} E F$ is $\qquad$
5. $\frac{5}{24}$.

Draw a line through $D_{1}$ parallel to $\mathrm{AF}$ intersecting the extension of $\mathrm{CD}$ at $G$. Then $G D_{1} / /$ plane $F A E$. Connect $F G, A G$, we get
$$
\begin{array}{l}
V_{A D_{1} E F} \\
=V_{D_{1}-A E F} \\
=V_{G-A E F} \\
=V_{F-A G E} .
\end{array}
$$

The base of the tetrahedron $F$ -
$A G E$ is $\triangle A G E$, and the height $F B=\frac{1}{2}$.
By $\triangle D_{1} D G \backsim$
$$
\triangle F Z A \Rightarrow \frac{G \Gamma}{A B}=\frac{D_{1} D}{F B} \Rightarrow G D=2 \Rightarrow G E=\frac{5}{2} \text {, }
$$

then $S_{\text {AATi }}=\frac{1}{2} \cdot A D \cdot G E=\frac{5}{4}$.
$$
V_{A D_{1} E F}=V_{F-A C E}=\frac{1}{3} \times \frac{5}{4} \times \frac{1}{2}=\frac{5}{24} .
$$