As said by John Rennie, it has to do with the shadows' fuzzyness. However, that alone doesn't quite explain it.

Let's do this with actual fuzzyness:

I've simulated shadow by blurring each shape and multiplying the brightness values^{1}. Here's the GIMP file, so you can see how exactly and move the shapes around yourself.

I don't think you'd say there's any bending going on, at least to me the book's edge still looks perfectly straight.

So what's happening in your experiment, then?

*Nonlinear response* is the answer. In particular in your video, the directly-sunlit wall is overexposed, i.e. regardless of the "exact brightness", the pixel-value is pure white. For dark shades, the camera's noise surpression clips the values to black. We can simulate this for the above picture:

Now that looks a lot like your video, doesn't it?

With bare eyes, you'll *normally* not notice this, because our eyes are kind of trained to compensate for the effect, which is why nothing looks bent in the unprocessed picture. This only fails at rather extreme light conditions: probably, most of your room is dark, with a rather narrow beam of light making for a very large luminocity range. Then, the eyes also behave too non-linear, and the brain cannot reconstruct how the shapes would have looked without the fuzzyness anymore.

Actually of course, the brightness topography is always the same, as seen by quantising the colour palette:

^{1}To simulate shadows properly, you need to use convolution of the whole aperture, with the sun's shape as a kernel. As Ilmari Karonen remarks, this does make a relevant difference: the convolution of a product of two sharp shadows $A$ and $B$ with blurring kernel $K$ is

$$\begin{aligned}
C(\mathbf{x}) =& \int_{\mathbb{R}^2}\!\mathrm{d}{\mathbf{x'}}\:
\Bigl(
A(\mathbf{x} - \mathbf{x}') \cdot B(\mathbf{x} - \mathbf{x'})
\Bigr) \cdot K(\mathbf{x}')
\\ =& \mathrm{IFT}\left(\backslash{\mathbf{k}} \to
\mathrm{FT}\Bigl(\backslash\mathbf{x}' \to
A(\mathbf{x}') \cdot B(\mathbf{x}')
\Bigr)(\mathbf{k})
\cdot \tilde{K}(\mathbf{k})
\right)(\mathbf{x})
\end{aligned}
$$

whereas seperate blurring yields

$$\begin{aligned}
D(\mathbf{x}) =& \left( \int_{\mathbb{R}^2}\!\mathrm{d}{\mathbf{x'}}\:
A(\mathbf{x} - \mathbf{x}')
\cdot K(\mathbf{x}') \right)
\cdot \int_{\mathbb{R}^2}\!\mathrm{d}{\mathbf{x'}}\:
B(\mathbf{x} - \mathbf{x'})
\cdot K(\mathbf{x}')
\\ =& \mathrm{IFT}\left(\backslash{\mathbf{k}} \to
\tilde{A}(\mathbf{k}) \cdot \tilde{K}(\mathbf{k})
\right)(\mathbf{x})
\cdot \mathrm{IFT}\left(\backslash{\mathbf{k}} \to
\tilde{B}(\mathbf{k}) \cdot \tilde{K}(\mathbf{k})
\right)(\mathbf{x}).
\end{aligned}
$$

If we carry this out for a narrow slit of width $w$ between two shadows (almost a Dirac peak), the product's Fourier transform can be approximated by a constant proportional to $w$, while the $\mathrm{FT}$ of each shadow remains $\mathrm{sinc}$-shaped, so if we take the Taylor-series for the narrow overlap it shows the brightness will only decay as $\sqrt{w}$, i.e. stay brighter at close distances, which of course surpresses the bulging.

And indeed, if we properly blur both shadows *together*, even without any nonlinearity, we get much more of a "bridging-effect":

But that still looks nowhere as "bulgy" as what's seen in your video.

To complement Floris's answer, here's a quick animation showing how the shadow changes with the size of the light source.
In this animation, I've forced the intensity of the light to vary inversely with the surface area, so the total power output is constant ($P \approx 62.83 \, \mathrm{W}$).
This is why the object (Suzanne) doesn't appear to get any brighter or darker, but the shadow sharpness does change:

In this scene, the spherical lamp is the only light source, except for the dim ambient lighting.
This makes the shadows very easily visible.
In a real-world scenario with other light sources (windows, for example), the effect would be less pronounced because the shadows would be more washed out.

The following animation shows the scenario Floris described, with a rotating long object:

## Best Answer

In order to cast a discernible shadow, an object has to block the entire sun (or else some light rays from the unblocked parts will wash out whatever shadow was created from the blocked parts). Since birds tend to be relatively thin, irregular-shaped objects, whereas the sun is a circle with a diameter of about half a degree, it doesn't seem particularly likely that even a bird with sufficient wingspan would be able to block the entire sun at once.