The animation is unfortunately misleading. The speed of light is constant and all photons, of any energy, travel at the same speed. Higher energy photons have smaller wavelengths (or, equivalently, higher frequency) but not a different speed.
Unfortunately, this is difficult to illustrate clearly. The reason the illustration shows the higher energy photons as faster is because of the naturalness of equating speed with kinetic energy of an object. It "makes sense" to most people that a more energetic particle would move faster, even if this isn't an accurate description of the phenomenon.
The formula you want is called Planck's Law. Copying Wikipedia:
The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it
gives off as radiation of different frequencies. It is measured in
terms of the power emitted per unit area of the body, per unit solid
angle that the radiation is measured over, per unit frequency.
$$ B_\nu(\nu, T) = \frac{ 2 h \nu^3}{c^2} \frac{1}{e^\frac{h\nu}{k_\mathrm{B}T} - 1} $$
Now to work out the total power emitted per unit area per solid angle by our lightbulb in the X-ray part of the EM spectrum we can integrate this to infinity:
$$P_{\mathrm{X-ray}} = \int_{\nu_{min}}^{\infty} \mathrm{B}_{\nu}d\nu,
$$
where $\nu_{min}$ is where we (somewhat arbitrarily) choose the lowest frequency photon that we would call an X-ray photon. Let's say that a photon with a 10 nm wavelength is our limit. Let's also say that 100W bulb has a surface temperature of 3,700 K, the melting temperature of tungsten. This is a very generous upper bound - it seems like a typical number might be 2,500 K.
We can simplify this to:
$$
P_{\mathrm{X-ray}} = 2\frac{k^4T^4}{h^3c^2} \sum_{n=1}^{\infty} \int_{x_{min}}^{\infty}x^3e^{-nx}dx,
$$
where $x = \frac{h\nu}{kT}$. wythagoras points out we can express this in terms of the incomplete gamma function, to get
$$
2\frac{k^4T^4}{h^3c^2}\sum_{n=1}^{\infty}\frac{1}{n^4} \Gamma(4, n\cdot x)
$$
Plugging in some numbers reveals that the n = 1 term dominates the other terms, so we can drop higher n terms, resulting in
$$
P \approx 10^{-154} \ \mathrm{Wm^{-2}}.
$$
This is tiny. Over the course of the lifetime of the universe you can expect on average no X-Ray photons to be emitted by the filament.
More exact treatments might get you more exact numbers (we've ignored the surface area of the filament and the solid angle factor for instance), but the order of magnitude is very telling - there are no X-ray photons emitted by a standard light bulb.
Best Answer
Yes, X-ray, UV, and even radio waves are made of photons.
The difference is the Energy (or equivalently, the wavelength). See the picture of the Electromagnetic spectrum . The different nomination comes from the time of the discovery. Your eyes can see the visible part. the radio waves can be observed with antennas, etc. The only difference is the way we observe it. But they are all the same (photons).
Here is a scale of the different wavelength for electromagnetic waves:
The Energy reads $E=mc^2$ for massive particles, which a photon is not. The good relation is: $$ E=pc = \hbar c k = h \nu $$ You have to use the complete Einstein relation: $$ E = \sqrt{p^2c^2 + m^2c^4}$$
And the energy is always finite.