What are some unique integral representations of Apery's constant - $\zeta(3)$?

I've been playing around with some integrals, and I started looking at Apery's constant.

There are some integral representation I've found online, such as:

$$\zeta(3)=\frac{16}{3}\int_{0}^{1}\frac{x\log^2(x)}{1+x^2}dx$$

and:

$$\zeta(3)=\frac{32}{7}\int_{0}^{1}\frac{x\log^2(x)}{1-x^4}dx$$

and these two monsters found on Wikipedia:

$$\zeta(3)=\frac{8\pi^2}{7}\int_{1}^{\infty}\frac{x(x^4-4x^2+1)\log\log x}{(1+x^2)^4}dx$$

$$\zeta(3)=\pi\int_{0}^{\infty}\frac{\cos(2\arctan x)}{(1+x^2)(\cosh \frac{\pi}{2}x)²}dx$$

Are there any other unique integrals that have this property? I would love to find other non-trivial ways of defining $$\zeta(3)$$.

Thanks in advance for any responses.

I do not know the reason of interest but here are few more: \begin{align} \zeta(3) &= \frac{1}{7}\int_0^\pi x(\pi-x)\csc(x) \,dx \tag{1}\\ &= \frac{4}{7}\int_{0}^{\pi/2} (4x-\pi)\ln(\sin x) \, dx \tag 1 \\ &=-\frac{1}{2}\int_0^1\int_0^1\frac{\ln(xy)}{1-xy}\,dx\,dy, \tag 2\end{align}

References: 1, 2.

It should be noted that a "unique" integral is not an easy thing to define. This is not necessarily an answer but a clarification on the identities listed.

Let us start with the identity: $$I(s) = \int_0^1 \frac{\ln^{s-1}(x)}{x} f(x) \: dx = (-1)^{s-1} \Gamma(s) \sum_{n=1}^{\infty}\frac{g(n)}{n^s}$$ where $$f(x)$$ is simply a power series with coefficients $$g(k)$$ $$f(x) = \sum_{k=1}^{\infty} g(k) \: x^k$$

It can then be seen that your first "unique" integral has $$f(x) = \frac{x^2}{1+x^2}$$ in the above identity and the second integral has $$f(x) = \frac{x^2}{1-x^4}$$. These yield the sums $$\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{(2 n)^3}$$ and $$\sum _{n=1}^{\infty } \frac{1}{(4 n - 2)^3}$$, respectively.

The third integral can be derived from the derivative of the above identity: $$\frac{d}{ds}I(s) = \int_0^1 \frac{\ln^{s-1}(x)}{x} \ln(\ln(x))f(x) \: dx = I(s)\cdot[\psi^{(0)}(s)+\pi i]+(-1)^{s} \Gamma(s) \sum_{n=1}^{\infty}\ln(n)\frac{g(n)}{n^s}$$ and noting that $$\int_{1}^{\infty}\frac{x(x^4-4x^2+1)\ln (\ln(x))}{(1+x^2)^4}dx = \int_{0}^{1}\frac{x(x^4-4x^2+1)\ln(\ln(x))}{(1+x^2)^4}dx$$

The third integral from VIVID can also be derived this way as well. It is up to you if you wish to consider all of these integrals as "unique".

Here is clean one $$\int_0^1 \frac{\ln^2(1-x^2)}xdx=\zeta(3)$$