Solving higher order logarithms integrals without the beta function Prove the following

$$\int^1_0 \frac{\log(x)^3}{\sqrt{x(1-x)}}dx= -\pi (12 \zeta(3) +
 \log^3(4) + \pi^2 \log(4))$$

The basic approach is differentiating the beta function three times which I am trying to avoid here.A contour method would be nice.
Question 1
Hence how to solve the integral without having to differentiate the beta function ? 
My question can be generalized to any problem involving higher order logarithms.
Question 2
Is there a general formula for
$$\frac{\partial^n \partial^m}{\partial x^n \partial y^m} B(x,y)$$
 A: Ok, let's see...
The integral in question reads

$$
I=\int_0^1dx\frac{\log^3(x)}{\sqrt{x}\sqrt{1-x}}
$$

The basic idea is to exploit the Fourier expansion of $\log(\sin(x))$ so first of all we have to transform in an appropriate form. Namely set $x=y^2=\sin^2(\theta)$ to get
$$
I=2^4\int_0^1\frac{\log^3(y)}{\sqrt{1-y^2}}dy=16\int_0^{\pi/2}d\theta\log^3(\sin{\theta})
$$
we now can make use of the aformentioned Fourier expansion, yielding

$$
-\frac{1}{16}I=\color{red}{I_1}+3\color{blue}{I_2}+3\color{green}{I_3}+\color{orange}{I_4}\quad (\star)
$$

with $$\color{red}{I_1}=\log^3(2)\int_0^{\pi/2}d\theta=\color{red}{\frac{\pi}{2}\log^3(2)}\\
\color{blue}{I_2}=\log^2(2)\sum_{n\geq1}\frac{1}{n}\int_0^{\pi/2}\cos(2n\theta)d\theta=\log^2(2)\sum_{n\geq1}\frac{\sin(\pi n)}{n^2}=\color{blue}{0}$$
the remaining terms need a bit more work, we have (we use the orthogonatity relations for the trigonometric functions)
$$
\color{green}{I_3}=\log(2)\sum_{n,m\geq1}\frac{1}{nm}\int_{0}^{\pi/2}\cos(2n\theta)\cos(2m\theta)d\theta=\\\frac{\log(2)}{4}\sum_{n,m\geq1}\frac{1}{mn}\left(\frac{\sin(\pi(m-n))}{m-n}+\frac{\sin(\pi(m+n))}{m+n}\right)
$$
since $m,n\in \mathbb{N}^+$ the $\sin(\pi (m\mp n))$ terms can only contribute if their argument is zero, where we can use the limit $\sin(ax)/x\sim_0a$. This is only possible if $m=n$ in the first term and impossible in the second one ($m+n\neq0$ !). This means
$$
\color{green}{I_3}=\frac{\pi\log(2)}{4}\sum_{n\geq 1}\frac{1}{n^2}=\color{green   }{\frac{\pi^3\log(2)}{24}}
$$
for the last remaining sum we obtain
$$
\color{orange}{I_4}=\sum_{n,m,l\geq1}\frac{1}{nml}\int_{0}^{\pi/2}\cos(2n\theta)\cos(2m\theta)\cos(2l\theta)d\theta=\\
\sum_{n,m,l\geq1}\frac{1}{8nml}\left(\frac{3\sin(\pi(l-m-n))}{l-m-n}+\frac{\sin(\pi(l+m+n))}{l+m+n}\right)
$$
where we used the symmetry of the first term under cyclic relabelings of $(l,m,n)$ .Now the considerations concerning $\color{green}{I_2}$ also apply here and we find
$$
\color{orange}{I_4}=\frac{3\pi}{8}\sum_{n,m \geq1}\frac{1}{nm(m+l)}=\frac{3\pi}{8}S
$$
now following the steps in @r9m nice answer we can finally state that
$$
\color{orange}{I_4}=\frac{3\pi}{8}2\zeta(3)=\color{orange}{\frac{3\pi}{4}\zeta(3)}
$$
Putting everything we have into $(\star)$ we obtain

$$
I=-16\left(\color{red}{\frac{\pi}{2}\log^3(2)}+3\cdot\color{blue}{0}+3\color{green}{\frac{\pi^3\log(2)}{24}}+\color{orange}{\frac{3\pi}{4}\zeta(3)}\right)=\\
-\pi(8\log^3(2)+2\pi^2\log(2)+12\zeta(3))
$$

as announced

For an alternative evaluation of the sum contained in $\color{orange}{I_4}$ we can use $\int_{0}^{\infty}e^{-(m+n)x}=\frac{1}{m+n}$ which yields
$$
S=\int_0^{\infty}\log^2(1-e^{-x})dx\underbrace{=}_{e^{-x}=z}\int_0^{1}\frac{\log^2(1-z)}{z}dz \quad(\star\star)
$$
Now, using integration by parts (alternativly we could recognize the connection to the generating function of the harmonic number which gives us another nice oppurtunity to evaluate this series) we find that
$$
S=-2\int_0^1dz\frac{\log(1-z)\log(z)}{1-z}
$$
another integration by parts yields, using the integral representation of the polylogarithm 
$$
S=2\int_0^1dz\frac{\text{Li}_2(1-z)}{1-z}=-2\text{Li}_3(1-z)|_0^1=2\text{Li}_3(1)=2\zeta(3)
$$

Recognized too late, since wasted: Embarking from $(\star\star)$, setting $1-z=q$ and expanding the resulting geometric series, gives a near trivial evalution of $S$ which might be generalized easily. This enables us, in the end, to evaluate the whole family of sums 
$$
S_r=\sum_{m_1,..m_r\geq1}\frac{1}{m_1...m_r}\frac{1}{m_1+...+m_r}
$$
and as a consequence the integrals
$$
I_{r+1}=\int_0^1dx\frac{\log^{r+1}(x)}{\sqrt{x}\sqrt{1-x}}
$$
by the technique shown above
