Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$ Here is a challenging one maybe some would like a go at.
Show that:
$$\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{\pi^{3}}{32}\ln2+\frac{\pi^{2}}{6}G-\frac{1}{1536}\left[\psi_{3}\left(\frac34\right)-\psi_{3}\left(\frac14\right)\right]$$
 A: Following the same idea of @Cody we have
$$I=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(1+x^2)}\ dx+\int_{0}^{1}\frac{\ln^{2}x\ln(1+x)}{1+x^2}\ dx=K+J$$

\begin{align}
K&=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(x^{2}+1)}\ dx\\
&=\frac12\int_0^1\frac{x\ln^3x}{1+x^2}\ dx-\frac12\int_0^1\frac{\ln^3x}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln^3x}{1-x}\ dx\\
&=\frac12\left(-\frac{21}{64}\zeta(4)\right)-\frac12\left(-6\beta(4)\right)+\frac12\left(-6\zeta(4)\right)\\
&=-\frac{405}{128}\zeta(4)+3\beta(4)
\end{align}

The integral $J$ is evaluated here in two methods 
$$J=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2}\ dx=\frac{\pi^3}{32}\ln2+\zeta(2)G-2\beta(4)$$
Combining the results of $K$ and $J$ we have 
$$I=-\frac{405}{128}\zeta(4)+\frac{\pi^3}{32}\ln2+\zeta(2)G+\beta(4)$$
Substituting $\beta(4)=\frac1{768}\left(\psi_3(1/4)-8\pi^4\right)$ along with $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}{6}$ we get

$$I=\frac{\pi^2}{6}G+\frac{\pi^{3}}{32}\ln2-\frac{35}{768}\pi^4+\frac{1}{768}\psi_{3}(1/4)$$

A: I am going to go ahead and post my method. It is similar to xpauls except I used digamma, which is related to the harmonic series anyway.
Break integral up:
$$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{(1-x)(x^{2}+1)}dx+\int_{1}^{\infty}\frac{\log^{2}(x)\log(1+x)}{(1-x)(x^{2}+1)}dx$$
In the right integral, make the sub $x=1/t$. This gives:
$$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{(x^{2}+1)}dx+\int_{0}^{1}\frac{x\log^{3}(x)}{(1-x)(x^{2}+1)}dx$$
The right integral:
Break up into  $$1/2\int_{0}^{1}\frac{x\log^{3}(x)}{x^{2}+1}dx-1/2\int_{0}^{1}\frac{\log^{3}(x)}{x^{2}+1}dx+1/2\int_{0}^{1}\frac{\log^{3}(x)}{1-x}dx$$
I am not going to work through each of these. But, suffice to say, they can be done without too much effort by using geometric series.  For instance, take the middle one:
$$1/2\int_{0}^{1}\log^{3}(x)\sum_{k=0}^{\infty}(-1)^{k}x^{2k}dx=3\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{4}}$$
Doing so to all three leads to series which evaluate in terms of $\zeta(4)$ and $\psi_{3}$. Summing them results in:
$$ \boxed{\displaystyle \int_{0}^{1}\frac{x\log^{3}(x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{1}{512}\left[\psi_{3}(1/4)-\psi_{3}(3/4)\right]}$$
The left integral up top is a little more difficult. At least I think so.
$$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{x^{2}+1}dx$$
Use the Taylor series for $\log(1+x)$:
$$\int_{0}^{1}\frac{\log^{2}(x)}{x^{2}+1}\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n}$$
Note the incomplete Beta function defined as:  $\displaystyle \int_{0}^{1}\frac{x^{a}}{x^{2}+1}dx=1/4\left[\psi \left(\frac{a+3}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]$.  
Diffing this twice w.r.t 'a' introduces the log-square term and gives:
$$\int_{0}^{1}\frac{x^{a+n}\log^{2}(x)}{x^{2}+1}dx=1/64\left[\psi_{2} \left(\frac{a+n+3}{4} \right)-\psi_{2} \left(\frac{a+n+1}{4} \right) \right]$$.
Thus, letting $a=0$, $$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{x^{2}+1}dx=1/64\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\left[\psi_{2}\left(\frac{n+3}{4}\right)-\psi_{2}\left(\frac{n+1}{4}\right)\right]$$
$$=\boxed{\displaystyle \frac{\pi^{2}}{6}G+\frac{\pi^{3}}{32}\log(2)-\frac{1}{768}\left[\psi_{3}\left(1/4\right)-\psi_{3}\left(3/4\right)\right]}$$
This series result, when combined with the other boxed result, gives the solution to the original integral. 
The only minor issue I have is evaluating this tetragamma series. As I said, The Flajolet-Salvy residue method may work, but finding the correct kernel is the first important task.  Since it alternates, I would assume something with $\pi \csc(\pi z)$
Of course, one could just say the heck with it and use this as a lemma. But, I would like to evaluate it though. 
