Here is my attempt:
Let $R>1>r$ and $C$ be the closed curve in $\mathbb{C}$ consists of the following pieces: $$C_1=\{Re^{it}: t\in(0,\pi)\},\quad C_2=[r,R],\quad C_3=\{re^{it}: t\in(0,\pi)\},\quad C_4=[-R,r]$$ all curves are oriented counterclockwise. It can be seen that $C$ is the boundary of the upper half of the annulus centred at $0$ with inner radius $r$ and outer radius $R$. Let $f(z)=\frac{(\text{Log } z)^3}{1+z^2},\quad Arg(z)\in\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$. Then $\int_Cf(z)d z=\sum_{i=1}^4\int_{C_i}f(z)d z$. Note that $f$ has a singularity $z=i$ inside $C$, thus by Cauchy's integral formula $$\int_Cf(z)dz=2\pi i\cdot\frac{(\text{Log } i)^3}{i+i}=\pi\left(\ln|i|+\frac{\pi}{2}i\right)^3=-\frac{\pi^4}{8}i$$ Now consider $\int_{C_i}f(z)d z$: $$\begin{aligned}&\left|\int_{C_1}f(z)d z\right|\leq\int_0^\pi\frac{(|\ln R|+|it|)^3}{|1-R^2e^{2it}|}|Re^{it}|dt\leq\int_0^\pi\frac{R(\ln R+t)^3}{R^2-1}|dt\leq \frac{\pi R(\ln R+\pi)^3}{R^2-1} \end{aligned}$$ as $RHS\to 0$ as $R\to+\infty$, we have $\lim_{R\to\infty}\int_{C_1}f(z)d z=0$.
$$\int_{C_2}f(z)d z=\int_r^R\frac{(\ln x)^3}{1+x^2}d x\quad\Rightarrow\quad\lim_{r\to 0^+, R\to+\infty}\int_{C_2}f(z)d z=\int_0^\infty\frac{(\ln x)^3}{1+x^2}d x$$
$$\begin{aligned}&\left|\int_{C_3}f(z)d z\right|\leq\int_0^\pi\frac{(|\ln r|+|it|)^3}{|1+r^2e^{2it}|}|re^{it}|d t\leq\int_0^\pi\frac{r(\ln r+t)^3}{1-r^2}|d t\leq \frac{\pi r(\ln r+\pi)^3}{1-r^2} \end{aligned}$$ as $RHS\to 0$ as $r\to0^+$, we have $\lim_{r\to 0^+}\int_{C_3}f(z)d z=0$
$$\begin{aligned}&\int_{C_4}f(z)d z=\int_{[-R,-r]}\frac{(\text{Log } z)^3}{1+z^2}d z=\int_{[-R,-r]}\frac{(\ln|z| +\pi i)^3}{1+z^2}d z=\int_r^R\frac{(\ln x +\pi i)^3}{1+x^2}d x\\ =&\int_r^R\frac{(\ln x)^3}{1+x^2}dx+3\pi i\int_r^R\frac{(\ln x)^2}{1+x^2}d x-3\pi^2\int_r^R\frac{\ln x}{1+x^2}d x-\pi^3 i\int_r^R\frac{1}{1+x^2}d x\end{aligned}$$
At this point I have to evaluate $\int_0^\infty\frac{(\ln x)^2}{1+x^2}d x$, which can be done by complex analysis again. However, this method is way too long and from my point of view, not the most efficient. Is there a shorter way to do this?
By the way, the answer is 0.
Using the contour I used here, we have that $$ \int_0^{\infty}\frac{\log^3(x)}{1+x^2}dx = \int_{\Gamma}f(z)dz + \int_{-\infty}^{-\epsilon}f(z)dz + \int_{\gamma}f(z)dz + \int_{\epsilon}^{\infty}f(z)dz \tag{1} $$ where $f(z) = \frac{\log^3(z)}{z^2 + 1}$, $\Gamma$ is large semi circle, and $\gamma$ is small semi circle. As with the linked post, we will have the branch cut from $(-\infty, 0)$. By the estimation lemma, the first and third integrals of $(1)$ are zero. Now $$ \log^3(z) = \log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert + i(3\arg(z)\log^2\lvert z\rvert - \arg^3(z)) $$ Additionally, we are taking the real Cauchy principal value of the integral. With that in mind and by the residue theorem, we have \begin{align} \int_{-\infty}^0\frac{\log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert}{z^2 + 1} dz + \int_0^{\infty}\frac{\log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert}{z^2 + 1}dz &= \text{Re}\left(2\pi i\sum\text{Res}\right)\\ 2\int_0^{\infty}\frac{\log^3(x)}{x^2 + 1}dz - 3\pi^2\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz & = 0\\ \int_0^{\infty}\frac{\log^3(x)}{x^2 + 1}dz&= 0\tag{2} \end{align} where $(2)$ occurs since $\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2 + 1}dz = 0$ and the residue was purely imaginary so the real principal value is zero. Using the same contour for $\frac{\log(z)}{z^2+1}$, we obtain $$ \int_0^{\infty}\frac{\log\lvert z\rvert}{z^2 + 1}dz = 2\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz + \int_0^{\infty}\frac{i\pi}{z^2+1}dz = \frac{i\pi^2}{2}\tag{3} $$ Again, we were seeking the real Cauchy principal value so $(3)$ becomes $$ \int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz = 0 $$