A deceivingly small integral

We can use beta functions another way, noting that the integral is equal to $f''''(0)$ , where $f(a) \; = \; \int_0^\infty \frac{x^a}{(1+x^2)^3}\,dx \; =\; \tfrac12B\big(\tfrac{a+1}{2},\tfrac{5-a}{2}\big) \; = \; \tfrac14\Gamma\big(\tfrac{a+1}{2}\big)\Gamma\big(\tfrac{5-a}{2}\big)$ obtaining, after much simplification that the integral is $\tfrac{3}{16}\pi^3 + \tfrac{15}{256}\pi^5$ which makes the answer $\boxed{298}$ .

First, we consider the substitution $x=\tan \theta$ so that $\frac{dx}{d\theta}=\sec^2\theta,$ and the integral becomes $I=\int^{\infty}_0\frac{\ln^4x}{(1+x^2)^3}\ dx=\int^{\frac{\pi}{2}}_0\ln^4(\tan \theta)\cos^4\theta\ d\theta.$ Then, since $\begin{aligned}\ln^4(\tan \theta)&=&(\ln(\sin\theta)-\ln(\cos\theta))^4\\&=&\ln^4(\sin\theta)+4\ln^3(\sin\theta)\ln(\cos\theta)+6\ln^2(\cos\theta)\ln^2(\sin\theta)+4\ln(\sin\theta)\ln^3(\cos\theta)+\ln^4(\cos\theta),\end{aligned}$ we have $\begin{aligned}I&=&\int^{\frac{\pi}{2}}_0\left(\ln^4(\sin\theta)+4\ln^3(\sin\theta)\ln(\cos\theta)+6\ln^2(\cos\theta)\ln^2(\sin\theta)+4\ln(\sin\theta)\ln^3(\cos\theta)+\ln^4(\cos\theta)\right)\cos^4\theta\ d\theta\\&=&\int^{\frac{\pi}{2}}_0\ln^4(\sin\theta)\cos^4\theta\ d\theta+\int^{\frac{\pi}{2}}_04\ln^3(\sin\theta)\ln(\cos\theta)\cos^4\theta\ d\theta+\int^{\frac{\pi}{2}}_06\ln^2(\cos\theta)\ln^2(\sin\theta)\cos^4\theta\ d\theta\\&&+\int^{\frac{\pi}{2}}_04\ln(\sin\theta)\ln^3(\cos\theta)\cos^4\theta\ d\theta+\int^{\frac{\pi}{2}}_0\ln^4(\cos\theta)\cos^4\theta\ d\theta.\end{aligned}$ Now, we look at each individual integrals: it seems that we can use Beta function to calculate it. However, this is a long and arduous process, so I will not show all the intermediate calculations. Anyways, by applying Beta function, the integral becomes $\begin{aligned}I&=&\frac{1}{32}\left[\frac{\partial^4}{\partial x^4}B(x,y)+4\times\frac{\partial^4}{\partial x^3\partial y}B(x,y)+6\times\frac{\partial^4}{\partial x^2\partial y^2}B(x,y)+4\times\frac{\partial^4}{\partial x\partial y^3}B(x,y)+\frac{\partial^4}{\partial y^4}B(x,y)\right]_{x=\frac{1}{2}, y=\frac{5}{2}}\\&=&\frac{3}{16}\pi^3+\frac{15}{256}\pi^5.\end{aligned}$

Thus, $A=3, B=3, C=16, D=15, E=5, F=256$ , giving the final answer as $\fbox{298}.$

Note: this solution is extremely tedious and time-consuming, any better and more elegant solutions than this are welcomed. Thanks.