Latest Edit
Thanks to @Claude Leibovici and @Gary for giving us the closed form of the integral.
Using MA, we have
$$\frac{d^{ n}}{d{x}^{n}}\left(\sec x\right)= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k-n) !} E_{2 k} x^{2 k-n} $$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln ^{2 n} x}{1+x^{2}} d x=&\left.\frac{\pi}{2} \frac{\partial^{2 n}}{\partial{a}^{2 n}}\left(\sec \left(\frac{a \pi}{2}\right)\right)\right|_{a=0} \\
=&\left.\left(\frac{\pi}{2}\right)^{2 n+1} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k-2 n) !} E_{2 k} \left(\frac{a\pi}{2} \right) ^{2 k-2 n}\right|_{a=0} \\
=&\left(\frac{\pi}{2}\right)^{2 n+1}(-1)^{n} E_{2 n} \\
=&\left(\frac{\pi}{2}\right)^{2 n+1}\left|E_{2 n}\right|
\end{aligned}
$$
where $E_{2 k}$ is an Euler number.
For example, $$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln ^{20} x}{1+x^{2}} d x &=\frac{\pi^{21}}{2^{21}}|E _{10}| =\frac{370371188237525 \pi^{21}}{2097152},
\end{aligned}
$$
checked by MA.
Noting that $$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln^{2n-1}x}{1+x^2} d x &=\int_{0}^{\infty} \frac{\ln \left(\frac{1}{x}\right)}{1+\frac{1}{x^{2}}} \cdot \frac{d x}{x^{2}} =-\int_{0}^{\infty} \frac{\ln ^{2 n-1} x}{x^{2}+1} d x,
\end{aligned}
$$
we have $$
I_{2n-1}=\int_{0}^{\infty} \frac{\ln ^{2 n-1} x}{1+x^{2}} d x=0
$$
How about
$$I_{2n}=\int_{0}^{\infty} \frac{\ln ^{2 n} x}{1+x^{2}} d x?$$
I like to switch the integration problem to a differentiation problem by defining a new integral.
$$
J(a):=\int_{0}^{\infty} \frac{x^{a}}{1+x^{2}} d x
$$
Then $$
I_{k}=J^{(k)}(0)
$$
By my post
$$\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m},$$
we have
$$
J(a)=\frac{\pi}{2} \csc \frac{(a+1) \pi}{2}=\frac{\pi}{2} \sec \left(\frac{a \pi}{2}\right)
$$
Then $$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln ^{2n} x}{1+x^{2}} d x &=\left.\int_{0}^{\infty} \frac{1}{1+x^{2}} \frac{\partial^{2n}}{\partial a^{2n}}\left(x^{a}\right) d x\right|_{a=0} \\
&=\boxed{\frac{\pi}{2} \frac{d^{2n}}{d a^{2n}}\left[\left.\sec \left(\frac{a \pi}{2}\right) \right]\right|_{a=0}}
\end{aligned}
$$
For examples, $$
\begin{aligned}
I_{2}&=\left.\frac{\pi}{2} \cdot \frac{d^{2}}{d a^{2}} \sec \left(\frac{a \pi}{2}\right)\right|_{a=0}=\frac{\pi^{3}}{8} \\
I_{4}&=\left.\frac{\pi}{2} \cdot \frac{d^{4}}{d a^{4}} \sec \left(\frac{a \pi}{2}\right)\right|_{a=0}=\frac{5 \pi^{5}}{32}\\ &\qquad \vdots
\end{aligned}
$$
Alternative method(By reduction formula)
Applying Leibniz’s Rule and differentiating the following equation w.r.t $a$ by $n$ times$$
J(a) \cos \frac{a \pi}{2}=\frac{\pi}{2} ,
$$
we have $$
\sum_{k=0}^{2 n}\left(\begin{array}{c}
2 n \\
k
\end{array}\right) \left[\cos \left(\frac{a \pi}{2}\right)\right]^{(2 n-k)} J^{(k)}(a)=0
$$
Putting $a=0$ yields $$
\sum_{k=0}^{2 n}\left(\begin{array}{c}
2 n \\
k
\end{array}\right) \left(\frac{\pi}{2}\right)^{2 n-k}\cos \left(\frac{(2 n-k) \pi}{2}\right)I_{k}=0
$$
Since $I_{2k-1}=0,$ we have found a reduction formula relating $I_{2k}$.$$
\boxed{\sum_{k=0}^{n}\left(\begin{array}{l}
2 n \\
2 k
\end{array}\right)\left(\frac{\pi^{2}}{4}\right)^{n-k} \cos (n-k) \pi I_{2 k}=0}
$$
For example, $$
\begin{aligned}
\left(\begin{array}{l}
4 \\
0
\end{array}\right)\left(\frac{\pi^{2}}{4}\right)^{2} I_{0}+\left(\begin{array}{l}
4 \\
2
\end{array}\right)\left(\frac{\pi^{2}}{4}\right) I_{2}+\left(\begin{array}{l}
4 \\
4
\end{array}\right) I_{4}=0 . \\
\frac{\pi^{4}}{16} \cdot \frac{\pi}{2}-6 \cdot \frac{\pi^{2}}{4}\left(\frac{\pi^{3}}{8}\right)+I_{4}=0 \\
I_{4}=-\frac{\pi^{5}}{32}+\frac{6 \pi^{5}}{3 \cdot 2}=\frac{5 \pi^{5}}{32}
\end{aligned}
$$
checked by WA.
My Question
Is there any formula for the $n^{th}$ derivative of $\sec x$?
Your opinions and alternative solutions are highly appreciated.
Best Answer
$$I_n=\int_{0}^{\infty} \frac{\log ^{2n} (x)}{1+x^{2}}\, d x=\left(\frac{\pi }{2}\right)^{2 n+1} \,\left|E_{2 n}\right|$$