Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$

Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values

*

*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$


*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.

The generalized results for the even and odd cases of $$I_n=\int_0^\infty \frac{\log(1+x^n)}{(1+x^2)^2}dx
$$
are respectively as follows
\begin{align}
I_{2m} =& -\frac{m\pi}4+\frac{m\pi}2\ln2+\pi\sum_{k=1}^{[\frac m2]}\ln \cos\frac{(m-2k+1)\pi}{4m}\\
I_{2m+1} =& -\frac{(2m+1)\pi}8+\frac{(4m+1)\pi}8\ln2+\frac{(-1)^m G}{2(2m+1)}\\
&\ +\frac\pi2\sum_{k=0}^{m-1}\left[\ln \cos\frac{(2k+1)\pi}{4(2m+1)}+\frac{(-1)^{m+k}(2k+1)}{2(2m+1)}\ln\tan\frac{(2k+1)\pi}{4(2m+1)}
\right]\\
\end{align}
Specifically
\begin{align}
 \int_0^\infty \frac{\log(1+x)}{(1+x^2)^2}dx
=& -\frac\pi8+\frac\pi8\ln2 +\frac12G\\
\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx
=& -\frac\pi4+\frac\pi2\ln2 \\
\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx
=& -\frac{3\pi}8+\frac{\pi}8\ln2
+\frac\pi3\ln(2+\sqrt3) -\frac16G\\
\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx
=& -\frac\pi2+\frac\pi2\ln(2+\sqrt2)\\
 \int_0^\infty \frac{\log(1+x^5)}{(1+x^2)^2}dx
=& -\frac{5\pi}8-\frac{3\pi}8\ln2+\frac\pi2\ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\
&\  +\frac\pi{20}\ln\tan\frac\pi{20}-\frac{3\pi}{20}\ln\tan\frac{3\pi}{20} +\frac1{10}G\\
 \int_0^\infty \frac{\log(1+x^6)}{(1+x^2)^2}dx
=& -\frac{3\pi}4+\frac\pi2\ln6\\
 \int_0^\infty \frac{\log(1+x^7)}{(1+x^2)^2}dx
=& -\frac{7\pi}8+\frac{13\pi}8\ln2+\frac\pi2\ln\left(\cos\frac\pi{28} \cos\frac{3\pi}{28} \cos\frac{5\pi}{28}\right)\\
& -\frac\pi{28}\ln\tan\frac\pi{28}+\frac{3\pi}{28}\ln\tan\frac{3\pi}{28} -\frac{5\pi}{28}\ln\tan\frac{5\pi}{28} -\frac1{14}G\\
 \int_0^\infty \frac{\log(1+x^8)}{(1+x^2)^2}dx
=& -\pi+\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\\
\end{align}
