I started like this: $$\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx\: =\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+\overset {x=\frac{1}{x}}{\int _1^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx}$$ $$=2\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+5G$$ where $G$ is Catalan's constant. But the integral left is very hard to calculate. As suggested by Zacky one can use the sub $x=\frac{1-t}{1+t}$ and get these integrals. $$\ln 2\int _0^1\frac{1}{1+t^2}\:dt+\ln 5\int _0^1\frac{1}{1+t^2}\:dt+\int _0^1\frac{\ln \left(t^2+1-\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt$$ $$+\int _0^1\frac{\ln \left(t^2+1+\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt-5\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$ To evaluate those one can use the identity $$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x^2}\:dx=\frac{\pi }{2}\ln \left(\sqrt{a}+\sqrt{b}\right)+\text{Ti}_2\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)-G$$
In the end, the integral evaluates to \begin{align} \int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx &= -\frac{3\pi }{8}\ln 2+\frac{\pi }{4}\ln 5-2G\\ &+\frac{\pi }{2}\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)\\ &+\frac{\pi }{2}\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right) \end{align}
But, I have no idea how to simplify the $\text{Ti}_2\left(z\right)$ terms, which seems possible because I found that a similar version can be expressed without these $$\int _0^{\infty }\frac{\ln \left(1+x^5\right)}{\left(1+x^2\right)^2}\:dx=-\frac{5\pi }{8}-\frac{7\pi }{40}\ln \left(2\right)+\frac{\pi }{5}\ln \left(4+\sqrt{10-2\sqrt{5}}\right) +\frac{\pi }{10}\ln \left(43+7\sqrt{5}+4\sqrt{130+38\sqrt{5}}\right)+\frac{G}{10}$$
NB:
$\displaystyle x\in\mathbb{R},\text{Ti}_2(x)=\int_0^x \frac{\arctan t}{t}dt$
Continue to reduce the result in the post as follows
\begin{align} & \text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)\\ =&\ \text{Ti}_2(\tan\frac\pi{20}) - \text{Ti}_2(\tan\frac{3\pi}{20}) =\int_{\tan\frac{3\pi}{20} }^{\tan\frac{\pi}{20} }\frac{\tan^{-1}x}xdx\\ \overset{ibp}=& \ \frac\pi{20}\ln \tan\frac\pi{20}- \frac{3\pi}{20}\ln \tan\frac{3\pi}{20}+\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt\\ =& \ \frac\pi{20}\ln \frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}- \frac{3\pi}{20}\ln \frac{1-\sqrt{1-\frac{2}{\sqrt{5}}}}{1+\sqrt{1-\frac{2}{\sqrt{5}}}}-\frac25G\\ =& -\frac25G -\frac\pi5 \ln\frac2{\sqrt5 } -\frac\pi5\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right) +\frac{3\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right) \end{align}
where $\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt=-\frac25G$. Then, plug into the result to arrive at the elementary expression for the integral
\begin{align} \int_0^\infty\frac{\ln(1+x^5)}{1+x^2} &= \frac15G -\frac{19\pi }{20}\ln 2+\frac{3\pi }{5}\ln 5\\ &\hspace{5mm}+\frac{4\pi}5 \ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\frac{8\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)\\ \end{align}