How to evaluate $int _0^{infty }frac{ln left(x^3+1right)}{left(x^2+1right)^2}:dx$ without complex analysis

With subbing $t=frac{1-x}{1+x}$ we have

$$int_0^1frac{ln(1+t^3)}{1+t^2}dt=int_0^1frac{lnleft(frac{2(1+3x^2)}{(1+x)^3}right)}{1+x^2}dx$$

$$=ln2int_0^1frac{dx}{1+x^2}+int_0^1frac{ln(1+3x^2)}{1+x^2}dx-3int_0^1frac{ln(1+x)}{1+x^2}dx$$

where the first integral

$$int_0^1frac{dx}{1+x^2}=arctan(1)=frac{pi}{4},$$

and the second integral is already calculated in the comments by Dennis or it can be found calculated in details by Zacky in this solution (see the integral $J$);

$$int_0^1frac{ln(1+3x^2)}{1+x^2}dx=frac{pi}{3}ln(2+sqrt 3)+frac{pi}{4}ln 2-frac53G.$$

For the third integral, let $xmapsto frac{1-x}{1+x}$

$$int_0^1frac{ln(1+x)}{1+x^2}dx=int_0^1frac{lnleft(frac{2}{1+x}right)}{1+x^2}dx=ln2int_0^1frac{dx}{1+x^2}-int_0^1frac{ln(1+x)}{1+x^2}dx$$

$$Longrightarrow int_0^1frac{ln(1+x)}{1+x^2}dx=frac{pi}{8}ln2$$

Combine the three results we have

$$int_0^1frac{ln(1+t^3)}{1+t^2}dt=frac{pi}{3}ln(2+sqrt 3)+frac{pi}{8}ln 2-frac53G.$$

All credit goes to Zacky and Dennis for evaluating the second integral as its the key to crack the integral in the question.

As said in comments, consider $$I(a)=int _0^{infty }frac{log left(ax^3+1right)}{left(x^2+1right)^2},dx$$ $$I'(a)=int _0^{infty }frac{x^3}{left(x^2+1right)^2 left(a x^3+1right)},dx$$ Using partial fraction decomposition, the integrand is $$dfrac{ a(2a^2-1)x^2+3a^2x-a(a^2+2)}{(a^2+1)^2left(ax^3+1right)}-dfrac{left(2a^2-1right)x+3a}{(a^2+1)^2left(x^2+1right)}-dfrac{x-a}{left(a^2+1right)left(x^2+1right)^2}$$ which is not too bad. This leads to $$I'(a)=frac{-8 sqrt{3} pi a^{8/3}+24 sqrt{3} pi a^{4/3}+16 sqrt{3} pi a^{2/3}+9 pi a^3-18 a^2+12 left(2 a^2-1right) log (a)-45 pi a-18}{36 left(a^2+1right)^2}$$ This is not the most pleasant part but it leads to the result.