I need to evaluate the following indefinite integral for some other definite integral $$\int\frac{x\arctan x}{x^4+1}dx$$ I found that $$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{4}-2I(1)$$ where $$I(t)=\int_0^1\frac{x\operatorname{Ti}_2(tx)}{x^4+1}dx$$ taking derivatives $$tI'(t)=\int_0^1\frac{x\arctan(tx)}{x^4+1}dx$$ $$(tI')'=\int_0^1\frac{x^2}{(t^2x^2+1)(x^4+1)}dx$$ The answer for the integral is the integrand in my indefinite integral plus some stuff I can integrate on my own, which is where my problem comes from. In reality, I need $$\int\frac{1}{t}\left(\int\frac{t\arctan{t}}{t^4+1}dt\right)dt$$ But it's fine if I just get an answer for the inside integral. Back to the indefinite integral. I wanted to use partial fractions $$\int\frac{x\arctan x}{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}dx$$ We would get four integrals of the form $\int\frac{\arctan{x}}{x+c}dx$. Using the logarithmic definition of arctangent, we get 8 integrals of the form $\int\frac{\ln{(a+bx)}}{x+c}$. That's 8 dilogarithms. I need this for another calculation, and having 8 dilogarithms is too numerous to be practical to work with. I want another way to evaluate this integral.
TO CLARIFY
I am fine with dilogarithms