Since beating my head against a brick wall is so fun, I kept working on this old integral $\int \frac{ \sec{x} \tan{x}}{3x+5}dx$ . I think I have finally found a way to do it. Here goes. $$ \int \frac{ \sec{x} \tan{x}}{3x+5} $$ $$\text {take the nat. log then} $$ $$ \int ln \frac{ \sec{x} \tan{x}}{3x+5}= \int ln \sec x+ \int ln \tan x - \int ln(3x+5) dx \rightarrow $$Wolfram shows the ugliest complex expression I have ever seen (yeah I cheated so sue me). $$ \int (\log(\sec(x))+\log(\tan(x))-log(3 x+5)) dx = 1/6 (-3 i Li_2(-e^{2 i x})-3 i (Li_2(-i \tan(x))+\log(1+i \tan(x)) \log(\tan(x)))+3 i (Li_2(i \tan(x))+\log(1-i \tan(x)) \log(\tan(x)))-3 i x^2+6 x+6 x \log(1+e^{2 i x})-2 (3 x+5) \log(3 x+5)+6 x \log(\sec(x))+10)+constant $$
Problem is I still have no idea how to reconcile the natural log after integrating the expression in order to get the integral of the original expression. Is there are way to do this? Or have I just found another dead end?