So close yet so far Finding $\int \frac {\sec x \tan x}{3x+5} dx$ Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts.  below is his work.
How can I calculate
$$
\int {\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x
$$
My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$
Now Using Integration by Parts::
We get
$$= \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$$
Here he hit his road block. 
I tried the opposite tactic 
Taking the other approach by parts. 
let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$  
As you can see I got no further than he did. 
So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way?

Finally figured this one out. 
${let: 3x+5= \sec \theta, x=\frac {\sec\theta+5}{3}, dx=\frac{1}{3}\sec\theta\tan\theta d\theta}$
then we have $${\frac {1}{3}\int\frac{\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}{\sec\theta}\sec\theta\tan\theta d\theta=}$$ $${\frac{1}{3}\int\cos\theta {\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}\sec\theta\tan\theta d\theta}$$ Then
${let: u=\frac{\sec\theta+5}{3}=, du=\frac{1}{3}\sec\theta\tan\theta d\theta}$ $${\frac {1}{9}\cos\theta\int\sec u \tan u du=}$$ $${\frac{1}{9}\cos\theta\sec u+c=}$$ $${\frac{1}{9}\frac{\sec x}{3x+5}+c}$$ 
${Edit}$
Nope back to the drawing board.
