So close yet so far Finding $\int \frac {\sec x \tan x}{3x+5} dx$

Question

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?

Answer 1

Integrating elementary functions in elementary terms is completely algorithmic. The algorithm is implemented in all major computer algebra systems, so the fact that Mathematica fails to integrate this in closed form can be viewed as (in effect) a proof that such a closed form does not exist in elementary terms.

To answer the comments:

1. You may or may not trust Mathematica (I don't always, but do for this). The fact that "its algorithm is not open for inspection" is not relevant -- it would take you much longer to figure out what the code does than to run the algorithm by hand (well, you need some linear algebra).

2. If you do want to try it in the privacy of your own home, you need to go no further than the late, lamented Manuel Bronstein's excellent book: http://www.amazon.com/Symbolic-Integration-Transcendental-Computation-Mathematics/dp/3540214933

3. I am quite sure that this particular integral is easy to show non-integrable in elementary terms by hand (if you understand the Risch algorithm).

Answer 2

$\int\dfrac{\sec x\tan x}{3x+5}~dx$

$=\int\dfrac{1}{3x+5}~d(\sec x)$

$=\dfrac{\sec x}{3x+5}-\int\sec x~d\left(\dfrac{1}{3x+5}\right)$

$=\dfrac{\sec x}{3x+5}+\int\dfrac{3\sec x}{(3x+5)^2}~dx$

$=\dfrac{\sec x}{3x+5}+\int\sum\limits_{n=0}^\infty\dfrac{12(-1)^n(2n+1)\pi}{(3x+5)^2((2n+1)^2\pi^2-4x^2)}~dx$ (use Mittag-Leffler Expansion of secant)

$=\dfrac{\sec x}{3x+5}+\int\sum\limits_{n=0}^\infty\dfrac{12(-1)^n(2n+1)\pi}{(3x+5)^2((2n+1)^2\pi^2-4x^2)}~dx$

$=\dfrac{\sec x}{3x+5}+\int\sum\limits_{n=0}^\infty\dfrac{24(-1)^n}{(3(2n+1)\pi-10)^2((2n+1)\pi+2x)}~dx+\int\sum\limits_{n=0}^\infty\dfrac{24(-1)^n}{(3(2n+1)\pi+10)^2((2n+1)\pi-2x)}~dx-\int\sum\limits_{n=0}^\infty\dfrac{4320(-1)^n(2n+1)\pi}{(9(2n+1)^2\pi^2-100)^2(3x+5)}~dx+\int\sum\limits_{n=0}^\infty\dfrac{108(-1)^n(2n+1)\pi}{(9(2n+1)^2\pi^2-100)(3x+5)^2}~dx$

$=\dfrac{\sec x}{3x+5}+\sum\limits_{n=0}^\infty\dfrac{12(-1)^n\ln((2n+1)\pi+2x)}{(3(2n+1)\pi-10)^2}-\sum\limits_{n=0}^\infty\dfrac{12(-1)^n\ln((2n+1)\pi-2x)}{(3(2n+1)\pi+10)^2}-\sum\limits_{n=0}^\infty\dfrac{1440(-1)^n(2n+1)\pi\ln(3x+5)}{(9(2n+1)^2\pi^2-100)^2}-\sum\limits_{n=0}^\infty\dfrac{36(-1)^n(2n+1)\pi}{(9(2n+1)^2\pi^2-100)(3x+5)}+C$

$=\dfrac{\sec x-\sec\dfrac{5}{3}}{3x+5}-\dfrac{\ln(3x+5)}{3}\sec\dfrac{5}{3}\tan\dfrac{5}{3}+\sum\limits_{n=0}^\infty\dfrac{12(-1)^n\ln((2n+1)\pi+2x)}{(3(2n+1)\pi-10)^2}-\sum\limits_{n=0}^\infty\dfrac{12(-1)^n\ln((2n+1)\pi-2x)}{(3(2n+1)\pi+10)^2}+C$

Answer 3

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.