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?

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    $\begingroup$ It's entirely possible that this function has no elementary antiderivative. In other words, there's no reason to expect that you can find the answer. $\endgroup$ Dec 25, 2013 at 22:18
  • $\begingroup$ duplicate of math.stackexchange.com/questions/395271/… $\endgroup$ Dec 25, 2013 at 22:19
  • $\begingroup$ @GregMartin Not exactly a duplicate. I did have to include the original post in my question or it would have been plagiaristic. Also to explain why I took my approach. Honestly it is a separate "challenge" to solve and old question I found interesting. $\endgroup$
    – Chris
    Dec 25, 2013 at 22:26
  • $\begingroup$ @GregMartin So for all intents and purposes this could be like trying to find the last digit of ${\pi}$? Can anyone find a way? Or prove their is no elementary integral? $\endgroup$
    – Chris
    Dec 25, 2013 at 22:30
  • $\begingroup$ "Finding the last digit of $\pi$" isn't a sensible task, since $\pi$ (being irrational) has a non-terminating decimal. It might be possible to prove that there is no elementary antiderivative (once one defines exactly what one means by "elementary"). $\endgroup$ Dec 25, 2013 at 22:40

3 Answers 3


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).

  • 1
    $\begingroup$ Take care. Indeed, from that page: "The Risch algorithm applied to general elementary functions is not an algorithm but a semi-algorithm because it needs to check [...] if certain expressions are equivalent to zero [...]. For expressions that involve only functions commonly taken to be elementary it is not known whether an algorithm performing such a check exists or not (current computer algebra systems use heuristics); moreover, if one adds the absolute value function to the list of elementary functions, it is known that no such algorithm exists; see Richardson's theorem." $\endgroup$ Dec 26, 2013 at 2:48
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    $\begingroup$ While Mathematica's failure to integrate is probably good enough for practical purposes, the fact that its algorithm is not open for inspection means that it cannot be considered a reliable proof. $\endgroup$ Dec 26, 2013 at 2:51

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

${let: u=\frac{\sec\theta+5}{3}=, du=\frac{1}{3}\sec\theta\tan\theta d\theta}$ $${\frac {1}{9}\cos\theta\int\sec u \tan u du=}$$ $${\frac{1}{9}\cos\theta\sec u+c=}$$ $${\frac{1}{9}\frac{\sec x}{3x+5}+c}$$


Nope back to the drawing board.

  • 1
    $\begingroup$ $\cos \theta$ cannot come out of the integral. $\endgroup$ Jan 20, 2014 at 1:20
  • $\begingroup$ You can express $\cos \theta$ as a function of $u$, and then you get back with $\int \frac{\sec (u) \tan (u) }{3u-5}du$. $\endgroup$
    – Olivier
    Jan 20, 2014 at 2:13
  • $\begingroup$ I can eliminate ${\cos\theta}$ before the second substitution that leaves me with $${\frac{1}{3}\int {\sec\frac{\sec\theta+5}{3}\tan\frac{\sec\theta+5}{3}}\tan\theta d\theta}$$ That still leaves me with $${\frac{1}{3}\int \sec u\tan u\cos\theta du}$$ so that gets me nowhere. Would a laplace transform help with this? $\endgroup$
    – Chris
    Jan 20, 2014 at 3:18

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

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

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

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

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

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

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

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

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


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