# $\int\arctan^4(x)\sec^3(x)dx$

I've tried just about every $$u$$-substitution I can think of, and I've tried integrating by parts, but I can't seem to get anywhere. I'd appreciate any help at all.

My most recent attempt involved using the identity $$1+\tan^2x=\sec^2x$$ But after expanding, and using the sum rule for integrals, I tried doing u-substitution by letting $$u = \arctan(x)$$. However I'm not sure if this is the best approach because I can't find a way to express $$\tan(x)$$ in terms of u.

• What did you try exactly? Please edit it into the question so we can see where you are stuck. – Ramanujan Feb 22 at 23:53
• Do you have any reason to think there's an answer, meisturtle? Most elementary functions don't have an elementary antiderivative. – Gerry Myerson Feb 22 at 23:57
• Wolfram doesn't know the answer. Have you tried using $\cos(x)^3 = \cos(3x) + 3 \cos(x)$ and series expansion? Also the fustian has countably infinitely many singularities. Why are you interested in the indefinite integral? – Ramanujan Feb 23 at 0:28
• I wonder if the integrand is supposed to be $\tan^{-4}x\sec^3x$. – Gerry Myerson Feb 23 at 11:20
• @GerryMyerson the function, sorry. – Ramanujan Feb 23 at 15:11

Even using special functions, I do not think that we could obtain the antiderivative. If you want to integrate over a small range (say $$0 \leq x \leq \frac \pi 4$$), I suppose that series expansion built around $$x=0$$ could be the only way.
Using a very few terms, $$\arctan^4(x)\sec^3(x)=x^4+\frac{x^6}{6}+\frac{101 x^8}{120}-\frac{2249 x^{10}}{15120}+\frac{1046861 x^{12}}{1814400}-\frac{479333 x^{14}}{1478400}+O\left(x^{16}\right)$$ and integrate termwise.
Jus a few result for $$\int_0^t \arctan^4(x)\sec^3(x)\,dx$$ $$\left( \begin{array}{ccc} t & \text{approximation} & \text{exact} \\ 0.2 & 0.0000643524 & 0.0000643524 \\ 0.3 & 0.000493031 & 0.000493031 \\ 0.4 & 0.00211123 & 0.00211124 \\ 0.5 & 0.00661682 & 0.00661699 \\ 0.6 & 0.0171597 & 0.0171634 \\ 0.7 & 0.0394086 & 0.0394559 \\ 0.8 & 0.083599 & 0.0840325 \\ 0.9 & 0.168305 & 0.171374 \\ 1.0 & 0.326574 & 0.344429 \end{array} \right)$$ As you can see, it is not good at all for $$t>0.8$$. For sure, we could improve using more terms.