Integral of the exponential function [closed]

I am searching the indefinite integral of this function: $\dfrac{\exp(x)}{(1+x)^{5/3}}$. Thank you alot.

closed as off-topic by GNUSupporter 8964民主女神 地下教會, Lord Shark the Unknown, José Carlos Santos, Vinyl_cape_jawa, SongFeb 28 at 15:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Lord Shark the Unknown, José Carlos Santos, Vinyl_cape_jawa, Song
If this question can be reworded to fit the rules in the help center, please edit the question.

• First thing I would do is make a change if variable from $x+1=u$ – Chinny84 Jul 25 '14 at 16:05
• It cannot be expressed in terms of elementary functions. See exponential integral and/or incomplete $\Gamma$ function for more details. – Lucian Jul 25 '14 at 18:15
• I have tried integration by parts, twice after that I find the integral of exp(x^3) dx. any ideas? – Ahmed Abrous Jul 25 '14 at 18:22
• You are looking for the holy grail since, as Lucian just said, functions like $e^{x^n}$ (with $n\geq 2$) do not have an "elementary" antiderivative. – Jack D'Aurizio Jul 26 '14 at 2:13
• Are you looking for a primitive or for the value of a specific integral of this function? – Did Jul 31 '14 at 20:52

First rewrite your integral: $$\int{\frac{e^x}{(x+1)^{5/3}}}{dx} = e^{-1}\int{\frac{e^{x+1}}{(x+1)^{5/3}}}{dx}$$ Make a substitution $u = x+1$ and $du = dx$ $$e^{-1}\int{\frac{e^{x+1}}{(x+1)^{5/3}}}{dx} = e^{-1}\int{\frac{e^{u}}{u^{5/3}}}{du}$$ Make another substitution: $s = u^{1/3}$ and $du = 3s^2ds$ $$e^{-1}\int{\frac{e^{u}}{u^{5/3}}}{du} = 3e^{-1}\int{\frac{e^{s^3}}{s^3}}{ds}$$ The last integral is not elementary (which can be proven by the Risch Algorithm). Thus you can conclude that your initial integral is not an elementary function.
However, your integral has a closed form in terms of special functions (using Mathematica): $$\int_{-\infty}^{\infty}{s(\omega)e^{\tau*I*\omega}}{d\omega} = \frac{\left(\frac{2}{3}\right)^{2/3} \pi \text{c1} |r|^{2/3} \left(-i \text{sgn}(r) \left(3 \left(\sqrt{-\text{c2}}-\sqrt{\text{c2}}\right) \cos \left(\frac{2 |r|}{3 \text{c2}}\right)-\sqrt{3} \left(\sqrt{-\text{c2}}+\sqrt{\text{c2}}\right) \sin \left(\frac{2 |r|}{3 \text{c2}}\right)\right)+3 \left(\sqrt{\text{c2}}-\sqrt{-\text{c2}}\right) \sin \left(\frac{2 |r|}{3 \text{c2}}\right)-\sqrt{3} \left(\sqrt{-\text{c2}}+\sqrt{\text{c2}}\right) \cos \left(\frac{2 |r|}{3 \text{c2}}\right)\right)}{3 \text{c2}^2 \Gamma \left(\frac{2}{3}\right)}$$
with the restrictions: $r \in \mathbb{R} \land c2 \in \mathbb{C}\backslash\mathbb{R} \lor c2 = 0$
• Of course. But you need to specify the constants first. What do you mean with $-\infty < \tau <\infty$? Is this a double integral? – Andrei Kh Aug 5 '14 at 20:34