Antiderivative of $e^{2\arctan{x}}$ Is there any way to integrate this: 
$$\int e^{2\arctan x}\, dx$$
I tried to solve it using integration by parts but I could not end the integral because it got very difficult.
Then I tried to solve it using Mathematica but it returned a really weird expression.
The expression is part of solving this equation
$$ y' + \frac{y}{1+x^2}=e^{\arctan x}\ $$
$$ V = e^{-\int f(x) dx} = e^{-\int \frac{dx}{1+x^2}} = e^{-\arctan x}$$
$$R = e^{\arctan x} $$
$$U = \int \frac{e^{\arctan x}}{ e^{-\arctan x}} dx$$
$$ U = \int e^{2\arctan x} dx $$
 A: The "really weird expression" is one involving Gaussian hypergeometric functions, which means that there is no way of representing this antiderivative by elementary functions only. Here is the result.
A: If you want a way to proceed, we can make use of the Taylor expansion of the arctangent function:
$$\arctan(x) = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{2k+1}\ x^{2k+1}$$
Hence you get:
$$\int e^{2\left(\sum_{k = 0}^{+\infty} \frac{(-1)^k}{2k+1}\ x^{2k+1}\right)}\ \text{d}x$$
Now we use the well known symbolic property:
$$e^{\sum X} = \prod e^X$$
So we obtain:
$$\int\prod_{k = 0}^{+\infty}\ e^{\frac{2 (-1)^k}{2k+1}\ x^{2k+1}}\ \text{d}x$$
It doesn't seem very suitable right? But then let's light the writing by recalling
$$\frac{2(-1)^k}{2k+1}  = A(k) ~~~~~~~ 2k+1 = B$$
Now we have:
$$\int\prod_{k = 0}^{+\infty}\ e^{A(k)\ x^B}\ \text{d}x$$
Using the Taylor expansion for the exponential we get:
$$\prod_{k = 0}^{+\infty}\int \sum_{j = 0}^{+\infty} \frac{(A(k)\ x^B)^j}{j!}\ \text{d}x = \prod_{k = 0}^{+\infty}\sum_{j = 0}^{+\infty}\frac{A^j(k)}{j!}\int x^{Bj}\ \text{d}x$$
Which is evaluable and you eventually get a solution in terms of a productory of a summatory:
$$ \prod_{k = 0}^{+\infty}\sum_{j = 0}^{+\infty} \frac{A^j(k)}{j! (Bj + 1)}\ x^{Bj + 1}$$
