closed form solution of an integral Here I stuck with this integral thingy and i cant evaluate it closed form.
does anybody knows how to attain a closed form(if possible finding the exact solution in terms of simple known functions and if not possible then in terms of complicated known function like erf(x), hypergeometric function or etc..)
the integral is:
$$I(a,b,\theta,r,h_{eff})=\int_a^b\frac{4\pi x}{3 \sqrt{3}r^2}\ e^{-1.2\cdot \ln(10)\cdot\left(\frac{\arctan\left(\frac{h_{eff}}{x}\right)}{\theta_{3dB}}\right)^2}\ \text{d}x$$
$\theta_{3dB}  \;  $  is  a known number
$h_{eff} \;, r \; $ are variables irrelevant to integral cause integral variable is $x$
any helpful tips for solving this monster is welcomed.
thanks alot for attention folks.
*******edit*******
I'm really sorry about simplifying constraints. I must do that next time. cause these questions arise in engineering and basically I'm newbie in math so please accept my apologies.
indeed the main question was like below:
$$I(a,b,\theta,r,h_{eff})=\int_a^b\frac{4\pi x^\beta}{3 \sqrt{3}r^2}\ e^{-1.2\cdot \ln(10)\cdot\left(\frac{\arctan\left(\frac{h_{eff}}{x}\right)}{\theta_{3dB}}\right)^2}\ \text{d}x$$
where beta is chosen from the set {1, 0 , (-2.8), (-3.1)}
I just want to see if it is solvable in simplest case or not. which now i know it's not solvable.
I add this just to say sorry for not simplifying i was concerned with bigger problem literally. 
 A: Since the integration is only in $\text{d}x$, it's better to rewrite the integral in a more comprehensive way. Set
$$A = \frac{1.2\ln(10)}{\theta^2}$$
$$\alpha = h_{\text{eff}}$$
$$B = \frac{4\pi}{r^2 3\sqrt{3}}$$
Thence the integral is now written as
$$B \int_a^b x\ e^{-A\arctan^2\left(\frac{\alpha}{x}\right)}\ \text{d}x$$
We now may perform a change of variable, setting
$$y = \frac{\alpha}{x} ~~~~~ \text{d}x = -\frac{\alpha}{y^2}\ \text{d}y ~~~~~ -\alpha^2 B = C$$
$$C\int_{\alpha/a}^{\alpha/b} \frac{e^{-A \arctan^2(y)}}{y^3}\ \text{d}y$$
For general extrema of integration, I don't know if a close form does exist. (Mathematica says nothing about). We may try a Series expansion at first, for the exponential:
$$C\int_{\alpha/a}^{\alpha/b} \sum_{k = 0}^{+\infty} \frac{(-A \arctan^2(y))^k}{y^3 k!}\ \text{d}y = C \sum_{k = 0}^{+\infty} \frac{(-A)^k}{k!}\int_{\alpha/a}^{\alpha/b} \frac{\arctan^{2k}(y)}{y^3}\ \text{d}y$$
But again, this integral is not doable in terms of simple functions, and also Mathematica says nothing here too.
Suggestions
1) One may try again the series expansion for the arctangent function
2) Some smart change of variable? If it does exist..
I will think more about this. However, I can assure you that there is no close solution in terms of simple function. The question is now: is there any solution in terms of any function?
A: Ignoring the irrelevant constants and after rescaling, your integral can be expressed as the undefinite
$$\int ta^{-\text{arccot}^2(t)}dt,$$ where you can change to $t=\cot(u)$
$$\int\frac{\cos(u)}{\sin^3(u)}a^{-u^2}du.$$
Chances of a closed form are below zero.
