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I am having quite a bit of difficulty integrating this integral...

$$ \int_{\theta}^{T} \frac{2 \lambda _R \left(x+\frac{x^2}{2}-\frac{1-e^{\mu (-n) x}}{\mu n}\right)}{2 \left(1-\frac{\lambda _R e^{\mu (-n) x} \left(\mu (-n) x+e^{\mu n x}-1\right)}{\mu n}\right)}\mu n e^{-n \mu x}\mathrm dx $$ the fraction is $$\color{blue}{ 2 \lambda _R \left(x+\frac{x^2}{2}-\frac{1-e^{\mu (-n) x}}{\mu n}\right)/2 \left(1-\frac{\lambda _R e^{\mu (-n) x} \left(\mu (-n) x+e^{\mu n x}-1\right)}{\mu n}\right) } $$

Any ideas on how to integrate this monster. Thanks.

I believe the term $\frac{x^2}{2}$ and $\frac{1-e^{-n\mu x}}{n\mu }$ in the numerator is causing the problem. Note that if the numerator changes to $2 \lambda _R x$, then this does integrate to:

$$\log (\mu n)-\log \left(\lambda _R \left(\mu n T-e^{\mu n T}+1\right)+\mu n e^{\mu n T}\right)+\mu n T$$

So, there might be some sort of change of variable that might work.

Basically, I am interested in how "experts" would handle this scenario. I am certainly NOT an expert.

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    what is $\mu$, a constant? – 2017-02-21
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    Have you tried anything? – 2017-02-21
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    Yes, $\mu$ is a constant. So is $\lambda_R$. – 2017-02-21
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    First of all, you could cancel the 2 in the numerator and denominator :p – 2017-02-21
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    Just to confirm: $\lambda_R, \mu, n, \theta, T$ are all constants and can be any real numbers? – 2017-02-21
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    I think the fastest way is just using some computer algebra system. I dislike strongly doing calculus by hand, Im not a computer or a robot, dont you? – 2017-02-21
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    @Masacroso I'd absolutely agree, but Mathematica isn't able to solve it. Perhaps a different CAS would fare better? – 2017-02-21
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    @diracdeltafunk If Mathematica can't solve it it's unlikely there's a closed form. – 2017-02-21
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    @diracdeltafunk the variable is real or complex? Ans the constant? Try to refine it in mathematica (or any other CAS) to see if it have a closable form. – 2017-02-21
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    @PMF Not only is your integral difficult to SOLVE, it's also difficult to READ. Some simple substitutions and scaling arguments can reduce your integral to an equivalent one with fewer parameters. Eliminating any unnecessary parameters would be a good first step towards making the problem more manageable. – 2017-02-21

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