TITLE: Prove $\int_{0}^{\infty}{\phi^3e^{4x}-\phi^4e^{3x}-\phi^3e^{2x}+e^x+2\over (\phi e^x)^5-1}\cdot2xdx=\left({\pi\over 5}\right)^2$
QUESTION [5 upvotes]: Prove the following equation, given that $\phi$ stands for the golden ratio:

$$I=\int_{0}^{\infty}{\phi^3e^{4x}-\phi^4e^{3x}-\phi^3e^{2x}+e^x+2\over (\phi e^x)^5-1}\cdot2xdx=\color{blue}{\left({\pi\over 5}\right)^2}\tag1$$


What I've tried:

$${1\over (\phi e^x)^5-1}={1\over \phi^5e^{5x}}\cdot{1\over 1-\phi^{-5}e^{-5x}}={1\over \phi^{5(n+1)}e^{5x}}\sum_{n=0}^{\infty}e^{-5nx}\tag2$$

Apply $(2)$ to $(1)$ 
Hence
Let $$S=\sum_{n=0}^{\infty}\phi^{-5(n+1)}$$

$$I=S\int_{0}^{\infty}(\phi^3 e^{-x(5n+1)}-\phi^4 e^{-x(5n+2)}-\phi^3e^{-x(5n+3)}+e^{-x(5n+4)}+2e^{-x(5n+5)})\cdot2xdx\tag3$$

integrate $(3)$ term by term using integration by parts 
Take an example from $(3)$

$$2\phi^3\int_{0}^{\infty}xe^{-(5n+1)x}dx= {2\phi^3\over (5n+1)^2}$$

This method it is definitely tedious. To evaluate the sum is another lengthy method.
Anyone with a clever technique of tackling (1)? (Avoid series if possible)

REPLY [4 votes]: We want to compute:
$$ I = \int_{0}^{+\infty}\frac{\phi^{-2}e^{-x}-\phi^{-1}e^{-2x}-\phi^{-2}e^{-3x}-\phi^{-5}e^{-4x}+2\phi^{-5}e^{-5x}}{1-\phi^{-5}e^{-5x}}\,2x\,dx $$
hence by expanding the integrand function as a (convergent!) geometric series we get:
$$\small I = 2\sum_{k\geq 0}\left(\frac{1}{\phi^{5k+2}(5k+1)^2}-\frac{1}{\phi^{5k+1}(5k+2)^2}-\frac{1}{\phi^{5k+2}(5k+3)^2}+\frac{1}{\phi^{5k+5}(5k+4)^2}+\frac{2}{\phi^{5k+5}(5k+5)^2}\right)$$
that is related with $\text{Li}_2$ evaluated at $\frac{1}{\phi}$. See Landen's identity.