Closed form of series involving Fibonacci numbers

Let $F_n$ denote the $n$-th Fibonacci number and $\phi$ be the golden ratio, that $\phi = \frac{1+\sqrt{5}}{2}$. Find a closed form for the sum:

$$\sum_{n=0}^{\infty} \frac{1}{(5\phi)^n(n+2)} \sum_{k=0}^{n} \frac{F_{k+1} F_{n-k+1}}{k+1}$$

I don't know how to tackle the sum. A closed form exists as the book suggests but there is no hind nor answer on how to tackle it. And I have not come up with some idea to kill this sum. Can anyone help me find a closed form?

By setting $\sigma=\frac{-1+\sqrt{5}}{2},\overline{\sigma}=\frac{-1-\sqrt{5}}{2}$ we have: $$\sum_{k\geq 1}F_k\, x^k = \frac{x}{1-x-x^2} = \frac{-\sigma}{x-\sigma}+\frac{\overline{\sigma}}{x-\overline{\sigma}}\tag{1}$$ hence: $$\sum_{k\geq 1}\frac{F_k}{k}\,x^k = \frac{1}{\sqrt{5}}\,\log\left(\frac{1+\sigma x}{1+\overline{\sigma}x}\right)\tag{2}$$ and: $$\sum_{n\geq 0}x^{n+1}\sum_{h=0}^{n}\frac{F_{h+1}F_{n-h+1}}{h+1}=\frac{1}{\sqrt{5}(1-x-x^2)}\,\log\left(\frac{1+\sigma x}{1+\overline{\sigma}x}\right) \tag{3}$$ so: $$\sum_{n\geq 0}\frac{t^{n+2}}{n+2}\sum_{h=0}^{n}\frac{F_{h+1}F_{n-h+1}}{h+1}=\int_{0}^{t}\frac{dt}{\sqrt{5}(1-x-x^2)}\,\log\left(\frac{1+\sigma x}{1+\overline{\sigma}x}\right) \tag{4}$$ and the problem boils down to finding the right $t$ and computing the integral in the RHS, through partial fraction decomposition and $\int\frac{\log x}{x}\,dx=\log^2 x$.
Can you take it from here? The final answer should be something like $\frac{5}{4} \left(3+\sqrt{5}\right) \log\left[\frac{2}{41} \left(13+\sqrt{5}\right)\right]^2$, but I have to check my computations.
• @Masacroso: there is no EGF involved, in order to go from $(1)$ to $(2)$ you just need to integrate termwise $(1)$. – Jack D'Aurizio Aug 3 '15 at 23:51
• @Tolaso: you just need to use the partial fraction decomposition appearing in $(1)$ to compute a primitive in terms of a squared logarithm. – Jack D'Aurizio Aug 4 '15 at 12:51