Logarithm integral with golden ratio $\int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$ How to evaluate the following integral 
$$ \int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$$
where 
$$\phi =\frac{\sqrt{5}+1}{2}$$
Is there a closed form for the integral ?
My first attempt would involve relating the integral to dilogarithms but that seems complicated !

\begin{align}
\int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x
&=
\sum_{\sigma = \pm}
\int_{0}^{1}{\ln\left(1 + {\rm i}\,\sigma\,\phi^{1/2}x\right) \over 1 + x}\,{\rm d}x
=
\sum_{\sigma = \pm}
\int_{1}^{2}{\ln\left(1 - {\rm i}\,\sigma\,\phi^{1/2} + {\rm i}\,\sigma\,\phi^{1/2}x\right) \over x}\,{\rm d}x
\\[2mm]&=
\sum_{\sigma = \pm}\left\lbrack\ln\left(1 - {\rm i}\,\sigma\,\phi^{1/2}\right)\ln\left(2\right)
+
\int_{1}^{2}{\ln\left(1 - z_{\sigma}x\right) \over x}\,{\rm d}x\right\rbrack
\end{align}
where $\displaystyle{z_{\sigma}
                    \equiv
                     -\,{{\rm i}\,\sigma\,\phi^{1/2} \over 1 - {\rm i}\,\sigma\,\phi^{1/2}}}$
\begin{align}
\int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x
&=
\ln\left(1 + \phi\right)\ln\left(2\right)
+
\sum_{\sigma = \pm}\left\lbrack%
\int_{0}^{1}{\ln\left(1 - 2z_{\sigma}x\right) \over x}\,{\rm d}x
-
\int_{0}^{1}{\ln\left(1 - z_{\sigma}x\right) \over x}\,{\rm d}x
\right\rbrack
\\[2mm]&=
\ln\left(1 + \phi\right)\ln\left(2\right)
+
\sum_{\sigma = \pm}\left\lbrack\vphantom{\Large A}%
-{\rm Li_{2}}\left(2z_{\sigma}\right) + {\rm Li_{2}}\left(z_{\sigma}\right)
\right\rbrack
\\[1cm]&
\end{align}
\begin{align}
\int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x
&=
\ln\left(1 + \phi\right)\ln\left(2\right)
+
2\Re\left\lbrack\vphantom{\Large A}%
{\rm Li_{2}}\left(z\right) - {\rm Li_{2}}\left(2z\right)
\right\rbrack
\\[2mm]&
\mbox{with}\quad z = -\,{{\rm i}\,\phi^{1/2} \over 1 - {\rm i}\,\phi^{1/2}}
\end{align}
See http://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm
