A double sum or a definite integral. I am trying to evaluate the following double sum
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty}   \frac{(-1)^{n+m}}{n(3n+m)}.
\end{eqnarray*}
Using the integral trick
\begin{eqnarray*}
\frac{1}{3n+m} =\int_0^1 y^{3n+m-1} dy,
\end{eqnarray*}
the sum can be transformed into integral
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y^3)}{1+y} dy.
\end{eqnarray*}
Now "half" of this is easy (IBP & rearrange)
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y)}{1+y} dy = \frac{1}{2} (\ln 2)^2.
\end{eqnarray*}
So we are left with
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1-y+y^2)}{1+y} dy = \int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy.
\end{eqnarray*}
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
 A: 
$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x^3)}{1+x}\color{blue}{=\frac 12\operatorname{Li}_2\left(-\frac 13\right)+\frac 14\log^23+\frac 12\log^22-\frac {\pi^2}{36}}$$Confirmed by Wolfram Alpha.

Call the integral $\mathfrak{I}$ and make the transformation $x\mapsto\tfrac {1-x}{1+x}$. What's left is$$\mathfrak{I}=\log^22+\int\limits_0^1\mathrm dx\,\frac {\log(1+3x^2)}{1+x}-3\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}$$
The last integral is trivially equal to$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}\color{red}{=\frac 12\log^22}$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$\mathfrak{I}(a)=\int\limits_0^1\mathrm dx\space\frac {\log(1+ax^2)}{1+x}$$Thus$$\begin{align*}\mathfrak{I}'(a) & =\int\limits_0^1\mathrm dx\space\left[\frac 1{(1+a)(1+x)}+\frac x{(1+a)(1+ax^2)}-\frac 1{(1+a)(1+ax^2)}\right]\\ & =\frac 1{1+a}\log 2+\frac {\log(1+a)}{2a(1+a)}-\frac 1{\sqrt a(1+a)}\arctan\sqrt a\end{align*}$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$\int\limits_0^1\mathrm dx\space\frac {\log(1+3x^2)}{1+x}\color{brown}{=\log^22+\frac 12\operatorname{Li}_2\left(-\frac 13\right)+\frac 14\log^23-\frac {\pi^2}{36}}$$Now add everything together and you should get the result I stated at the beginning of my answer.
A: If it helps: Maple 7 computes the expression 
$$ - \mathrm{ln}(2)\,\mathrm{ln}(3) - \mathrm{ln}(2)\,\mathrm{
ln}({\displaystyle \frac {1}{3}} \,I) - \mathrm{ln}(2)\,\mathrm{
ln}({\displaystyle \frac {-1}{3}} \,I) 
\\- \mathrm{dilog}( -
{\displaystyle \frac {1}{3 + I\,\sqrt{3}}}  + {\displaystyle
\frac {I\,\sqrt{3}}{3 + I\,\sqrt{3}}} ) 
\\ - \mathrm{dilog}({\displaystyle \frac {1}{ - 3 + I\,
\sqrt{3}}}  + {\displaystyle \frac {I\,\sqrt{3}}{ - 3 + I\,\sqrt{
3}}} )
\\ + \mathrm{dilog}( - {\displaystyle \frac {1}{ - 3 + I\,
\sqrt{3}}}  + {\displaystyle \frac {I\,\sqrt{3}}{ - 3 + I\,\sqrt{
3}}} ) \\
 + \mathrm{dilog}({\displaystyle \frac {1}{3 + I\,\sqrt{3}
}}  + {\displaystyle \frac {I\,\sqrt{3}}{3 + I\,\sqrt{3}}} ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$\operatorname{dilog(x)} = \operatorname{Li}_2(1-x)$$
A: This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $\operatorname {Li_2}(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$\int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy = -2\int_0^1 \frac{ (y-\frac12)\ln(1+y)}{(y-\frac12)^2 +\frac34} dy.$$ 
Then apply the substitution $x=y-\frac12$.
$$\int_{-\frac 12}^{\frac 12} \frac{x \ln(x+ \frac32)}{x^2+ \frac 34}dx = \int_{-\frac 12}^{\frac 12}  \overbrace{\ln(x+ \frac32)}^{u} \cdot \underbrace {\frac {x}{x^2+ \frac 34} dx }_{dv}$$ Integration by parts show the original integral is equivalent to $\displaystyle \int_{-\frac 12}^{\frac12} \frac{\ln(x^2+\frac34)}{x+\frac 32}dx$. Now we can introduce yet another substitution, say $z=x+\frac 32$. Now we get:
$$\int_1^2 \frac {\ln \bigl((z- \frac 32)^2 + \frac 34\big)}{z}dz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$\int_1^2 \frac {\ln\big((z- \frac 32 + \frac {\sqrt 3}{2}i)(z- \frac 32 - \frac {\sqrt 3}{2}i)\big)}zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$\int_1^2 \frac {\ln(z- \frac 32 + \frac {\sqrt 3}{2}i)}zdz + \int_1^2 \frac {\ln (z- \frac 32 - \frac {\sqrt 3}{2}i)}zdz \tag{1}$$
Now lets look for a general solution to $\displaystyle \int_1^2 \frac {\ln (x+a)}{x}dx $, where $a$ is any constant.
$$\int_1^2 \frac {\ln (x+a)}{x}dx = \int_1^2 \frac {\ln\big(a(\frac xa +1)\big)}{x}dx = \ln(a) \int_1^2 \frac {dx}x+ \int_1^2 \frac {\ln(\frac xa +1)}{x} dx.$$ Apply the substitution $u = - \frac xa$. This changes it to:
$$ \ln(a)\ln(2)+ \int \frac {\ln(1-u)}{u} du = \Big[\ln(a)\ln(x)+ \operatorname {Li_2} (- \frac xa)+ C\Big]_1^2$$
$$ \int_1^2 \frac {\ln (x+a)}{x}dx =\ln(a)\ln(2)+ \operatorname {Li_2} (- \frac 2a)- \operatorname {Li_2} (-\frac 1a) \tag{2}$$
Now, putting (2) into (1) yields:
$$ \ln(- \frac 32 + \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} (- \frac 2{- \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (-\frac 1{- \frac 32 + \frac {\sqrt 3}{2}i}) + \ln(- \frac 32 - \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} ( \frac 2{ \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (\frac 1{ \frac 32 + \frac {\sqrt 3}{2}i}) $$ 
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
