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.

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.
