deriving general form of rational log integral Is there a nice/clever method to derive a general closed form for:
$$\displaystyle \int_0^1 \frac{\ln(1+x^a)}{1+x}dx, \;\ a>1\quad?$$
I thought maybe start with differentiating w.r.t. $a$.
This gives $\displaystyle \int_0^1 \frac{x^{a}\ln(x)}{(1+x^{a})(1+x)}dx$.
Maybe even use $\ln(1+x^{a})=\int_0^{x^{a}}\frac{1}{1+t}dt$ and/or series somehow. 
But, now is there some way to link it to digamma, incomplete beta function, polylog, or some other advanced function?.  
I just got to wondering about this one. If a general from can be derived, it would be 
handy for many values of $a$.   Thanks very much.
 A: Expanding $\log(1+x^a) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} x^{a k}$ and integrating term-wise:
$$
  \int_0^1 \frac{\log(1+x^a)}{1+x} \mathrm{d} x = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 k} \left( \psi\left( \frac{a k}{2} + 1 \right) - \psi\left( \frac{a k}{2} + \frac{1}{2} \right) \right)  
$$ 
Integrate by parts:
$$
  \log(1+x^a) \mathrm{d}\log(1+x) = \mathrm{d} \left( \log(1+x) \log(1+x^a) \right) - \log(1+x) \mathrm{d} \log(1+x^a)
$$
Therefore
$$ \begin{eqnarray}
\int_0^1 \frac{\log(1+x^a)}{1+x} \mathrm{d} x &=& \log^2(2) - a \int_0^1 \frac{\log(1+x) x^{a-1}}{1+x^a} \mathrm{d} x \\
&=& \log^2(2) - \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 k} \left( \psi\left( \frac{k}{2 a} + 1 \right) - \psi\left( \frac{k}{2a } + \frac{1}{2} \right) \right)  
\end{eqnarray}
$$
Notice that this implies, that for $a=1$, the result is $\frac{1}{2} \log^2(2)$.
These sums can not be evaluated in closed forms, I am afraid, unless $a$ is a rational number.
Notice that $a$ need not be greater than $1$ in order to assure convergence of the integral. It can be any real number.
A: There's an interesting series expansion around $a=0$, which I think must be an asymptotic series:
$$ \int_0^1 \frac{\ln(1+x^a)}{1+x}\ dx = \ln(2)^2 - \frac{\pi^2}{24} a + \sum_{k=1}^\infty
\left(4^k + 4^{-k} - 2\right) \frac{B_{2k}\ \zeta(1+2k)}{2k} a^{2k}$$
where $B_{2k}$ are Bernoulli numbers.
A: I have noticed some other interesting relationships with log integrals and $2+\sqrt{3}$.  $\tan(\frac{\pi}{12})=2-\sqrt{3}$.  For instance, $\displaystyle \frac{3}{2}\int_{2+\sqrt{3}}^{\infty}\frac{\ln(x)}{1+x^{2}}\mathrm{d}x=\text{Catalan constant}$.   There is an interesting identity that can be shown using the addition formula for tangent:  $\displaystyle 2\int_{0}^{\frac{\pi}{12}}\ln(\tan(3x))dx=\int_{0}^{\frac{\pi}{12}}\ln(\tan(x))\mathrm{d}x$.  Let $y=3x$ and we get  $\displaystyle\frac{-3}{2}\int_{0}^{\frac{\pi}{12}}\ln(\tan(x))\mathrm{d}x=-\int_{0}^{\frac{\pi}{4}}\ln(\tan(y))\mathrm{d}y$.  By letting $y=1/x$ we can also show $\displaystyle \frac{3}{2}\int_{2+\sqrt{3}}^{\infty}\frac{\ln(x)}{1+x^{2}}\mathrm{d}x$.   There is that $2+\sqrt{3}$ again.  I have a feeling something is in there somewhere. 
