On the integral $I(a)=\int_0^1\frac{\log(a+t^2)}{1+t^2}\mathrm dt$ Consider the parameter integral

$$I(a)=\int_0^1\frac{\log(a+t^2)}{1+t^2}\,{\rm d}t\tag1$$

Where $a\in\mathbb{C}$. I am struggling to evaluate this integral in a closed-form.
However, first of all lets just concentrate on some particular values of $a$ for which I was actually able to evaluate the integral exactly

$$\begin{align}
&a=0:&&\int_0^1\frac{\log(t^2)}{1+t^2}\,{\rm d}t=-2G\\
&a=1:&&\int_0^1\frac{\log(1+t^2)}{1+t^2}\,{\rm d}t=\frac{\pi}2\log(2)-G
\end{align}$$

Here $G$ denotes Catalan's Constant. The first case is just one of many integral definitions of Catalan's Constant whereas the second case can be reduced to integrals of this type by the substitution $t=\tan(y)$. Furthermore WolframAlpha is capable of providing a closed-form for the case $a=-1$

$$a=-1:\int_0^1\frac{\log(t^2-1)}{t^2+1}\,{\rm d}t=\frac{\pi}4\log(2)+\frac{i\pi^2}4-G$$

It seems like the general anti-derivative of the case $a=-1$ can be expressed in terms of the Polylogarithm (the term can be found within the given link but is far to complicated to be included here).
For other values of $a$ I was not able to get anything done. I tried to expand the $\log$ and respectively the denominator as a series which ended up in an infinite summation of Hypergeometric Functions $($of the kind $_2F_1(1,k+1;k+2;-1/3)$ paired with a denominator depending on $k$$)$ I was not able to express explicit. Furthermore I tried to apply Feynman's Trick, i.e. differentiate w.r.t. to $a$ in order to get rid of the $\log$. The so occurring integral was easily evaluated by using partial fraction decomposition. Anyway I did not managed to find suitable borders for the integration w.r.t. $a$ afterwards. Applying a trigonometric substitution $($to be precise $t=\tan(x)$$)$ lead to the logarithmic term $\log(1+\cos^2(x))$ which I was not sure how to handle without invoking several powers of the cosine function $($i.e. by using the Taylor series expansion of the natural logarithm$)$.
The first approach aswell as the last one resulted in an infinite double summation. My knowledge about double sums, especially their evaluation, is quite weak. Maybe someone else is able to finish this up.

I have doubts that it is possible to derive an explicit closed-form expression for $I(a)$. Nevertheless for the case that the upper bound is given by $\infty$ instead of $1$ there actually exists a closed-form expression which makes me curious

$$I(a,b,c,g)=\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+g^2x^2}\,{\rm d}x = \frac{\pi}{cg}\log\left(\frac{ag+bc}{g}\right)\tag2$$

I am not familiar with the way this elegant relation was deduced since I just stumbled upon this one within this post.


I would highly appreciate an explicit expression for $I(a)$, maybe similar to the one given for $(2)$, even though I am not sure whether such a term exists. However, I am especially interested in the case $a=3$ for another integral I am working on  right now.

Thanks in advance!
 A: $$\mathcal J(a,t)=\int_0^1 \frac{\ln(a+t(1+x^2))}{1+x^2}\mathrm dx\Rightarrow I(a)=\int_0^1\frac{\ln(a+x^2)}{1+x^2}\mathrm dx=\mathcal J(a-1,1)$$
$$ \frac{\mathrm d}{\mathrm dt}\mathcal J(a,t)=\int_0^1 \frac{\mathrm dx}{a+t+tx^2}=\frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)$$
$$\mathcal J(a,0)=\frac{\pi\ln a}{4}\Rightarrow \mathcal J(a,1)=\underbrace{\int_0^1 \frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)\mathrm dt}_{=J}+\frac{\pi\ln a}{4}$$
Now via the substitution $\displaystyle{\sqrt{\frac{t}{t+a}}=x\Rightarrow  \frac{\mathrm dt}{\sqrt{t(t+a)}}=\frac{2}{1-x^2}dx}$ we get:
$$J=2\int_0^\frac{1}{\sqrt{1+a}}\frac{\arctan x}{1-x^2}\mathrm dx \overset{x=\frac{1-y}{1+y}}=\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\arctan\left(\frac{1-y}{1+y}\right)}{y}\mathrm dy$$
$$=\frac{\pi}{4}\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\mathrm dy}{y}-\int_0^1 \frac{\arctan y}{y}\mathrm dy+\int^{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}_0\frac{\arctan y}{y}\mathrm dy$$
$$\Rightarrow \mathcal J(a,1)=\frac{\pi}{4} \ln\left(\frac{\sqrt{a+1}+1}{\sqrt{a+1}-1}\right)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a+1}-1}{\sqrt{a+1}+1}\right)+\frac{\pi}{4}\ln a$$
$$\Rightarrow \boxed{I(a)=\int_0^1 \frac{\ln(a+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln(\sqrt a+1)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)}$$
Where $\operatorname{Ti}_2(x)$ is the inverse tangent integral and $\mathrm G$ is Catalan's constant.

Some nice results that follows:
$$\boxed{I(3)=\int_0^1\frac{\ln(3+x^2)}{1+x^2}\mathrm dx=\frac{\pi}{4}\ln  2+\frac{\pi}{6}\ln(2+\sqrt 3)-\frac13\mathrm G}$$
$$\boxed{I\left(\frac13\right)=\int_0^1 \frac{\ln\left(\frac13 +x^2\right)}{1+x^2}\mathrm dx=\frac{\pi}4 \ln \left(\frac23\right)+\frac{\pi}{3}\ln(2+\sqrt 3)-\frac53\mathrm G}$$
A: I get 
$$ I'(a) = 
\int_0^1 \dfrac{dt}{(t^2+a)(t^2+1)} = {\frac {\pi\,\sqrt {a}-4\,\arctan \left( {\frac {1}{\sqrt {a}}}
 \right) }{ 4 \left( a-1 \right) \sqrt {a}}}
$$
Integrating this using Maple produces a rather complicated expression, which seems to work for $0 < a < 1$ (a different branch should be used after $a=1$):
$$\frac{i}{2}{\it dilog} \left( {\frac {-2\,\sqrt {a}+1-i+ \left( 1+i \right) a
}{a+1}} \right) -\frac{i}{2}{\it dilog} \left( {\frac {2\,\sqrt {a}+1+i+
 \left( 1-i \right) a}{a+1}} \right) +\frac{\pi}{4}\,\ln  \left( 1-\sqrt {a
} \right) +\frac{\pi}{4}\,\ln  \left( 1+\sqrt {a} \right) -\frac{1}{2}\,\arctan
 \left( {\frac {1}{\sqrt {a}}} \right) \ln  \left( -\sqrt {2} \left( 1
+\sqrt {a} \right) \sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( {
\frac {1}{\sqrt {a}}} \right) \ln  \left( \sqrt {2} \left( -1+\sqrt {a
} \right) \sqrt {a+1}+2\,a+2 \right) -\frac{1}{2}\,\arctan \left( {\frac {1}{
\sqrt {a}}} \right) \ln  \left( \sqrt {2} \left( 1+\sqrt {a} \right) 
\sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( {\frac {1}{\sqrt {a}}}
 \right) \ln  \left( -\sqrt {2} \left( -1+\sqrt {a} \right) \sqrt {a+1
}+2\,a+2 \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) 
\arctan \left( {\frac {-1+\sqrt {a}}{-\sqrt {2}\sqrt {a+1}+\sqrt {a}+1
}} \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \arctan
 \left( {\frac {1+\sqrt {a}}{\sqrt {2}\sqrt {a+1}+\sqrt {a}-1}}
 \right) -i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \pi-\frac{i}{4}{\pi}^
{2}-{\it Catalan}
$$
A: Let $a>-1$ be a real number. Then
$$\int_0^1 \frac{\log(1+a^2x^2)}{1+x^2}\textrm{d}x=\frac{\pi}{2}\log(1+a)-G+\text{Ti}_2\left(\frac{1-a}{1+a}\right),$$
where $G$ is the Catalan's constant and $\displaystyle \text{Ti}_2(x)=\int_0^x \frac{\arctan(t)}{t}\textrm{d}t$ is the inverse tangent integral.
Thanks to Cornel for this way of writing the closed-form of the integral.
A: $$\int_0^1\frac{\ln(a^2+x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx}_{x\to 1/x}$$
$$=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\int_0
^1\frac{\ln(\frac{1+a^2x^2}{x^2})}{1+x^2}dx$$
$$=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\int_0
^1\frac{\ln(1+a^2x^2)}{1+x^2}dx-2\int_0^1\frac{-\ln(x)}{1+x^2}dx$$
$$=\pi\ln(1+a)-\int_0
^1\frac{\ln(1+a^2x^2)}{1+x^2}dx-2G$$
$$=\pi\ln(1+a)-I-2G,$$
where the first integral was evaluated by differentiation under the integral sign and the third integral is the integral representation of the Catalan constant. To find $I$, let
$$I(b)=\int_0
^1\frac{\ln(1+b^2x^2)}{1+x^2}dx$$
and notice that $I(a)=I$ and $I(0)=0,$
$$I'(b)=\int_0^1\frac{2bx^2}{(1+x^2)(1+b^2x^2)}dx=\frac{4\arctan(b)-b\pi}{2(1-b^2)}.$$
Now integrate both sides from $b=0$ to $b=a,$
$$\int_0^a I'(b)db=I(b)|_0^a=I(a)-I(0)=I-0=\int_0^a \frac{4\arctan(b)-b\pi}{2(1-b^2)}db$$
$$\overset{b=(1-x)/(1+x)}{=}\frac{\pi}{2}\int_{\frac{1-a}{1+a}}^1\frac{dx}{1+x}dx-\int_{\frac{1-a}{1+a}}^1\frac{\arctan(x)}{x}dx$$
$$I=\frac{\pi}{2}\ln(1+a)-\left(\int_0^1-\int_0^\frac{1-a}{1+a}\right)\frac{\arctan(x)}{x}dx$$
$$I=\frac{\pi}{2}\ln(1+a)-G+\text{Ti}\left(\frac{1-a}{1+a}\right),$$
and so
$$\int_0^1\frac{\ln(a^2+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln(1+a)-G-\text{Ti}\left(\frac{1-a}{1+a}\right).$$
Note that $\text{Ti}(-x)=-\text{Ti}(x)$ since $\arctan(-x)=-\arctan(x).$
