Methods to evaluate $ \int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx +p}}{dx} $ Today I saw a question with an answer that made me rethink of the following question, since it's not the first time I try to find an answer to it. If you look at the answer of Mhenni Benghorbal
here you'll see $2$ interesting integrals, namely:
$$  \int _{0 }^{\infty }\!{\frac {\ln  \left( u \right) }{2+{u}^{2}-
2\,u}}{du} ; \int _{0}^{\infty }\!{\frac {\ln  \left( z \right) }{2+{z}^{2}+2\,z}}dz $$
I try to find out if  there is a well defined strategy to tackle such integrals.   In a more general sense, we have to deal with:
$$  \int _{a }^{b }\!{\frac {\ln  \left( tx + u \right) }{m{x}^{2}+nx
+p}}{dx}   $$
Could you help here? Thanks.
 A: Have you seen http://www.recreatiimatematice.ro/arhiva/articole/RM12011DICU.pdf
For partial response
A: 
The answer below will restrict itself to the evaluation of integrals of the form $\int_{a}^{b}\mathrm{d}x\,\frac{\ln{\left(cx+d\right)}}{px^{2}+qx+r}$ for the case in which $q^{2}-4pr<0$ since the alternative cases are more than adequately addressed elsewhere.

Define the function $\mathcal{I}:\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the definite integral
$$\begin{align}
\mathcal{I}{\left(a,b,c,z\right)}
&:=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}.\\
\end{align}$$
Also, define the Clausen function (of order 2) for real arguments by the integral representation
$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$
We show below that the integral $\mathcal{I}$ can be evaluated in terms of Clausen functions and elementary functions.

Suppose $\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$, and set
$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\beta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$
Then,
$$\begin{align}
\mathcal{I}{\left(a,b,c,z\right)}
&=\left[\arctan{\left(\frac{z+a}{b}\right)}-\arctan{\left(\frac{a}{b}\right)}\right]\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}.\\
\end{align}$$
Proof:
$$\begin{align}
\mathcal{I}{\left(a,b,c,z\right)}
&=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}\\
&=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(\left|x+c\right|\right)}}{\left(x+a\right)^{2}+b^{2}}\\
&=\int_{\frac{a}{b}}^{\frac{z+a}{b}}\mathrm{d}y\,\frac{2b^{2}\ln{\left(\left|by-a+c\right|\right)}}{\left(by-a+a\right)^{2}+b^{2}};~~~\small{\left[x=by-a\right]}\\
&=\int_{\frac{a}{b}}^{\frac{z+a}{b}}\mathrm{d}y\,\frac{2\ln{\left(\left|by+c-a\right|\right)}}{y^{2}+1}\\
&=\int_{\tan{\left(\alpha\right)}}^{\tan{\left(\beta\right)}}\mathrm{d}y\,\frac{2\ln{\left(\left|by+b\tan{\left(\gamma\right)}\right|\right)}}{y^{2}+1}\\
&=\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\frac{2\sec^{2}{\left(\varphi\right)}\ln{\left(\left|b\tan{\left(\varphi\right)}+b\tan{\left(\gamma\right)}\right|\right)}}{\tan^{2}{\left(\varphi\right)}+1};~~~\small{\left[y=\tan{\left(\varphi\right)}\right]}\\
&=2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|b\tan{\left(\varphi\right)}+b\tan{\left(\gamma\right)}\right|\right)}\\
&=2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\tan{\left(\varphi\right)}+\tan{\left(\gamma\right)}\right|\right)}\\
&=2\ln{\left(b\right)}\int_{\alpha}^{\beta}\mathrm{d}\varphi+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi\right)}\cos{\left(\gamma\right)}+\cos{\left(\varphi\right)}\sin{\left(\gamma\right)}}{\cos{\left(\varphi\right)}\cos{\left(\gamma\right)}}\right|\right)}\\
&=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi+\gamma\right)}}{\cos{\left(\varphi\right)}\cos{\left(\gamma\right)}}\right|\right)}\\
&=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{1}{\cos{\left(\gamma\right)}}\right|\right)}\\
&~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi+\gamma\right)}}{\cos{\left(\varphi\right)}}\right|\right)}\\
&=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\left(\beta-\alpha\right)\ln{\left(\left|\frac{1}{\cos{\left(\gamma\right)}}\right|\right)}\\
&~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{2\sin{\left(\varphi+\gamma\right)}}{2\cos{\left(\varphi\right)}}\right|\right)}\\
&=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(\frac{1}{\cos^{2}{\left(\gamma\right)}}\right)}\\
&~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\frac{\left|2\sin{\left(\varphi+\gamma\right)}\right|}{\left|2\cos{\left(\varphi\right)}\right|}\right)}\\
&=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(\sec^{2}{\left(\gamma\right)}\right)}\\
&~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\varphi+\gamma\right)}\right|\right)}\\
&~~~~~-2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|2\cos{\left(\varphi\right)}\right|\right)}\\
&=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(1+\tan^{2}{\left(\gamma\right)}\right)}\\
&~~~~~+2\int_{\alpha+\gamma}^{\beta+\gamma}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\varphi\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\varphi-\gamma\right]}\\
&~~~~~+2\int_{\frac{\pi}{2}-\beta}^{\frac{\pi}{2}-\alpha}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\varphi\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\frac{\pi}{2}-\varphi\right]}\\
&=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(1+\left(\frac{c-a}{b}\right)^{2}\right)}\\
&~~~~~+\int_{2\beta+2\gamma}^{2\alpha+2\gamma}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)}\\
&~~~~~+\int_{\pi-2\beta}^{\pi-2\alpha}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\frac{\varphi}{2}\right]}\\
&=\left(\beta-\alpha\right)\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}\\
&=\left[\arctan{\left(\frac{z+a}{b}\right)}-\arctan{\left(\frac{a}{b}\right)}\right]\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\
&~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}\blacksquare.\\
\end{align}$$

A: Related problems: (I), (II). You can use the partial fraction technique combined with the use the dilogarithm function $\operatorname{Li}_{2}(x)$, which is defined by

$$\operatorname{Li}_{2}(x) = \int_{1}^{x} \frac{\ln(t)}{1-t} \,dt \,.$$

Here is an example,
$$ \int_{a}^{b} \frac{\ln(x)}{cx+d}dx =- \frac{1}{d}\left( \operatorname{Li}_{2}\left( {\frac {c+da}{c}} \right) +\ln  \left(a\right) \ln  \left( {\frac {c+da}{c}} \right) -\operatorname{Li}_{2} \left( {
\frac {c+bd}{c}} \right) -\ln  \left( b \right) \ln  \left( {\frac {c+
bd}{c}} \right)  \right) 
$$
Note that the above integral is undefined for 
$$ \left(a < -\frac{c}{d}, -\frac{c}{d} < b \right)  $$
