Integrand size = 50, antiderivative size = 85 \[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \operatorname {PolyLog}\left (2,1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d}-\frac {\operatorname {PolyLog}\left (3,1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d} \]
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Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2588, 6745} \[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{b c-a d}-\frac {\operatorname {PolyLog}\left (3,\frac {b c-a d}{d (a+b x)}+1\right )}{b c-a d} \]
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Rule 2588
Rule 6745
Rubi steps \begin{align*} \text {integral}& = \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \text {Li}_2\left (1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d}+\int \frac {\text {Li}_2\left (1-\frac {-b c+a d}{d (a+b x)}\right )}{(a+b x) (c+d x)} \, dx \\ & = \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \text {Li}_2\left (1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d}-\frac {\text {Li}_3\left (1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )-\operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b c-a d} \]
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Time = 3.80 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.84
method | result | size |
default | \(\frac {\frac {\ln \left (-\frac {\frac {e \left (d x +c \right ) b}{b x +a}-d e}{d e}\right ) \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2}}{2}-\frac {\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2} \ln \left (1-\frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )}{2}-\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) \operatorname {Li}_{2}\left (\frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )+\operatorname {Li}_{3}\left (\frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )}{a d -c b}\) | \(156\) |
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\[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {b c - a d}{{\left (b x + a\right )} d}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {b \int \frac {\log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}^{2}}{a + b x}\, dx}{2 \left (a d - b c\right )} + \frac {\log {\left (\frac {a d - b c}{d \left (a + b x\right )} \right )} \log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )}^{2}}{2 a d - 2 b c} \]
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\[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {b c - a d}{{\left (b x + a\right )} d}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {b c - a d}{{\left (b x + a\right )} d}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {\log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int \frac {\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\,\ln \left (\frac {a\,d-b\,c}{d\,\left (a+b\,x\right )}\right )}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]
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