Integrand size = 55, antiderivative size = 150 \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=-\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 \operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g} \]
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Time = 0.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {2588, 2590, 6745} \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=-\frac {\operatorname {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}+\frac {2 \operatorname {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}-\frac {2 \operatorname {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]
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Rule 2588
Rule 2590
Rule 6745
Rubi steps \begin{align*} \text {integral}& = -\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \int \frac {\log \left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g} \\ & = -\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 \int \frac {\text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g} \\ & = -\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 \text {Li}_4\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.73 \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\frac {-\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )+2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \operatorname {PolyLog}\left (3,\frac {d (a+b x)}{b (c+d x)}\right )-2 \operatorname {PolyLog}\left (4,\frac {d (a+b x)}{b (c+d x)}\right )}{(b c-a d) g} \]
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Time = 4.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.33
method | result | size |
default | \(-\frac {\frac {\ln \left (-\frac {\frac {e \left (b x +a \right ) d}{d x +c}-b e}{b e}\right ) \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{3}}{3}-\frac {\ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{3} \ln \left (1-\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )}{3}-\ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )^{2} \operatorname {Li}_{2}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )+2 \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right ) \operatorname {Li}_{3}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )-2 \,\operatorname {Li}_{4}\left (\frac {d \left (b x +a \right )}{b \left (d x +c \right )}\right )}{g \left (a d -c b \right )}\) | \(200\) |
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\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=- \frac {d \int \frac {\log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}^{3}}{c + d x}\, dx}{3 g \left (a d - b c\right )} - \frac {\log {\left (\frac {- a d + b c}{b \left (c + d x\right )} \right )} \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}^{3}}{3 a d g - 3 b c g} \]
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\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int { \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx=\int \frac {{\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}^2\,\ln \left (-\frac {a\,d-b\,c}{b\,\left (c+d\,x\right )}\right )}{\left (a\,g+b\,g\,x\right )\,\left (c+d\,x\right )} \,d x \]
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