Integrand size = 62, antiderivative size = 109 \[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \operatorname {PolyLog}\left (2,1+\frac {(b c-a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}-\frac {\operatorname {PolyLog}\left (3,1+\frac {(b c-a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d} \]
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Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {2588, 6745} \[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \operatorname {PolyLog}\left (2,\frac {(b c-a d) (e+f x)}{(d e-c f) (a+b x)}+1\right )}{b c-a d}-\frac {\operatorname {PolyLog}\left (3,\frac {(b c-a d) (e+f x)}{(d e-c f) (a+b x)}+1\right )}{b c-a d} \]
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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) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}+\int \frac {\text {Li}_2\left (1-\frac {(-b c+a d) (e+f x)}{(d e-c f) (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) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}-\frac {\text {Li}_3\left (1+\frac {(b c-a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \operatorname {PolyLog}\left (2,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )-\operatorname {PolyLog}\left (3,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{b c-a d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(422\) vs. \(2(109)=218\).
Time = 25.00 (sec) , antiderivative size = 423, normalized size of antiderivative = 3.88
method | result | size |
default | \(\frac {\frac {\ln \left (-\frac {\frac {e \left (d x +c \right ) a f}{b x +a}-\frac {e^{2} \left (d x +c \right ) b}{b x +a}-c e f +d \,e^{2}}{e \left (c f -d e \right )}\right ) \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2}}{2}-\frac {a f \left (\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2} \ln \left (1-\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )+2 \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) \operatorname {Li}_{2}\left (\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )-2 \,\operatorname {Li}_{3}\left (\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )\right )}{2 \left (a f -b e \right )}+\frac {b e \left (\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2} \ln \left (1-\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )+2 \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) \operatorname {Li}_{2}\left (\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )-2 \,\operatorname {Li}_{3}\left (\frac {\left (a f -b e \right ) e \left (d x +c \right )}{\left (b x +a \right ) \left (c e f -d \,e^{2}\right )}\right )\right )}{2 a f -2 b e}}{a d -c b}\) | \(423\) |
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\[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {{\left (b c - a d\right )} {\left (f x + e\right )}}{{\left (d e - c f\right )} {\left (b x + a\right )}}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\frac {\log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )}^{2} \log {\left (\frac {\left (e + f x\right ) \left (a d - b c\right )}{\left (a + b x\right ) \left (- c f + d e\right )} \right )}}{2 a d - 2 b c} - \frac {\left (a f - b e\right ) \int \frac {\log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}^{2}}{a e + a f x + b e x + b f x^{2}}\, dx}{2 \left (a d - b c\right )} \]
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\[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {{\left (b c - a d\right )} {\left (f x + e\right )}}{{\left (d e - c f\right )} {\left (b x + a\right )}}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac {{\left (b c - a d\right )} {\left (f x + e\right )}}{{\left (d e - c f\right )} {\left (b x + a\right )}}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \log \left (\frac {(-b c+a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{(a+b x) (c+d x)} \, dx=\int \frac {\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\,\ln \left (-\frac {\left (e+f\,x\right )\,\left (a\,d-b\,c\right )}{\left (c\,f-d\,e\right )\,\left (a+b\,x\right )}\right )}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]
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