Integrand size = 22, antiderivative size = 54 \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {\operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)} \]
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Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2221, 2317, 2438} \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]
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Rule 2221
Rule 2317
Rule 2438
Rubi steps \begin{align*} \text {integral}& = \frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac {\int \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{b d \log (F)} \\ & = \frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{b d^2 \log ^2(F)} \\ & = \frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {\text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {\operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(54)=108\).
Time = 0.04 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.85
method | result | size |
risch | \(-\frac {c x}{d b}-\frac {c^{2}}{2 d^{2} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x}{d \ln \left (F \right ) b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c}{d^{2} \ln \left (F \right ) b}+\frac {\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{d^{2} \ln \left (F \right )^{2} b}-\frac {c \ln \left (F^{c} F^{d x} b +a \right )}{d^{2} \ln \left (F \right ) b}+\frac {c \ln \left (F^{d x} F^{c}\right )}{d^{2} \ln \left (F \right ) b}\) | \(154\) |
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Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=-\frac {c \log \left (F^{d x + c} b + a\right ) \log \left (F\right ) - {\left (d x + c\right )} \log \left (F\right ) \log \left (\frac {F^{d x + c} b + a}{a}\right ) - {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right )}{b d^{2} \log \left (F\right )^{2}} \]
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\[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\int \frac {F^{c + d x} x}{F^{c + d x} b + a}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\frac {d x \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right )}{b d^{2} \log \left (F\right )^{2}} \]
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\[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\int { \frac {F^{d x + c} x}{F^{d x + c} b + a} \,d x } \]
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Timed out. \[ \int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx=\int \frac {F^{c+d\,x}\,x}{a+F^{c+d\,x}\,b} \,d x \]
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