Integrand size = 15, antiderivative size = 107 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=-\frac {x}{a^2 d}+\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2} \]
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Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2216, 2215, 2221, 2317, 2438, 2222, 2320, 36, 29, 31} \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac {x}{a^2 d}+\frac {x^2}{2 a^2}+\frac {x}{a d \left (a+b e^{c+d x}\right )} \]
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Rule 29
Rule 31
Rule 36
Rule 2215
Rule 2216
Rule 2221
Rule 2222
Rule 2317
Rule 2320
Rule 2438
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x}{a+b e^{c+d x}} \, dx}{a}-\frac {b \int \frac {e^{c+d x} x}{\left (a+b e^{c+d x}\right )^2} \, dx}{a} \\ & = \frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {b \int \frac {e^{c+d x} x}{a+b e^{c+d x}} \, dx}{a^2}-\frac {\int \frac {1}{a+b e^{c+d x}} \, dx}{a d} \\ & = \frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,e^{c+d x}\right )}{a d^2}+\frac {\int \log \left (1+\frac {b e^{c+d x}}{a}\right ) \, dx}{a^2 d} \\ & = \frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^2}+\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx,x,e^{c+d x}\right )}{a^2 d^2}+\frac {b \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,e^{c+d x}\right )}{a^2 d^2} \\ & = -\frac {x}{a^2 d}+\frac {x}{a d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a^2}+\frac {\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}-\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {\frac {d x \left (a d x+b e^{c+d x} (-2+d x)\right )}{a+b e^{c+d x}}-2 (-1+d x) \log \left (1+\frac {b e^{c+d x}}{a}\right )-2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{2 a^2 d^2} \]
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Time = 0.03 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {-\frac {\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-\frac {\left (d x +c \right ) \ln \left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}-\frac {b \left (d x +c \right ) {\mathrm e}^{d x +c}}{a^{2} \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\left (d x +c \right )^{2}}{2 a^{2}}-c \left (-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}\right )}{d^{2}}\) | \(154\) |
default | \(\frac {-\frac {\operatorname {dilog}\left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}-\frac {\left (d x +c \right ) \ln \left (\frac {a +b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}-\frac {b \left (d x +c \right ) {\mathrm e}^{d x +c}}{a^{2} \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\left (d x +c \right )^{2}}{2 a^{2}}-c \left (-\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{2}}\right )}{d^{2}}\) | \(154\) |
risch | \(\frac {x}{a d \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {x^{2}}{2 a^{2}}+\frac {c x}{d \,a^{2}}+\frac {c^{2}}{2 d^{2} a^{2}}-\frac {x \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d}-\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c}{d^{2} a^{2}}-\frac {\operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{2}}+\frac {\ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d^{2}}-\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a^{2}}+\frac {c \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{d^{2} a^{2}}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a^{2}}\) | \(186\) |
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Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.64 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {a d^{2} x^{2} - a c^{2} - 2 \, a c - 2 \, {\left (b e^{\left (d x + c\right )} + a\right )} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) + {\left (b d^{2} x^{2} - b c^{2} - 2 \, b d x - 2 \, b c\right )} e^{\left (d x + c\right )} + 2 \, {\left (a c + {\left (b c + b\right )} e^{\left (d x + c\right )} + a\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \, {\left (a d x + a c + {\left (b d x + b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right )}{2 \, {\left (a^{2} b d^{2} e^{\left (d x + c\right )} + a^{3} d^{2}\right )}} \]
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\[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x}{a^{2} d + a b d e^{c + d x}} + \frac {\int \frac {d x}{a + b e^{c} e^{d x}}\, dx + \int \left (- \frac {1}{a + b e^{c} e^{d x}}\right )\, dx}{a d} \]
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Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac {x^{2}}{2 \, a^{2}} - \frac {x}{a^{2} d} - \frac {d x \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right )}{a^{2} d^{2}} + \frac {\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{2} d^{2}} \]
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\[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\int { \frac {x}{{\left (b e^{\left (d x + c\right )} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x}{\left (a+b e^{c+d x}\right )^2} \, dx=\int \frac {x}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^2} \,d x \]
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