Integrand size = 11, antiderivative size = 42 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {1}{a \left (a+b e^{p x}\right ) p}+\frac {x}{a^2}-\frac {\log \left (a+b e^{p x}\right )}{a^2 p} \]
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Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2320, 46} \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=-\frac {\log \left (a+b e^{p x}\right )}{a^2 p}+\frac {x}{a^2}+\frac {1}{a p \left (a+b e^{p x}\right )} \]
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Rule 46
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^2} \, dx,x,e^{p x}\right )}{p} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {b}{a (a+b x)^2}-\frac {b}{a^2 (a+b x)}\right ) \, dx,x,e^{p x}\right )}{p} \\ & = \frac {1}{a \left (a+b e^{p x}\right ) p}+\frac {x}{a^2}-\frac {\log \left (a+b e^{p x}\right )}{a^2 p} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {\frac {a}{a+b e^{p x}}+\log \left (e^{p x}\right )-\log \left (a+b e^{p x}\right )}{a^2 p} \]
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Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right )}+\frac {\ln \left ({\mathrm e}^{p x}\right )}{a^{2}}}{p}\) | \(43\) |
default | \(\frac {-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right )}+\frac {\ln \left ({\mathrm e}^{p x}\right )}{a^{2}}}{p}\) | \(43\) |
risch | \(\frac {x}{a^{2}}+\frac {1}{a \left (a +b \,{\mathrm e}^{p x}\right ) p}-\frac {\ln \left ({\mathrm e}^{p x}+\frac {a}{b}\right )}{a^{2} p}\) | \(43\) |
norman | \(\frac {\frac {x}{a}+\frac {b x \,{\mathrm e}^{p x}}{a^{2}}-\frac {b \,{\mathrm e}^{p x}}{a^{2} p}}{a +b \,{\mathrm e}^{p x}}-\frac {\ln \left (a +b \,{\mathrm e}^{p x}\right )}{a^{2} p}\) | \(59\) |
parallelrisch | \(-\frac {-b^{2} {\mathrm e}^{p x} x p +\ln \left (a +b \,{\mathrm e}^{p x}\right ) {\mathrm e}^{p x} b^{2}-x a b p +\ln \left (a +b \,{\mathrm e}^{p x}\right ) a b -a b}{a^{2} b p \left (a +b \,{\mathrm e}^{p x}\right )}\) | \(73\) |
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Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {b p x e^{\left (p x\right )} + a p x - {\left (b e^{\left (p x\right )} + a\right )} \log \left (b e^{\left (p x\right )} + a\right ) + a}{a^{2} b p e^{\left (p x\right )} + a^{3} p} \]
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Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {1}{a^{2} p + a b p e^{p x}} + \frac {x}{a^{2}} - \frac {\log {\left (\frac {a}{b} + e^{p x} \right )}}{a^{2} p} \]
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Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {x}{a^{2}} + \frac {1}{{\left (a b e^{\left (p x\right )} + a^{2}\right )} p} - \frac {\log \left (b e^{\left (p x\right )} + a\right )}{a^{2} p} \]
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Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {b {\left (\frac {\log \left ({\left | -\frac {a}{b e^{\left (p x\right )} + a} + 1 \right |}\right )}{a^{2} b} + \frac {1}{{\left (b e^{\left (p x\right )} + a\right )} a b}\right )}}{p} \]
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Time = 0.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (a+b e^{p x}\right )^2} \, dx=\frac {\frac {x}{a}+\frac {b\,x\,{\mathrm {e}}^{p\,x}}{a^2}-\frac {b\,{\mathrm {e}}^{p\,x}}{a^2\,p}}{a+b\,{\mathrm {e}}^{p\,x}}-\frac {\ln \left (a+b\,{\mathrm {e}}^{p\,x}\right )}{a^2\,p} \]
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