Integrand size = 17, antiderivative size = 37 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {a}{2 b^2 \left (a+b e^{2 x}\right )}+\frac {\log \left (a+b e^{2 x}\right )}{2 b^2} \]
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Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2280, 45} \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {a}{2 b^2 \left (a+b e^{2 x}\right )}+\frac {\log \left (a+b e^{2 x}\right )}{2 b^2} \]
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Rule 45
Rule 2280
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(a+b x)^2} \, dx,x,e^{2 x}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^2}+\frac {1}{b (a+b x)}\right ) \, dx,x,e^{2 x}\right ) \\ & = \frac {a}{2 b^2 \left (a+b e^{2 x}\right )}+\frac {\log \left (a+b e^{2 x}\right )}{2 b^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {\frac {a}{a+b e^{2 x}}+\log \left (b \left (a+b e^{2 x}\right )\right )}{2 b^2} \]
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Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\) | \(32\) |
norman | \(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\) | \(32\) |
risch | \(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {a}{b}\right )}{2 b^{2}}\) | \(34\) |
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none
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {{\left (b e^{\left (2 \, x\right )} + a\right )} \log \left (b e^{\left (2 \, x\right )} + a\right ) + a}{2 \, {\left (b^{3} e^{\left (2 \, x\right )} + a b^{2}\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {a}{2 a b^{2} + 2 b^{3} e^{2 x}} + \frac {\log {\left (\frac {a}{b} + e^{2 x} \right )}}{2 b^{2}} \]
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none
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {a}{2 \, {\left (b^{3} e^{\left (2 \, x\right )} + a b^{2}\right )}} + \frac {\log \left (b e^{\left (2 \, x\right )} + a\right )}{2 \, b^{2}} \]
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Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {\log \left ({\left | b e^{\left (2 \, x\right )} + a \right |}\right )}{2 \, b^{2}} + \frac {a}{2 \, {\left (b e^{\left (2 \, x\right )} + a\right )} b^{2}} \]
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Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 x}}{\left (a+b e^{2 x}\right )^2} \, dx=\frac {\ln \left (a+b\,{\mathrm {e}}^{2\,x}\right )}{2\,b^2}-\frac {{\mathrm {e}}^{2\,x}}{2\,b\,\left (a+b\,{\mathrm {e}}^{2\,x}\right )} \]
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