Integrand size = 29, antiderivative size = 39 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=e^x+\frac {e^{2 x}}{2}-\arctan \left (e^x\right )+\log \left (1-e^x\right )-\frac {1}{2} \log \left (1+e^{2 x}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2320, 2099, 649, 209, 266} \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=-\arctan \left (e^x\right )+e^x+\frac {e^{2 x}}{2}+\log \left (1-e^x\right )-\frac {1}{2} \log \left (e^{2 x}+1\right ) \]
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Rule 209
Rule 266
Rule 649
Rule 2099
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-1-x^4}{1-x+x^2-x^3} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (1+\frac {1}{-1+x}+x+\frac {-1-x}{1+x^2}\right ) \, dx,x,e^x\right ) \\ & = e^x+\frac {e^{2 x}}{2}+\log \left (1-e^x\right )+\text {Subst}\left (\int \frac {-1-x}{1+x^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {e^{2 x}}{2}+\log \left (1-e^x\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {e^{2 x}}{2}-\arctan \left (e^x\right )+\log \left (1-e^x\right )-\frac {1}{2} \log \left (1+e^{2 x}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=e^x+\frac {e^{2 x}}{2}-\arctan \left (e^x\right )+\log \left (-1+e^x\right )-\frac {1}{2} \log \left (1+e^{2 x}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{2}-\arctan \left ({\mathrm e}^{x}\right )+\ln \left (-1+{\mathrm e}^{x}\right )+\frac {{\mathrm e}^{2 x}}{2}+{\mathrm e}^{x}\) | \(29\) |
risch | \(\frac {{\mathrm e}^{2 x}}{2}+{\mathrm e}^{x}-\frac {\ln \left ({\mathrm e}^{x}-i\right )}{2}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}+\ln \left (-1+{\mathrm e}^{x}\right )\) | \(49\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=-\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \]
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Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=\frac {e^{2 x}}{2} + e^{x} + \log {\left (e^{x} - 1 \right )} + \operatorname {RootSum} {\left (2 z^{2} + 2 z + 1, \left ( i \mapsto i \log {\left (\frac {4 i^{2}}{5} - \frac {6 i}{5} + e^{x} - \frac {3}{5} \right )} \right )\right )} \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=-\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=-\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left ({\left | e^{x} - 1 \right |}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx=\frac {{\mathrm {e}}^{2\,x}}{2}-\frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )}{2}-\mathrm {atan}\left ({\mathrm {e}}^x\right )+\ln \left ({\mathrm {e}}^x-1\right )+{\mathrm {e}}^x \]
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