Integrand size = 14, antiderivative size = 52 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=-\frac {4 e^{x/2}}{3}-\frac {4}{9} e^{3 x/2}+\frac {4}{3} \text {arctanh}\left (e^{x/2}\right )+\frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right ) \]
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Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2225, 2634, 12, 2280, 308, 213} \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {4}{3} \text {arctanh}\left (e^{x/2}\right )-\frac {4 e^{x/2}}{3}-\frac {4}{9} e^{3 x/2}+\frac {2}{3} e^{3 x/2} \log \left (e^x-1\right ) \]
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Rule 12
Rule 213
Rule 308
Rule 2225
Rule 2280
Rule 2634
Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\int \frac {2 e^{5 x/2}}{3 \left (-1+e^x\right )} \, dx \\ & = \frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac {2}{3} \int \frac {e^{5 x/2}}{-1+e^x} \, dx \\ & = \frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac {4}{3} \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,e^{x/2}\right ) \\ & = \frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac {4}{3} \text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,e^{x/2}\right ) \\ & = -\frac {4 e^{x/2}}{3}-\frac {4}{9} e^{3 x/2}+\frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right )-\frac {4}{3} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{x/2}\right ) \\ & = -\frac {4 e^{x/2}}{3}-\frac {4}{9} e^{3 x/2}+\frac {4}{3} \text {arctanh}\left (e^{x/2}\right )+\frac {2}{3} e^{3 x/2} \log \left (-1+e^x\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {2}{9} \left (6 \text {arctanh}\left (e^{x/2}\right )+e^{x/2} \left (-2 \left (3+e^x\right )+3 e^x \log \left (-1+e^x\right )\right )\right ) \]
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Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{\frac {3 x}{2}} \ln \left (-1+{\mathrm e}^{x}\right )}{3}-\frac {4 \,{\mathrm e}^{\frac {3 x}{2}}}{9}-\frac {4 \,{\mathrm e}^{\frac {x}{2}}}{3}+\frac {2 \ln \left ({\mathrm e}^{\frac {x}{2}}+1\right )}{3}-\frac {2 \ln \left (-1+{\mathrm e}^{\frac {x}{2}}\right )}{3}\) | \(43\) |
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} - 1\right ) \]
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Timed out. \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {2}{3} \, e^{\left (\frac {3}{2} \, x\right )} \log \left (e^{x} - 1\right ) - \frac {4}{9} \, e^{\left (\frac {3}{2} \, x\right )} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x\right )} + \frac {2}{3} \, \log \left (e^{\left (\frac {1}{2} \, x\right )} + 1\right ) - \frac {2}{3} \, \log \left ({\left | e^{\left (\frac {1}{2} \, x\right )} - 1 \right |}\right ) \]
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Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int e^{3 x/2} \log \left (-1+e^x\right ) \, dx=\frac {4\,\mathrm {atanh}\left (\sqrt {{\mathrm {e}}^x}\right )}{3}-\frac {4\,{\mathrm {e}}^{\frac {3\,x}{2}}}{9}-\frac {4\,{\mathrm {e}}^{x/2}}{3}+\frac {2\,{\mathrm {e}}^{\frac {3\,x}{2}}\,\ln \left ({\mathrm {e}}^x-1\right )}{3} \]
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