Integrand size = 8, antiderivative size = 18 \[ \int e^t \log (1+t) \, dt=-\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \]
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Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2225, 2634, 2209} \[ \int e^t \log (1+t) \, dt=e^t \log (t+1)-\frac {\operatorname {ExpIntegralEi}(t+1)}{e} \]
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Rule 2209
Rule 2225
Rule 2634
Rubi steps \begin{align*} \text {integral}& = e^t \log (1+t)-\int \frac {e^t}{1+t} \, dt \\ & = -\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^t \log (1+t) \, dt=-\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \]
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Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
risch | \({\mathrm e}^{t} \ln \left (1+t \right )+{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-t \right )\) | \(19\) |
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none
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int e^t \log (1+t) \, dt={\left (e^{\left (t + 1\right )} \log \left (t + 1\right ) - {\rm Ei}\left (t + 1\right )\right )} e^{\left (-1\right )} \]
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\[ \int e^t \log (1+t) \, dt=\int e^{t} \log {\left (t + 1 \right )}\, dt \]
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none
Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^t \log (1+t) \, dt=e^{\left (-1\right )} E_{1}\left (-t - 1\right ) + e^{t} \log \left (t + 1\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^t \log (1+t) \, dt=-{\rm Ei}\left (t + 1\right ) e^{\left (-1\right )} + e^{t} \log \left (t + 1\right ) \]
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Timed out. \[ \int e^t \log (1+t) \, dt=\int \ln \left (t+1\right )\,{\mathrm {e}}^t \,d t \]
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