Integrand size = 14, antiderivative size = 15 \[ \int \frac {e^{-t}}{-1-a+t} \, dt=e^{-1-a} \operatorname {ExpIntegralEi}(1+a-t) \]
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Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2209} \[ \int \frac {e^{-t}}{-1-a+t} \, dt=e^{-a-1} \operatorname {ExpIntegralEi}(a-t+1) \]
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Rule 2209
Rubi steps \begin{align*} \text {integral}& = e^{-1-a} \operatorname {ExpIntegralEi}(1+a-t) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-t}}{-1-a+t} \, dt=e^{-1-a} \operatorname {ExpIntegralEi}(1+a-t) \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13
method | result | size |
default | \(-{\mathrm e}^{-1-a} \operatorname {Ei}_{1}\left (-1-a +t \right )\) | \(17\) |
risch | \(-{\mathrm e}^{-1-a} \operatorname {Ei}_{1}\left (-1-a +t \right )\) | \(17\) |
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none
Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-t}}{-1-a+t} \, dt={\rm Ei}\left (a - t + 1\right ) e^{\left (-a - 1\right )} \]
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\[ \int \frac {e^{-t}}{-1-a+t} \, dt=\int \frac {e^{- t}}{- a + t - 1}\, dt \]
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none
Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-t}}{-1-a+t} \, dt=-e^{\left (-a - 1\right )} E_{1}\left (-a + t - 1\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-t}}{-1-a+t} \, dt={\rm Ei}\left (a - t + 1\right ) e^{\left (-a - 1\right )} \]
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Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-t}}{-1-a+t} \, dt={\mathrm {e}}^{-a-1}\,\mathrm {ei}\left (a-t+1\right ) \]
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