Integrand size = 14, antiderivative size = 12 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]
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Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2437, 2338} \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]
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Rule 2338
Rule 2437
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\log (t)}{t} \, dt,t,1-t\right ) \\ & = -\frac {1}{2} \log ^2(1-t) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \log ^2(1-t) \]
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Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(-\frac {\ln \left (1-t \right )^{2}}{2}\) | \(11\) |
default | \(-\frac {\ln \left (1-t \right )^{2}}{2}\) | \(11\) |
norman | \(-\frac {\ln \left (1-t \right )^{2}}{2}\) | \(11\) |
risch | \(-\frac {\ln \left (1-t \right )^{2}}{2}\) | \(11\) |
parts | \(-\ln \left (-1+t \right ) \ln \left (1-t \right )+\frac {\ln \left (-1+t \right )^{2}}{2}\) | \(22\) |
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none
Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]
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Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {\log (1-t)}{1-t} \, dt=- \frac {\log {\left (1 - t \right )}^{2}}{2} \]
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none
Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]
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none
Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {1}{2} \, \log \left (-t + 1\right )^{2} \]
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Time = 0.37 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {\log (1-t)}{1-t} \, dt=-\frac {{\ln \left (1-t\right )}^2}{2} \]
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