Integrand size = 9, antiderivative size = 12 \[ \int -\frac {24}{-20+5 x} \, dx=5-\frac {24}{5} (-4+\log (-4+x)) \]
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Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 31} \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24}{5} \log (4-x) \]
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Rule 12
Rule 31
Rubi steps \begin{align*} \text {integral}& = -\left (24 \int \frac {1}{-20+5 x} \, dx\right ) \\ & = -\frac {24}{5} \log (4-x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24}{5} \log (20-5 x) \]
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Time = 0.33 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58
method | result | size |
default | \(-\frac {24 \ln \left (x -4\right )}{5}\) | \(7\) |
risch | \(-\frac {24 \ln \left (x -4\right )}{5}\) | \(7\) |
parallelrisch | \(-\frac {24 \ln \left (x -4\right )}{5}\) | \(7\) |
norman | \(-\frac {24 \ln \left (5 x -20\right )}{5}\) | \(9\) |
meijerg | \(-\frac {24 \ln \left (-\frac {x}{4}+1\right )}{5}\) | \(9\) |
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none
Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24}{5} \, \log \left (x - 4\right ) \]
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Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int -\frac {24}{-20+5 x} \, dx=- \frac {24 \log {\left (5 x - 20 \right )}}{5} \]
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none
Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24}{5} \, \log \left (x - 4\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58 \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24}{5} \, \log \left ({\left | x - 4 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.50 \[ \int -\frac {24}{-20+5 x} \, dx=-\frac {24\,\ln \left (x-4\right )}{5} \]
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