Integrand size = 18, antiderivative size = 21 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=x \left (-1+2 \left (4-\frac {1}{25} (-5-2 x)^2+\log (x)\right )\right ) \]
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Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2332} \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8 x^3}{25}-\frac {8 x^2}{5}+5 x+2 x \log (x) \]
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Rule 12
Rule 2332
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx \\ & = 7 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 \int \log (x) \, dx \\ & = 5 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 x \log (x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=5 x-\frac {8 x^2}{5}-\frac {8 x^3}{25}+2 x \log (x) \]
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Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\) | \(20\) |
norman | \(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\) | \(20\) |
risch | \(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\) | \(20\) |
parallelrisch | \(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\) | \(20\) |
parts | \(-\frac {8 x^{3}}{25}-\frac {8 x^{2}}{5}+5 x +2 x \ln \left (x \right )\) | \(20\) |
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=- \frac {8 x^{3}}{25} - \frac {8 x^{2}}{5} + 2 x \log {\left (x \right )} + 5 x \]
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Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {8}{25} \, x^{3} - \frac {8}{5} \, x^{2} + 2 \, x \log \left (x\right ) + 5 \, x \]
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Time = 9.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{25} \left (175-80 x-24 x^2+50 \log (x)\right ) \, dx=-\frac {x\,\left (40\,x-50\,\ln \left (x\right )+8\,x^2-125\right )}{25} \]
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