Integrand size = 33, antiderivative size = 23 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=\frac {2 \left (-3+x+\frac {4+x}{9}\right )}{5 (-16+2 x+\log (2))} \]
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Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 2006, 27, 32} \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {2 (57-\log (32))}{45 (-2 x+16-\log (2))} \]
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Rule 12
Rule 27
Rule 32
Rule 2006
Rubi steps \begin{align*} \text {integral}& = -\left ((4 (57-\log (32))) \int \frac {1}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx\right ) \\ & = -\left ((4 (57-\log (32))) \int \frac {1}{180 x^2-180 x (16-\log (2))+45 (16-\log (2))^2} \, dx\right ) \\ & = -\left ((4 (57-\log (32))) \int \frac {1}{45 (-16+2 x+\log (2))^2} \, dx\right ) \\ & = -\left (\frac {1}{45} (4 (57-\log (32))) \int \frac {1}{(-16+2 x+\log (2))^2} \, dx\right ) \\ & = -\frac {2 (57-\log (32))}{45 (16-2 x-\log (2))} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {2 (-57+\log (32))}{45 (-16+2 x+\log (2))} \]
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Time = 0.48 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74
method | result | size |
norman | \(\frac {\frac {38}{15}-\frac {2 \ln \left (2\right )}{9}}{\ln \left (2\right )+2 x -16}\) | \(17\) |
gosper | \(-\frac {2 \left (5 \ln \left (2\right )-57\right )}{45 \left (\ln \left (2\right )+2 x -16\right )}\) | \(18\) |
default | \(-\frac {20 \ln \left (2\right )-228}{90 \left (\ln \left (2\right )+2 x -16\right )}\) | \(18\) |
parallelrisch | \(-\frac {20 \ln \left (2\right )-228}{90 \left (\ln \left (2\right )+2 x -16\right )}\) | \(18\) |
risch | \(\frac {38}{15 \left (\ln \left (2\right )+2 x -16\right )}-\frac {2 \ln \left (2\right )}{9 \left (\ln \left (2\right )+2 x -16\right )}\) | \(26\) |
meijerg | \(\frac {38 x}{15 \left (\frac {\ln \left (2\right )}{2}-8\right ) \left (-\ln \left (2\right )+16\right ) \left (1-\frac {2 x}{-\ln \left (2\right )+16}\right )}-\frac {2 \ln \left (2\right ) x}{9 \left (\frac {\ln \left (2\right )}{2}-8\right ) \left (-\ln \left (2\right )+16\right ) \left (1-\frac {2 x}{-\ln \left (2\right )+16}\right )}\) | \(72\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {2 \, {\left (5 \, \log \left (2\right ) - 57\right )}}{45 \, {\left (2 \, x + \log \left (2\right ) - 16\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=- \frac {-228 + 20 \log {\left (2 \right )}}{180 x - 1440 + 90 \log {\left (2 \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {2 \, {\left (5 \, \log \left (2\right ) - 57\right )}}{45 \, {\left (2 \, x + \log \left (2\right ) - 16\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {2 \, {\left (5 \, \log \left (2\right ) - 57\right )}}{45 \, {\left (2 \, x + \log \left (2\right ) - 16\right )}} \]
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Time = 11.91 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx=-\frac {\frac {2\,\ln \left (2\right )}{9}-\frac {38}{15}}{2\,x+\ln \left (2\right )-16} \]
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