Integrand size = 39, antiderivative size = 25 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=1-\frac {15}{\log (2) \left (25-\frac {5 (3-x)}{3+\log (4)}\right )} \]
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Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2006, 27, 32} \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {3 (3+\log (4))}{\log (2) (x+12+5 \log (4))} \]
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Rule 12
Rule 27
Rule 32
Rule 2006
Rubi steps \begin{align*} \text {integral}& = (9+3 \log (4)) \int \frac {1}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx \\ & = (9+3 \log (4)) \int \frac {1}{x^2 \log (2)+2 x \log (2) (12+5 \log (4))+\log (2) (12+5 \log (4))^2} \, dx \\ & = (9+3 \log (4)) \int \frac {1}{\log (2) (12+x+5 \log (4))^2} \, dx \\ & = \frac {(9+3 \log (4)) \int \frac {1}{(12+x+5 \log (4))^2} \, dx}{\log (2)} \\ & = -\frac {3 (3+\log (4))}{\log (2) (12+x+5 \log (4))} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {3 (3+\log (4))}{\log (2) (12+x+5 \log (4))} \]
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Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
gosper | \(-\frac {3 \left (2 \ln \left (2\right )+3\right )}{\ln \left (2\right ) \left (10 \ln \left (2\right )+x +12\right )}\) | \(22\) |
default | \(-\frac {6 \ln \left (2\right )+9}{\ln \left (2\right ) \left (10 \ln \left (2\right )+x +12\right )}\) | \(22\) |
norman | \(-\frac {3 \left (2 \ln \left (2\right )+3\right )}{\ln \left (2\right ) \left (10 \ln \left (2\right )+x +12\right )}\) | \(22\) |
parallelrisch | \(-\frac {6 \ln \left (2\right )+9}{\ln \left (2\right ) \left (10 \ln \left (2\right )+x +12\right )}\) | \(22\) |
risch | \(-\frac {6}{10 \ln \left (2\right )+x +12}-\frac {9}{\ln \left (2\right ) \left (10 \ln \left (2\right )+x +12\right )}\) | \(28\) |
meijerg | \(\frac {3 x}{\left (10 \ln \left (2\right )+12\right ) \left (5 \ln \left (2\right )+6\right ) \left (1+\frac {x}{10 \ln \left (2\right )+12}\right )}+\frac {9 x}{2 \ln \left (2\right ) \left (10 \ln \left (2\right )+12\right ) \left (5 \ln \left (2\right )+6\right ) \left (1+\frac {x}{10 \ln \left (2\right )+12}\right )}\) | \(74\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {3 \, {\left (2 \, \log \left (2\right ) + 3\right )}}{{\left (x + 12\right )} \log \left (2\right ) + 10 \, \log \left (2\right )^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=- \frac {6 \log {\left (2 \right )} + 9}{x \log {\left (2 \right )} + 10 \log {\left (2 \right )}^{2} + 12 \log {\left (2 \right )}} \]
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Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {3 \, {\left (2 \, \log \left (2\right ) + 3\right )}}{x \log \left (2\right ) + 10 \, \log \left (2\right )^{2} + 12 \, \log \left (2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {3 \, {\left (2 \, \log \left (2\right ) + 3\right )}}{{\left (x + 10 \, \log \left (2\right ) + 12\right )} \log \left (2\right )} \]
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Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {9+3 \log (4)}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx=-\frac {\ln \left (64\right )+9}{\ln \left (2\right )\,\left (x+10\,\ln \left (2\right )+12\right )} \]
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