Integrand size = 37, antiderivative size = 22 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x \left (5-9 x+\frac {x}{\log (4 x)}\right )}{5 \log (4)} \]
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Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {12, 6874, 2343, 2346, 2209} \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^2}{5 \log (4) \log (4 x)}-\frac {9 x^2}{5 \log (4)}+\frac {x}{\log (4)} \]
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Rule 12
Rule 2209
Rule 2343
Rule 2346
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{\log ^2(4 x)} \, dx}{5 \log (4)} \\ & = \frac {\int \left (5-18 x-\frac {x}{\log ^2(4 x)}+\frac {2 x}{\log (4 x)}\right ) \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}-\frac {\int \frac {x}{\log ^2(4 x)} \, dx}{5 \log (4)}+\frac {2 \int \frac {x}{\log (4 x)} \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)}+\frac {\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (4 x)\right )}{40 \log (4)}-\frac {2 \int \frac {x}{\log (4 x)} \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {\text {Ei}(2 \log (4 x))}{40 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)}-\frac {\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (4 x)\right )}{40 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {5 x-9 x^2+\frac {x^2}{\log (4 x)}}{5 \log (4)} \]
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Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {-9 x^{2}+5 x +\frac {x^{2}}{\ln \left (4 x \right )}}{10 \ln \left (2\right )}\) | \(26\) |
derivativedivides | \(\frac {-144 x^{2}+80 x +\frac {16 x^{2}}{\ln \left (4 x \right )}}{160 \ln \left (2\right )}\) | \(27\) |
risch | \(-\frac {9 x^{2}}{10 \ln \left (2\right )}+\frac {x}{2 \ln \left (2\right )}+\frac {x^{2}}{10 \ln \left (2\right ) \ln \left (4 x \right )}\) | \(33\) |
parallelrisch | \(\frac {-9 x^{2} \ln \left (4 x \right )+x^{2}+5 x \ln \left (4 x \right )}{10 \ln \left (2\right ) \ln \left (4 x \right )}\) | \(33\) |
norman | \(\frac {\frac {x^{2}}{10 \ln \left (2\right )}+\frac {x \ln \left (4 x \right )}{2 \ln \left (2\right )}-\frac {9 x^{2} \ln \left (4 x \right )}{10 \ln \left (2\right )}}{\ln \left (4 x \right )}\) | \(42\) |
parts | \(-\frac {9 x^{2}-5 x}{10 \ln \left (2\right )}-\frac {-\frac {x^{2}}{\ln \left (4 x \right )}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (4 x \right )\right )}{8}}{10 \ln \left (2\right )}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (4 x \right )\right )}{80 \ln \left (2\right )}\) | \(59\) |
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Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^{2} - {\left (9 \, x^{2} - 5 \, x\right )} \log \left (4 \, x\right )}{10 \, \log \left (2\right ) \log \left (4 \, x\right )} \]
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Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=- \frac {9 x^{2}}{10 \log {\left (2 \right )}} + \frac {x^{2}}{10 \log {\left (2 \right )} \log {\left (4 x \right )}} + \frac {x}{2 \log {\left (2 \right )}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=-\frac {72 \, x^{2} - 40 \, x - {\rm Ei}\left (2 \, \log \left (4 \, x\right )\right ) + \Gamma \left (-1, -2 \, \log \left (4 \, x\right )\right )}{80 \, \log \left (2\right )} \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=-\frac {9 \, x^{2} - 5 \, x - \frac {x^{2}}{\log \left (4 \, x\right )}}{10 \, \log \left (2\right )} \]
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Time = 10.64 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^2}{10\,\ln \left (4\,x\right )\,\ln \left (2\right )}-\frac {x\,\left (9\,x-5\right )}{10\,\ln \left (2\right )} \]
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