Integrand size = 59, antiderivative size = 25 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=\frac {1}{9} (-4+x) x \log \left (\left (-x+\frac {1}{9} (-4+4 x)\right ) \log (x)\right ) \]
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Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 22, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6874, 78, 2367, 2335, 2346, 2209, 2635, 12} \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=\frac {1}{9} x^2 \log \left (-\frac {1}{9} (5 x+4) \log (x)\right )-\frac {4}{9} x \log \left (-\frac {1}{9} (5 x+4) \log (x)\right ) \]
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Rule 12
Rule 78
Rule 2209
Rule 2335
Rule 2346
Rule 2367
Rule 2635
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-4+x) (4+5 x+5 x \log (x))}{9 (4+5 x) \log (x)}+\frac {2}{9} (-2+x) \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )\right ) \, dx \\ & = \frac {1}{9} \int \frac {(-4+x) (4+5 x+5 x \log (x))}{(4+5 x) \log (x)} \, dx+\frac {2}{9} \int (-2+x) \log \left (-\frac {1}{9} (4+5 x) \log (x)\right ) \, dx \\ & = -\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} \int \left (\frac {5 (-4+x) x}{4+5 x}+\frac {-4+x}{\log (x)}\right ) \, dx-\frac {2}{9} \int \frac {(4-x) (-4-5 x-5 x \log (x))}{2 (4+5 x) \log (x)} \, dx \\ & = -\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} \int \frac {-4+x}{\log (x)} \, dx-\frac {1}{9} \int \frac {(4-x) (-4-5 x-5 x \log (x))}{(4+5 x) \log (x)} \, dx+\frac {5}{9} \int \frac {(-4+x) x}{4+5 x} \, dx \\ & = -\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \int \left (\frac {5 (-4+x) x}{4+5 x}+\frac {-4+x}{\log (x)}\right ) \, dx+\frac {1}{9} \int \left (-\frac {4}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {5}{9} \int \left (-\frac {24}{25}+\frac {x}{5}+\frac {96}{25 (4+5 x)}\right ) \, dx \\ & = -\frac {8 x}{15}+\frac {x^2}{18}+\frac {32}{75} \log (4+5 x)-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \int \frac {-4+x}{\log (x)} \, dx+\frac {1}{9} \int \frac {x}{\log (x)} \, dx-\frac {4}{9} \int \frac {1}{\log (x)} \, dx-\frac {5}{9} \int \frac {(-4+x) x}{4+5 x} \, dx \\ & = -\frac {8 x}{15}+\frac {x^2}{18}+\frac {32}{75} \log (4+5 x)-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {4 \operatorname {LogIntegral}(x)}{9}-\frac {1}{9} \int \left (-\frac {4}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+\frac {1}{9} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {5}{9} \int \left (-\frac {24}{25}+\frac {x}{5}+\frac {96}{25 (4+5 x)}\right ) \, dx \\ & = \frac {1}{9} \operatorname {ExpIntegralEi}(2 \log (x))-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {4 \operatorname {LogIntegral}(x)}{9}-\frac {1}{9} \int \frac {x}{\log (x)} \, dx+\frac {4}{9} \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{9} \operatorname {ExpIntegralEi}(2 \log (x))-\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )-\frac {1}{9} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {4}{9} x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right )+\frac {1}{9} x^2 \log \left (-\frac {1}{9} (4+5 x) \log (x)\right ) \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=\frac {1}{9} (-4+x) x \log \left (-\frac {1}{9} (4+5 x) \log (x)\right ) \]
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Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
norman | \(-\frac {4 x \ln \left (\frac {\left (-5 x -4\right ) \ln \left (x \right )}{9}\right )}{9}+\frac {x^{2} \ln \left (\frac {\left (-5 x -4\right ) \ln \left (x \right )}{9}\right )}{9}\) | \(30\) |
parallelrisch | \(-\frac {4 x \ln \left (\frac {\left (-5 x -4\right ) \ln \left (x \right )}{9}\right )}{9}+\frac {x^{2} \ln \left (\frac {\left (-5 x -4\right ) \ln \left (x \right )}{9}\right )}{9}\) | \(30\) |
risch | \(\frac {2 i \pi x \,\operatorname {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )}{9}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )}{18}-\frac {4 x \ln \left (5\right )}{9}+\frac {8 x \ln \left (3\right )}{9}+\frac {x^{2} \ln \left (5\right )}{9}-\frac {2 x^{2} \ln \left (3\right )}{9}+\frac {x^{2} \ln \left (\ln \left (x \right )\right )}{9}-\frac {4 x \ln \left (\ln \left (x \right )\right )}{9}-\frac {2 i \pi x \,\operatorname {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}+\frac {i \pi \,x^{2}}{9}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )}{18}-\frac {2 i \pi x \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )}{9}+\frac {4 i \pi x \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{3}}{18}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2}}{9}-\frac {4 i \pi x}{9}-\frac {2 i \pi x \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{3}}{9}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (x +\frac {4}{5}\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (x +\frac {4}{5}\right )\right )^{2}}{18}+\left (\frac {1}{9} x^{2}-\frac {4}{9} x \right ) \ln \left (x +\frac {4}{5}\right )\) | \(284\) |
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=\frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-\frac {1}{9} \, {\left (5 \, x + 4\right )} \log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=\left (\frac {x^{2}}{9} - \frac {4 x}{9} - \frac {16}{225}\right ) \log {\left (\left (- \frac {5 x}{9} - \frac {4}{9}\right ) \log {\left (x \right )} \right )} + \frac {16 \log {\left (225 x + 180 \right )}}{225} + \frac {16 \log {\left (\log {\left (x \right )} \right )}}{225} \]
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Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (16) = 32\).
Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=-\frac {2}{9} \, x^{2} \log \left (3\right ) + \frac {8}{9} \, x \log \left (3\right ) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-5 \, x - 4\right ) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (\log \left (x\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=-\frac {2}{9} \, x^{2} \log \left (3\right ) + \frac {8}{9} \, x \log \left (3\right ) + \frac {1}{9} \, {\left (x^{2} - 4 \, x\right )} \log \left (-5 \, x \log \left (x\right ) - 4 \, \log \left (x\right )\right ) \]
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Time = 7.97 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-16-16 x+5 x^2+\left (-20 x+5 x^2\right ) \log (x)+\left (-16-12 x+10 x^2\right ) \log (x) \log \left (\frac {1}{9} (-4-5 x) \log (x)\right )}{(36+45 x) \log (x)} \, dx=-\ln \left (-\frac {\ln \left (x\right )\,\left (5\,x+4\right )}{9}\right )\,\left (\frac {4\,x}{9}-\frac {x^2}{9}\right ) \]
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