Integrand size = 85, antiderivative size = 25 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\frac {4 (3-x) x (1+\log (2)) \log (-1+x)}{3-\frac {x}{25}} \]
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Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6820, 12, 6874, 147, 2465, 2436, 2332, 2442, 36, 31} \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=7300 (1+\log (2)) \log (1-x)-100 (1-x) (1+\log (2)) \log (x-1)-\frac {540000 (1+\log (2)) \log (x-1)}{75-x} \]
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Rule 12
Rule 31
Rule 36
Rule 147
Rule 2332
Rule 2436
Rule 2442
Rule 2465
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {100 (1+\log (2)) \left (-x \left (225-78 x+x^2\right )-\left (-225+375 x-151 x^2+x^3\right ) \log (-1+x)\right )}{(1-x) (75-x)^2} \, dx \\ & = (100 (1+\log (2))) \int \frac {-x \left (225-78 x+x^2\right )-\left (-225+375 x-151 x^2+x^3\right ) \log (-1+x)}{(1-x) (75-x)^2} \, dx \\ & = (100 (1+\log (2))) \int \left (\frac {(-3+x) x}{(-75+x) (-1+x)}+\frac {\left (225-150 x+x^2\right ) \log (-1+x)}{(-75+x)^2}\right ) \, dx \\ & = (100 (1+\log (2))) \int \frac {(-3+x) x}{(-75+x) (-1+x)} \, dx+(100 (1+\log (2))) \int \frac {\left (225-150 x+x^2\right ) \log (-1+x)}{(-75+x)^2} \, dx \\ & = (100 (1+\log (2))) \int \left (1+\frac {2700}{37 (-75+x)}+\frac {1}{37 (-1+x)}\right ) \, dx+(100 (1+\log (2))) \int \left (\log (-1+x)-\frac {5400 \log (-1+x)}{(-75+x)^2}\right ) \, dx \\ & = 100 x (1+\log (2))+\frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)+(100 (1+\log (2))) \int \log (-1+x) \, dx-(540000 (1+\log (2))) \int \frac {\log (-1+x)}{(-75+x)^2} \, dx \\ & = 100 x (1+\log (2))+\frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x}+(100 (1+\log (2))) \text {Subst}(\int \log (x) \, dx,x,-1+x)-(540000 (1+\log (2))) \int \frac {1}{(-75+x) (-1+x)} \, dx \\ & = \frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)-100 (1-x) (1+\log (2)) \log (-1+x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x}-\frac {1}{37} (270000 (1+\log (2))) \int \frac {1}{-75+x} \, dx+\frac {1}{37} (270000 (1+\log (2))) \int \frac {1}{-1+x} \, dx \\ & = 7300 (1+\log (2)) \log (1-x)-100 (1-x) (1+\log (2)) \log (-1+x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(25)=50\).
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\frac {100}{37} (1+\log (2)) \left (5400 \text {arctanh}\left (\frac {1}{37} (-38+x)\right )+\log (1-x)+2700 \log (75-x)+\frac {199800 \log (-1+x)}{-75+x}+37 (-1+x) \log (-1+x)\right ) \]
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Time = 0.39 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36
method | result | size |
norman | \(\frac {\left (100 \ln \left (2\right )+100\right ) x^{2} \ln \left (-1+x \right )+\left (-300-300 \ln \left (2\right )\right ) x \ln \left (-1+x \right )}{x -75}\) | \(34\) |
parallelrisch | \(\frac {100 \ln \left (2\right ) \ln \left (-1+x \right ) x^{2}-300 \ln \left (2\right ) \ln \left (-1+x \right ) x +100 \ln \left (-1+x \right ) x^{2}-300 \ln \left (-1+x \right ) x}{x -75}\) | \(44\) |
risch | \(\frac {100 \left (x^{2} \ln \left (2\right )-75 x \ln \left (2\right )+x^{2}+5400 \ln \left (2\right )-75 x +5400\right ) \ln \left (-1+x \right )}{x -75}+7200 \ln \left (2\right ) \ln \left (-1+x \right )+7200 \ln \left (-1+x \right )\) | \(50\) |
derivativedivides | \(100 \ln \left (2\right ) \left (\left (-1+x \right ) \ln \left (-1+x \right )+\frac {2700 \ln \left (-1+x \right ) \left (-1+x \right )}{37 \left (x -75\right )}+\frac {\ln \left (-1+x \right )}{37}\right )+100 \left (-1+x \right ) \ln \left (-1+x \right )+\frac {270000 \ln \left (-1+x \right ) \left (-1+x \right )}{37 \left (x -75\right )}+\frac {100 \ln \left (-1+x \right )}{37}\) | \(64\) |
default | \(100 \ln \left (2\right ) \left (\left (-1+x \right ) \ln \left (-1+x \right )+\frac {2700 \ln \left (-1+x \right ) \left (-1+x \right )}{37 \left (x -75\right )}+\frac {\ln \left (-1+x \right )}{37}\right )+100 \left (-1+x \right ) \ln \left (-1+x \right )+\frac {270000 \ln \left (-1+x \right ) \left (-1+x \right )}{37 \left (x -75\right )}+\frac {100 \ln \left (-1+x \right )}{37}\) | \(64\) |
parts | \(100 \left (1+\ln \left (2\right )\right ) \left (x +\frac {\ln \left (-1+x \right )}{37}+\frac {2700 \ln \left (x -75\right )}{37}\right )+100 \left (-1+x \right ) \ln \left (-1+x \right )-100 x +100+100 \ln \left (2\right ) \left (\left (-1+x \right ) \ln \left (-1+x \right )-x +1\right )+100 \left (-5400-5400 \ln \left (2\right )\right ) \left (\frac {\ln \left (x -75\right )}{74}-\frac {\ln \left (-1+x \right ) \left (-1+x \right )}{74 \left (x -75\right )}\right )\) | \(81\) |
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\frac {100 \, {\left (x^{2} + {\left (x^{2} - 3 \, x\right )} \log \left (2\right ) - 3 \, x\right )} \log \left (x - 1\right )}{x - 75} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\left (7200 \log {\left (2 \right )} + 7200\right ) \log {\left (x - 1 \right )} + \frac {\left (100 x^{2} \log {\left (2 \right )} + 100 x^{2} - 7500 x - 7500 x \log {\left (2 \right )} + 540000 \log {\left (2 \right )} + 540000\right ) \log {\left (x - 1 \right )}}{x - 75} \]
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Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\frac {100 \, {\left (x^{2} {\left (\log \left (2\right ) + 1\right )} - 3 \, x {\left (\log \left (2\right ) + 1\right )}\right )} \log \left (x - 1\right )}{x - 75} \]
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Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=100 \, {\left (x {\left (\log \left (2\right ) + 1\right )} + \frac {5400 \, {\left (\log \left (2\right ) + 1\right )}}{x - 75}\right )} \log \left (x - 1\right ) + 7200 \, {\left (\log \left (2\right ) + 1\right )} \log \left (x - 1\right ) \]
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Time = 0.40 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {22500 x-7800 x^2+100 x^3+\left (22500 x-7800 x^2+100 x^3\right ) \log (2)+\left (-22500+37500 x-15100 x^2+100 x^3+\left (-22500+37500 x-15100 x^2+100 x^3\right ) \log (2)\right ) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx=\frac {100\,x\,\ln \left (x-1\right )\,\left (\ln \left (2\right )+1\right )\,\left (x-3\right )}{x-75} \]
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