Integrand size = 179, antiderivative size = 27 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \]
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Time = 0.49 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6820, 2458, 2395, 2334, 2335, 2339, 30, 2436, 6874, 6818} \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x+\frac {15 (8-x)}{\log (x-8)}-\frac {120}{\log (x-8)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \]
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Rule 30
Rule 2334
Rule 2335
Rule 2339
Rule 2395
Rule 2436
Rule 2458
Rule 6818
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {15 x}{(-8+x) \log ^2(-8+x)}-\frac {15}{\log (-8+x)}+\frac {4+x^2 \log ^2\left (1-\frac {\log (2 x)}{x}\right )-\log (2 x) \left (4+x \log ^2\left (1-\frac {\log (2 x)}{x}\right )\right )}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )}\right ) \, dx \\ & = 15 \int \frac {x}{(-8+x) \log ^2(-8+x)} \, dx-15 \int \frac {1}{\log (-8+x)} \, dx+\int \frac {4+x^2 \log ^2\left (1-\frac {\log (2 x)}{x}\right )-\log (2 x) \left (4+x \log ^2\left (1-\frac {\log (2 x)}{x}\right )\right )}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )} \, dx \\ & = 15 \text {Subst}\left (\int \frac {8+x}{x \log ^2(x)} \, dx,x,-8+x\right )-15 \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-8+x\right )+\int \left (1-\frac {4 (-1+\log (2 x))}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )}\right ) \, dx \\ & = x-15 \operatorname {LogIntegral}(-8+x)-4 \int \frac {-1+\log (2 x)}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )} \, dx+15 \text {Subst}\left (\int \left (\frac {1}{\log ^2(x)}+\frac {8}{x \log ^2(x)}\right ) \, dx,x,-8+x\right ) \\ & = x+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}-15 \operatorname {LogIntegral}(-8+x)+15 \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-8+x\right )+120 \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-8+x\right ) \\ & = x+\frac {15 (8-x)}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}-15 \operatorname {LogIntegral}(-8+x)+15 \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-8+x\right )+120 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-8+x)\right ) \\ & = x-\frac {120}{\log (-8+x)}+\frac {15 (8-x)}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \]
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Time = 15.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52
method | result | size |
default | \(x -\frac {15 \left (-8+x \right )}{\ln \left (-8+x \right )}-\frac {120}{\ln \left (-8+x \right )}+\frac {4}{\ln \left (\frac {-\ln \left (2\right )-\ln \left (x \right )+x}{x}\right )}\) | \(41\) |
parts | \(x -\frac {15 \left (-8+x \right )}{\ln \left (-8+x \right )}-\frac {120}{\ln \left (-8+x \right )}+\frac {4}{\ln \left (\frac {-\ln \left (2\right )-\ln \left (x \right )+x}{x}\right )}\) | \(41\) |
parallelrisch | \(\frac {16 x \ln \left (-8+x \right ) \ln \left (-\frac {\ln \left (2 x \right )-x}{x}\right )-240 \ln \left (-\frac {\ln \left (2 x \right )-x}{x}\right ) x +64 \ln \left (-\frac {\ln \left (2 x \right )-x}{x}\right ) \ln \left (-8+x \right )+64 \ln \left (-8+x \right )}{16 \ln \left (-8+x \right ) \ln \left (-\frac {\ln \left (2 x \right )-x}{x}\right )}\) | \(90\) |
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Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right ) + 4 \, \log \left (x - 8\right )}{\log \left (x - 8\right ) \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x - \frac {15 x}{\log {\left (x - 8 \right )}} + \frac {4}{\log {\left (\frac {x - \log {\left (2 x \right )}}{x} \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (27) = 54\).
Time = 0.37 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (x - \log \left (2\right ) - \log \left (x\right )\right ) - {\left (x \log \left (x\right ) - 4\right )} \log \left (x - 8\right ) + 15 \, x \log \left (x\right )}{\log \left (x - \log \left (2\right ) - \log \left (x\right )\right ) \log \left (x - 8\right ) - \log \left (x - 8\right ) \log \left (x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (27) = 54\).
Time = 0.62 (sec) , antiderivative size = 235, normalized size of antiderivative = 8.70 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x + \frac {4 \, {\left ({\left (x - 8\right )} \log \left (2\right ) - \log \left (2\right ) \log \left (2 \, x\right ) + {\left (x - 8\right )} \log \left (x\right ) - \log \left (2 \, x\right ) \log \left (x\right ) - x + 8 \, \log \left (2\right ) + \log \left (2 \, x\right ) + 8 \, \log \left (x\right )\right )}}{{\left (x - 8\right )} \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - \log \left (2\right ) \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - {\left (x - 8\right )} \log \left (2 \, x\right ) \log \left (x\right ) + \log \left (2\right ) \log \left (2 \, x\right ) \log \left (x\right ) - \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) \log \left (x\right ) + \log \left (2 \, x\right ) \log \left (x\right )^{2} - {\left (x - 8\right )} \log \left (x - \log \left (2 \, x\right )\right ) + \log \left (2\right ) \log \left (x - \log \left (2 \, x\right )\right ) + 8 \, \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) + {\left (x - 8\right )} \log \left (x\right ) - \log \left (2\right ) \log \left (x\right ) - 8 \, \log \left (2 \, x\right ) \log \left (x\right ) + \log \left (x - \log \left (2 \, x\right )\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 8 \, \log \left (x - \log \left (2 \, x\right )\right ) + 8 \, \log \left (x\right )} - \frac {15 \, x}{\log \left (x - 8\right )} - 8 \]
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Time = 15.99 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {4}{\ln \left (\frac {x-\ln \left (2\,x\right )}{x}\right )}-\frac {15\,x-\ln \left (x-8\right )\,\left (15\,x-120\right )}{\ln \left (x-8\right )}-14\,x \]
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