Integrand size = 111, antiderivative size = 23 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \]
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\[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{5 (3-x) (5 x-\log (4 (-3+x)))} \, dx \\ & = \frac {1}{5} \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{(3-x) (5 x-\log (4 (-3+x)))} \, dx \\ & = \frac {1}{5} \int \frac {32 x-10 x^2-e^x x (-16+5 x)-(-3+x) \left (2+e^x (1+x)\right ) (5 x-\log (4 (-3+x))) \log (5 x-\log (4 (-3+x)))}{(3-x) (5 x-\log (4 (-3+x)))} \, dx \\ & = \frac {1}{5} \int \left (\frac {2 \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))+5 x^2 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {e^x \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))-10 x^2 \log (5 x-\log (4 (-3+x)))+5 x^3 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))+2 x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x^2 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))}\right ) \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))-10 x^2 \log (5 x-\log (4 (-3+x)))+5 x^3 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))+2 x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x^2 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {2}{5} \int \frac {-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))+5 x^2 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (-\frac {x (-16+5 x)}{5 x-\log (4 (-3+x))}-\left (-3-2 x+x^2\right ) \log (5 x-\log (4 (-3+x)))\right )}{3-x} \, dx+\frac {2}{5} \int \left (\frac {x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))}+\log (5 x-\log (4 (-3+x)))\right ) \, dx \\ & = \frac {1}{5} \int \left (\frac {e^x x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))}+e^x (1+x) \log (5 x-\log (4 (-3+x)))\right ) \, dx+\frac {2}{5} \int \frac {x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx \\ & = \frac {1}{5} \int \frac {e^x x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {1}{5} \int e^x (1+x) \log (5 x-\log (4 (-3+x))) \, dx+\frac {2}{5} \int \left (-\frac {3}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {5 x}{5 x-\log (4 (-3+x))}+\frac {1}{-5 x+\log (4 (-3+x))}\right ) \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx \\ & = \frac {1}{5} \int \left (-\frac {3 e^x}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {5 e^x x}{5 x-\log (4 (-3+x))}+\frac {e^x}{-5 x+\log (4 (-3+x))}\right ) \, dx+\frac {1}{5} \int \left (e^x \log (5 x-\log (4 (-3+x)))+e^x x \log (5 x-\log (4 (-3+x)))\right ) \, dx+\frac {2}{5} \int \frac {1}{-5 x+\log (4 (-3+x))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx-\frac {6}{5} \int \frac {1}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+2 \int \frac {x}{5 x-\log (4 (-3+x))} \, dx \\ & = \frac {1}{5} \int \frac {e^x}{-5 x+\log (4 (-3+x))} \, dx+\frac {1}{5} \int e^x \log (5 x-\log (4 (-3+x))) \, dx+\frac {1}{5} \int e^x x \log (5 x-\log (4 (-3+x))) \, dx+\frac {2}{5} \int \frac {1}{-5 x+\log (4 (-3+x))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx-\frac {3}{5} \int \frac {e^x}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx-\frac {6}{5} \int \frac {1}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+2 \int \frac {x}{5 x-\log (4 (-3+x))} \, dx+\int \frac {e^x x}{5 x-\log (4 (-3+x))} \, dx \\ \end{align*}
Time = 1.91 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \]
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Time = 1.88 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\left (\frac {2 x}{5}+\frac {{\mathrm e}^{x} x}{5}\right ) \ln \left (-\ln \left (4 x -12\right )+5 x \right )\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{x} \ln \left (-\ln \left (4 x -12\right )+5 x \right ) x}{5}+\frac {2 \ln \left (-\ln \left (4 x -12\right )+5 x \right ) x}{5}-\frac {3 \ln \left (-\frac {\ln \left (4 x -12\right )}{5}+x \right )}{5}+\frac {3 \ln \left (-\ln \left (4 x -12\right )+5 x \right )}{5}\) | \(64\) |
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 10.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {x e^{x} \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )}}{5} + \left (\frac {2 x}{5} - \frac {3}{5}\right ) \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )} + \frac {3 \log {\left (- 5 x + \log {\left (4 x - 12 \right )} \right )}}{5} \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - 2 \, \log \left (2\right ) - \log \left (x - 3\right )\right ) \]
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Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {1}{5} \, x e^{x} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) + \frac {2}{5} \, x \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \]
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Time = 7.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx=\frac {x\,\ln \left (5\,x-\ln \left (4\,x-12\right )\right )\,\left ({\mathrm {e}}^x+2\right )}{5} \]
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