Integrand size = 110, antiderivative size = 29 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}-\frac {2}{5-x}\right ) x\right ) \]
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\[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{5 x} \, dx \\ & = \frac {1}{5} \int \frac {\frac {\left (-10+3 e^{4 x} (-5+x)^2 (1+4 x)\right ) \log (x)}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)}+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx \\ & = \frac {1}{5} \int \left (-\frac {2 (-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )}+\frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx \\ & = \frac {1}{5} \int \frac {\log (x)+4 x \log (x)+\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx-\frac {2}{5} \int \frac {(-19+4 x) \log (x)}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx \\ & = \frac {1}{5} \int \left (\frac {(1+4 x) \log (x)}{x}+\frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x}\right ) \, dx+\frac {2}{5} \int \frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx+\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx \\ & = \frac {1}{5} \int \frac {(1+4 x) \log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \left (\frac {4 \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x}+\frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x}\right ) \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx \\ & = \frac {1}{5} \int \frac {\log (x)}{x} \, dx+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {4}{5} \int \log (x) \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx \\ & = -\frac {4 x}{5}+\frac {4}{5} x \log (x)+\frac {\log ^2(x)}{10}+\frac {1}{5} \int \frac {\log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right )}{x} \, dx+\frac {2}{5} \int \frac {\int \frac {1}{\left (2+3 e^{4 x} (-5+x)\right ) (-5+x)} \, dx}{x} \, dx+\frac {8}{5} \int \frac {\int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx}{x} \, dx-\frac {1}{5} (2 \log (x)) \int \frac {1}{(-5+x) \left (2-15 e^{4 x}+3 e^{4 x} x\right )} \, dx-\frac {1}{5} (8 \log (x)) \int \frac {1}{2+3 e^{4 x} (-5+x)} \, dx \\ \end{align*}
Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \log (x) \log \left (e^e \left (3 e^{4 x}+\frac {2}{-5+x}\right ) x\right ) \]
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Time = 10.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right ) \ln \left (\frac {\left (\left (3 x^{2}-15 x \right ) {\mathrm e}^{4 x}+2 x \right ) {\mathrm e}^{{\mathrm e}}}{-5+x}\right )}{5}\) | \(33\) |
risch | \(\frac {\ln \left (x \right ) \ln \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{5}-\frac {\ln \left (x \right ) \ln \left (-5+x \right )}{5}+\frac {\ln \left (x \right )^{2}}{5}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}-\frac {i \pi \ln \left (x \right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{3}}{10}-\frac {i \pi \ln \left (x \right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{3}}{10}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}{10}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{4 x}+\frac {2}{3}\right )}{-5+x}\right )}^{2}}{10}+\frac {\ln \left (3\right ) \ln \left (x \right )}{5}+\frac {{\mathrm e} \ln \left (x \right )}{5}\) | \(405\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, \log \left (x\right ) \log \left (\frac {{\left (3 \, {\left (x^{2} - 5 \, x\right )} e^{\left (4 \, x\right )} + 2 \, x\right )} e^{e}}{x - 5}\right ) \]
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Time = 0.51 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {\left (2 x + \left (3 x^{2} - 15 x\right ) e^{4 x}\right ) e^{e}}{x - 5} \right )}}{5} \]
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Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \]
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Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {1}{5} \, e \log \left (x\right ) + \frac {1}{5} \, \log \left (3 \, x e^{\left (4 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 2\right ) \log \left (x\right ) - \frac {1}{5} \, \log \left (x - 5\right ) \log \left (x\right ) + \frac {1}{5} \, \log \left (x\right )^{2} \]
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Time = 12.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-10+e^{4 x} \left (75+270 x-117 x^2+12 x^3\right )\right ) \log (x)+\left (-10+2 x+e^{4 x} \left (75-30 x+3 x^2\right )\right ) \log \left (\frac {e^e \left (2 x+e^{4 x} \left (-15 x+3 x^2\right )\right )}{-5+x}\right )}{-50 x+10 x^2+e^{4 x} \left (375 x-150 x^2+15 x^3\right )} \, dx=\frac {\ln \left (x\right )\,\left (\ln \left (\frac {2\,x-{\mathrm {e}}^{4\,x}\,\left (15\,x-3\,x^2\right )}{x-5}\right )+\mathrm {e}\right )}{5} \]
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