Integrand size = 76, antiderivative size = 27 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\frac {3+\log \left (-\frac {e^x}{4 x}\right )}{5-\frac {x^3}{\log (x)}} \]
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\[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x \left (x^3-5 \log (x)\right )^2} \, dx \\ & = \int \left (\frac {-1+x}{5 x}+\frac {x^2 \left (-5+3 x^3\right ) \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{5 \left (x^3-5 \log (x)\right )^2}-\frac {x^2 \left (8+x+3 \log \left (-\frac {e^x}{4 x}\right )\right )}{5 \left (x^3-5 \log (x)\right )}\right ) \, dx \\ & = \frac {1}{5} \int \frac {-1+x}{x} \, dx+\frac {1}{5} \int \frac {x^2 \left (-5+3 x^3\right ) \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{\left (x^3-5 \log (x)\right )^2} \, dx-\frac {1}{5} \int \frac {x^2 \left (8+x+3 \log \left (-\frac {e^x}{4 x}\right )\right )}{x^3-5 \log (x)} \, dx \\ & = \frac {1}{5} \int \left (1-\frac {1}{x}\right ) \, dx+\frac {1}{5} \int \left (-\frac {5 x^2 \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{\left (x^3-5 \log (x)\right )^2}+\frac {3 x^5 \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{\left (x^3-5 \log (x)\right )^2}\right ) \, dx-\frac {1}{5} \int \left (\frac {8 x^2}{x^3-5 \log (x)}+\frac {x^3}{x^3-5 \log (x)}+\frac {3 x^2 \log \left (-\frac {e^x}{4 x}\right )}{x^3-5 \log (x)}\right ) \, dx \\ & = \frac {x}{5}-\frac {\log (x)}{5}-\frac {1}{5} \int \frac {x^3}{x^3-5 \log (x)} \, dx+\frac {3}{5} \int \frac {x^5 \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{\left (x^3-5 \log (x)\right )^2} \, dx-\frac {3}{5} \int \frac {x^2 \log \left (-\frac {e^x}{4 x}\right )}{x^3-5 \log (x)} \, dx-\frac {8}{5} \int \frac {x^2}{x^3-5 \log (x)} \, dx-\int \frac {x^2 \left (3+\log \left (-\frac {e^x}{4 x}\right )\right )}{\left (x^3-5 \log (x)\right )^2} \, dx \\ & = \frac {x}{5}-\frac {\log (x)}{5}-\frac {1}{5} \int \frac {x^3}{x^3-5 \log (x)} \, dx+\frac {3}{5} \int \left (\frac {3 x^5}{\left (x^3-5 \log (x)\right )^2}+\frac {x^5 \log \left (-\frac {e^x}{4 x}\right )}{\left (x^3-5 \log (x)\right )^2}\right ) \, dx-\frac {3}{5} \int \frac {x^2 \log \left (-\frac {e^x}{4 x}\right )}{x^3-5 \log (x)} \, dx-\frac {8}{5} \int \frac {x^2}{x^3-5 \log (x)} \, dx-\int \left (\frac {3 x^2}{\left (x^3-5 \log (x)\right )^2}+\frac {x^2 \log \left (-\frac {e^x}{4 x}\right )}{\left (x^3-5 \log (x)\right )^2}\right ) \, dx \\ & = \frac {x}{5}-\frac {\log (x)}{5}-\frac {1}{5} \int \frac {x^3}{x^3-5 \log (x)} \, dx+\frac {3}{5} \int \frac {x^5 \log \left (-\frac {e^x}{4 x}\right )}{\left (x^3-5 \log (x)\right )^2} \, dx-\frac {3}{5} \int \frac {x^2 \log \left (-\frac {e^x}{4 x}\right )}{x^3-5 \log (x)} \, dx-\frac {8}{5} \int \frac {x^2}{x^3-5 \log (x)} \, dx+\frac {9}{5} \int \frac {x^5}{\left (x^3-5 \log (x)\right )^2} \, dx-3 \int \frac {x^2}{\left (x^3-5 \log (x)\right )^2} \, dx-\int \frac {x^2 \log \left (-\frac {e^x}{4 x}\right )}{\left (x^3-5 \log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\frac {(-3+x) x^3-x^3 \log \left (-\frac {e^x}{4 x}\right )-x \left (5+x^2\right ) \log (x)+5 \log ^2(x)}{5 \left (x^3-5 \log (x)\right )} \]
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Time = 1.62 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
parallelrisch | \(\frac {-36 x^{3}-60 \ln \left (-\frac {{\mathrm e}^{x}}{4 x}\right ) \ln \left (x \right )}{60 x^{3}-300 \ln \left (x \right )}\) | \(31\) |
default | \(-\frac {\ln \left (x \right )^{2}+\left (-3-\ln \left (-\frac {{\mathrm e}^{x}}{4 x}\right )+x -\ln \left (x \right )\right ) \ln \left (x \right )-x \ln \left (x \right )}{-x^{3}+5 \ln \left (x \right )}\) | \(45\) |
risch | \(-\frac {x^{3} \ln \left ({\mathrm e}^{x}\right )}{5 \left (x^{3}-5 \ln \left (x \right )\right )}+\frac {2 i \pi \,x^{3} \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}-i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}+i \pi \,x^{3} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-i \pi \,x^{3} \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )-i \pi \,x^{3} \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{3}-2 i \pi \,x^{3}+4 x^{3} \ln \left (2\right )+2 x^{4}-6 x^{3}+10 \ln \left (x \right )^{2}-10 x \ln \left (x \right )}{10 x^{3}-50 \ln \left (x \right )}\) | \(182\) |
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Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\frac {{\left (i \, \pi + 2 \, \log \left (2\right )\right )} x^{3} - 3 \, x^{3} - 5 \, x \log \left (x\right ) + 5 \, \log \left (x\right )^{2}}{5 \, {\left (x^{3} - 5 \, \log \left (x\right )\right )}} \]
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Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=- \frac {x^{3}}{25} + \frac {x}{5} - \frac {\log {\left (x \right )}}{5} + \frac {- x^{6} + 5 x^{4} - 10 x^{3} \log {\left (2 \right )} + 15 x^{3} + 5 i \pi x^{3}}{- 25 x^{3} + 125 \log {\left (x \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\frac {x^{4} + x^{3} {\left (2 \, \log \left (2\right ) - 3\right )} - x^{3} \log \left (-e^{x}\right ) - 5 \, x \log \left (x\right ) + 5 \, \log \left (x\right )^{2}}{5 \, {\left (x^{3} - 5 \, \log \left (x\right )\right )}} \]
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Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=-\frac {1}{25} \, x^{3} + \frac {1}{5} \, x + \frac {x^{6} - 5 i \, \pi x^{3} - 5 \, x^{4} + 10 \, x^{3} \log \left (2\right ) - 15 \, x^{3}}{25 \, {\left (x^{3} - 5 \, \log \left (x\right )\right )}} - \frac {1}{5} \, \log \left (x\right ) \]
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Time = 13.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 5.59 \[ \int \frac {-3 x^3+\left (10 x^3-x^4\right ) \log (x)+(-5+5 x) \log ^2(x)+\log \left (-\frac {e^x}{4 x}\right ) \left (-x^3+3 x^3 \log (x)\right )}{x^7-10 x^4 \log (x)+25 x \log ^2(x)} \, dx=\frac {x^3}{25}-\frac {\ln \left (x\right )}{5}-\frac {\frac {x\,\left (25\,x^2\,\left (\ln \left (-\frac {1}{4\,x}\right )+\ln \left (x\right )\right )+75\,x^2+25\,x^3-5\,x^5+5\,x^6-3\,x^8\right )}{25\,\left (3\,x^3-5\right )}-\frac {x\,\ln \left (x\right )\,\left (15\,x^2\,\left (\ln \left (-\frac {1}{4\,x}\right )+\ln \left (x\right )\right )+45\,x^2+20\,x^3-6\,x^5\right )}{5\,\left (3\,x^3-5\right )}}{5\,\ln \left (x\right )-x^3}-\frac {20\,x+15\,\ln \left (-\frac {1}{4\,x}\right )+15\,\ln \left (x\right )+35}{45\,x^3-75}-\frac {x}{15} \]
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