Integrand size = 124, antiderivative size = 27 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=\frac {1+\log \left (3 e^{\frac {4 x^2 (-5+x+\log (x))^2}{\log (x)}}\right )}{x} \]
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Time = 0.43 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89, number of steps used = 47, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6820, 2357, 2367, 2335, 2346, 2209, 2403, 2332, 30, 2635, 12} \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+\frac {1}{x} \]
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Rule 12
Rule 30
Rule 2209
Rule 2332
Rule 2335
Rule 2346
Rule 2357
Rule 2367
Rule 2403
Rule 2635
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (-76-\frac {1}{x^2}+24 x-\frac {4 (-5+x)^2}{\log ^2(x)}+\frac {8 \left (25-15 x+2 x^2\right )}{\log (x)}+8 \log (x)-\frac {\log \left (3 \exp \left (\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}\right ) x^{4 x^2}\right )}{x^2}\right ) \, dx \\ & = \frac {1}{x}-76 x+12 x^2-4 \int \frac {(-5+x)^2}{\log ^2(x)} \, dx+8 \int \frac {25-15 x+2 x^2}{\log (x)} \, dx+8 \int \log (x) \, dx-\int \frac {\log \left (3 \exp \left (\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}\right ) x^{4 x^2}\right )}{x^2} \, dx \\ & = \frac {1}{x}-84 x+12 x^2+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+8 \int \left (\frac {25}{\log (x)}-\frac {15 x}{\log (x)}+\frac {2 x^2}{\log (x)}\right ) \, dx-12 \int \frac {(-5+x)^2}{\log (x)} \, dx-40 \int \frac {-5+x}{\log (x)} \, dx+\int 4 \left (19-6 x+\frac {(-5+x)^2}{\log ^2(x)}+\frac {-50+30 x-4 x^2}{\log (x)}-2 \log (x)\right ) \, dx \\ & = \frac {1}{x}-84 x+12 x^2+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+4 \int \left (19-6 x+\frac {(-5+x)^2}{\log ^2(x)}+\frac {-50+30 x-4 x^2}{\log (x)}-2 \log (x)\right ) \, dx-12 \int \left (\frac {25}{\log (x)}-\frac {10 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx+16 \int \frac {x^2}{\log (x)} \, dx-40 \int \left (-\frac {5}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx-120 \int \frac {x}{\log (x)} \, dx+200 \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{x}-8 x+\frac {4 (5-x)^2 x}{\log (x)}+8 x \log (x)+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+200 \text {li}(x)+4 \int \frac {(-5+x)^2}{\log ^2(x)} \, dx+4 \int \frac {-50+30 x-4 x^2}{\log (x)} \, dx-8 \int \log (x) \, dx-12 \int \frac {x^2}{\log (x)} \, dx+16 \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-40 \int \frac {x}{\log (x)} \, dx+120 \int \frac {x}{\log (x)} \, dx-120 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+200 \int \frac {1}{\log (x)} \, dx-300 \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{x}-120 \text {Ei}(2 \log (x))+16 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+100 \text {li}(x)+4 \int \left (-\frac {50}{\log (x)}+\frac {30 x}{\log (x)}-\frac {4 x^2}{\log (x)}\right ) \, dx+12 \int \frac {(-5+x)^2}{\log (x)} \, dx-12 \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \int \frac {-5+x}{\log (x)} \, dx-40 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+120 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {1}{x}-40 \text {Ei}(2 \log (x))+4 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+100 \text {li}(x)+12 \int \left (\frac {25}{\log (x)}-\frac {10 x}{\log (x)}+\frac {x^2}{\log (x)}\right ) \, dx-16 \int \frac {x^2}{\log (x)} \, dx+40 \int \left (-\frac {5}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+120 \int \frac {x}{\log (x)} \, dx-200 \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{x}-40 \text {Ei}(2 \log (x))+4 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}-100 \text {li}(x)+12 \int \frac {x^2}{\log (x)} \, dx-16 \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \int \frac {x}{\log (x)} \, dx-120 \int \frac {x}{\log (x)} \, dx+120 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-200 \int \frac {1}{\log (x)} \, dx+300 \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{x}+80 \text {Ei}(2 \log (x))-12 \text {Ei}(3 \log (x))+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x}+12 \text {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )+40 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-120 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {1}{x}+\frac {\log \left (3 e^{\frac {4 (5-x)^2 x^2}{\log (x)}} x^{4 x^2-\frac {8 (5-x) x^2}{\log (x)}}\right )}{x} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=\frac {1+\log \left (3 e^{\frac {4 (-5+x) x^2 (-5+x+2 \log (x))}{\log (x)}} x^{4 x^2}\right )}{x} \]
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Time = 1.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}^{\frac {4 \left (\ln \left (x \right )+x -5\right )^{2} x^{2}}{\ln \left (x \right )}}\right )}{x}+\frac {\ln \left (3\right )}{x}+\frac {1}{x}\) | \(33\) |
parallelrisch | \(-\frac {-1-\ln \left (3 \,{\mathrm e}^{\frac {4 x^{2} \left (\ln \left (x \right )^{2}+2 x \ln \left (x \right )+x^{2}-10 \ln \left (x \right )-10 x +25\right )}{\ln \left (x \right )}}\right )}{x}\) | \(44\) |
default | \(\frac {4 x^{3}}{\ln \left (x \right )}-\frac {40 x^{2}}{\ln \left (x \right )}+\frac {100 x}{\ln \left (x \right )}+8 x^{2}-40 x +\frac {1}{x}+4 x \ln \left (x \right )+\frac {\ln \left (3 \,{\mathrm e}^{\frac {4 x^{2} \ln \left (x \right )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \left (x \right )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \left (x \right )}}\right )-\frac {4 x^{2} \ln \left (x \right )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \left (x \right )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \left (x \right )}}{x}\) | \(141\) |
parts | \(\frac {4 x^{3}}{\ln \left (x \right )}-\frac {40 x^{2}}{\ln \left (x \right )}+\frac {100 x}{\ln \left (x \right )}+8 x^{2}-40 x +\frac {1}{x}+4 x \ln \left (x \right )+\frac {\ln \left (3 \,{\mathrm e}^{\frac {4 x^{2} \ln \left (x \right )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \left (x \right )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \left (x \right )}}\right )-\frac {4 x^{2} \ln \left (x \right )^{2}+\left (8 x^{3}-40 x^{2}\right ) \ln \left (x \right )+4 x^{4}-40 x^{3}+100 x^{2}}{\ln \left (x \right )}}{x}\) | \(141\) |
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Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=\frac {4 \, x^{4} + 4 \, x^{2} \log \left (x\right )^{2} - 40 \, x^{3} + 100 \, x^{2} + {\left (8 \, x^{3} - 40 \, x^{2} + \log \left (3\right ) + 1\right )} \log \left (x\right )}{x \log \left (x\right )} \]
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=8 x^{2} + 4 x \log {\left (x \right )} - 40 x + \frac {4 x^{3} - 40 x^{2} + 100 x}{\log {\left (x \right )}} + \frac {1 + \log {\left (3 \right )}}{x} \]
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\[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=\int { \frac {8 \, x^{2} \log \left (x\right )^{3} - 4 \, x^{4} + 40 \, x^{3} + {\left (24 \, x^{3} - 76 \, x^{2} - 1\right )} \log \left (x\right )^{2} - \log \left (x\right )^{2} \log \left (3 \, e^{\left (\frac {4 \, {\left (x^{4} + x^{2} \log \left (x\right )^{2} - 10 \, x^{3} + 25 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (x\right )\right )}}{\log \left (x\right )}\right )}\right ) - 100 \, x^{2} + 8 \, {\left (2 \, x^{4} - 15 \, x^{3} + 25 \, x^{2}\right )} \log \left (x\right )}{x^{2} \log \left (x\right )^{2}} \,d x } \]
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Time = 0.54 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=8 \, x^{2} + 4 \, x \log \left (x\right ) - 40 \, x + \frac {\log \left (3\right ) + 1}{x} + \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )}}{\log \left (x\right )} \]
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Time = 12.86 (sec) , antiderivative size = 122, normalized size of antiderivative = 4.52 \[ \int \frac {-100 x^2+40 x^3-4 x^4+\left (200 x^2-120 x^3+16 x^4\right ) \log (x)+\left (-1-76 x^2+24 x^3\right ) \log ^2(x)-\log \left (3 e^{\frac {100 x^2-40 x^3+4 x^4+\left (-40 x^2+8 x^3\right ) \log (x)+4 x^2 \log ^2(x)}{\log (x)}}\right ) \log ^2(x)+8 x^2 \log ^3(x)}{x^2 \log ^2(x)} \, dx=60\,x+\frac {4\,x\,{\left (x-5\right )}^2-4\,x\,\ln \left (x\right )\,\left (3\,x^2-20\,x+25\right )}{\ln \left (x\right )}+4\,x\,\ln \left (x\right )-72\,x^2+12\,x^3+\frac {\ln \left (3\,x^{4\,x^2}\,{\mathrm {e}}^{8\,x^3}\,{\mathrm {e}}^{-40\,x^2}\,{\mathrm {e}}^{\frac {4\,x^4}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {40\,x^3}{\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {100\,x^2}{\ln \left (x\right )}}\right )-\frac {4\,x^2\,{\left (x+\ln \left (x\right )-5\right )}^2}{\ln \left (x\right )}+1}{x} \]
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