Integrand size = 100, antiderivative size = 25 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=-1+\left (e^{-4+x}-\frac {4+\frac {x^3}{\log (x)}}{x}\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(25)=50\).
Time = 0.97 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24, number of steps used = 18, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6874, 2225, 2326, 6820, 2343, 2346, 2209, 2367, 2334, 2335} \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\frac {x^4}{\log ^2(x)}+\frac {16}{x^2}-\frac {2 e^{x-4} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+e^{2 x-8}+\frac {8 x}{\log (x)} \]
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Rule 2209
Rule 2225
Rule 2326
Rule 2334
Rule 2335
Rule 2343
Rule 2346
Rule 2367
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{-8+2 x}-\frac {2 e^{-4+x} \left (-x^3+2 x^3 \log (x)+x^4 \log (x)-4 \log ^2(x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+\frac {2 \left (-x^6-4 x^3 \log (x)+2 x^6 \log (x)+4 x^3 \log ^2(x)-16 \log ^3(x)\right )}{x^3 \log ^3(x)}\right ) \, dx \\ & = 2 \int e^{-8+2 x} \, dx-2 \int \frac {e^{-4+x} \left (-x^3+2 x^3 \log (x)+x^4 \log (x)-4 \log ^2(x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)} \, dx+2 \int \frac {-x^6-4 x^3 \log (x)+2 x^6 \log (x)+4 x^3 \log ^2(x)-16 \log ^3(x)}{x^3 \log ^3(x)} \, dx \\ & = e^{-8+2 x}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+2 \int \left (-\frac {16}{x^3}-\frac {x^3}{\log ^3(x)}+\frac {2 \left (-2+x^3\right )}{\log ^2(x)}+\frac {4}{\log (x)}\right ) \, dx \\ & = e^{-8+2 x}+\frac {16}{x^2}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}-2 \int \frac {x^3}{\log ^3(x)} \, dx+4 \int \frac {-2+x^3}{\log ^2(x)} \, dx+8 \int \frac {1}{\log (x)} \, dx \\ & = e^{-8+2 x}+\frac {16}{x^2}+\frac {x^4}{\log ^2(x)}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+8 \operatorname {LogIntegral}(x)+4 \int \left (-\frac {2}{\log ^2(x)}+\frac {x^3}{\log ^2(x)}\right ) \, dx-4 \int \frac {x^3}{\log ^2(x)} \, dx \\ & = e^{-8+2 x}+\frac {16}{x^2}+\frac {x^4}{\log ^2(x)}+\frac {4 x^4}{\log (x)}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+8 \operatorname {LogIntegral}(x)+4 \int \frac {x^3}{\log ^2(x)} \, dx-8 \int \frac {1}{\log ^2(x)} \, dx-16 \int \frac {x^3}{\log (x)} \, dx \\ & = e^{-8+2 x}+\frac {16}{x^2}+\frac {x^4}{\log ^2(x)}+\frac {8 x}{\log (x)}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+8 \operatorname {LogIntegral}(x)-8 \int \frac {1}{\log (x)} \, dx+16 \int \frac {x^3}{\log (x)} \, dx-16 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = e^{-8+2 x}+\frac {16}{x^2}-16 \operatorname {ExpIntegralEi}(4 \log (x))+\frac {x^4}{\log ^2(x)}+\frac {8 x}{\log (x)}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)}+16 \text {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right ) \\ & = e^{-8+2 x}+\frac {16}{x^2}+\frac {x^4}{\log ^2(x)}+\frac {8 x}{\log (x)}-\frac {2 e^{-4+x} \left (x^4 \log (x)+4 x \log ^2(x)\right )}{x^2 \log ^2(x)} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\frac {\left (e^4 x^3+\left (4 e^4-e^x x\right ) \log (x)\right )^2}{e^8 x^2 \log ^2(x)} \]
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Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{2 x -8}-8 x \,{\mathrm e}^{x -4}+16}{x^{2}}+\frac {x \left (x^{3}-2 x \,{\mathrm e}^{x -4} \ln \left (x \right )+8 \ln \left (x \right )\right )}{\ln \left (x \right )^{2}}\) | \(48\) |
parallelrisch | \(\frac {-2 \ln \left (x \right ) {\mathrm e}^{x -4} x^{4}+x^{6}+\ln \left (x \right )^{2} {\mathrm e}^{2 x -8} x^{2}-8 \ln \left (x \right )^{2} {\mathrm e}^{x -4} x +8 x^{3} \ln \left (x \right )+16 \ln \left (x \right )^{2}}{\ln \left (x \right )^{2} x^{2}}\) | \(62\) |
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\frac {x^{6} + {\left (x^{2} e^{\left (2 \, x - 8\right )} - 8 \, x e^{\left (x - 4\right )} + 16\right )} \log \left (x\right )^{2} - 2 \, {\left (x^{4} e^{\left (x - 4\right )} - 4 \, x^{3}\right )} \log \left (x\right )}{x^{2} \log \left (x\right )^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\frac {x^{4} + 8 x \log {\left (x \right )}}{\log {\left (x \right )}^{2}} + \frac {x e^{2 x - 8} \log {\left (x \right )} + \left (- 2 x^{3} - 8 \log {\left (x \right )}\right ) e^{x - 4}}{x \log {\left (x \right )}} + \frac {16}{x^{2}} \]
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\[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\int { -\frac {2 \, {\left (x^{6} - {\left (x^{3} e^{\left (2 \, x - 8\right )} - 4 \, {\left (x^{2} - x\right )} e^{\left (x - 4\right )} - 16\right )} \log \left (x\right )^{3} - {\left (4 \, x^{3} - {\left (x^{5} + 2 \, x^{4}\right )} e^{\left (x - 4\right )}\right )} \log \left (x\right )^{2} - {\left (2 \, x^{6} + x^{4} e^{\left (x - 4\right )} - 4 \, x^{3}\right )} \log \left (x\right )\right )}}{x^{3} \log \left (x\right )^{3}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx=\frac {{\left (x^{6} e^{12} - 2 \, x^{4} e^{\left (x + 8\right )} \log \left (x\right ) + 8 \, x^{3} e^{12} \log \left (x\right ) + x^{2} e^{\left (2 \, x + 4\right )} \log \left (x\right )^{2} - 8 \, x e^{\left (x + 8\right )} \log \left (x\right )^{2} + 16 \, e^{12} \log \left (x\right )^{2}\right )} e^{\left (-12\right )}}{x^{2} \log \left (x\right )^{2}} \]
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Time = 15.98 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {-2 x^6+\left (-8 x^3+2 e^{-4+x} x^4+4 x^6\right ) \log (x)+\left (8 x^3+e^{-4+x} \left (-4 x^4-2 x^5\right )\right ) \log ^2(x)+\left (-32+2 e^{-8+2 x} x^3+e^{-4+x} \left (8 x-8 x^2\right )\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx={\mathrm {e}}^{2\,x-8}+\frac {8\,x}{\ln \left (x\right )}-\frac {8\,{\mathrm {e}}^{x-4}}{x}+\frac {x^4}{{\ln \left (x\right )}^2}+\frac {16}{x^2}-\frac {2\,x^2\,{\mathrm {e}}^{x-4}}{\ln \left (x\right )} \]
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