Integrand size = 48, antiderivative size = 34 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=1+x+\frac {1}{3} \left (3-\frac {5+x}{x}\right ) \left (x+e^{-x} \left (-4+\frac {\log (\log (3))}{x}\right )\right ) \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.38 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.03, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6874, 2230, 2225, 2208, 2209} \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=-\frac {2}{3} (10+\log (\log (3))) \operatorname {ExpIntegralEi}(-x)+\frac {5}{3} \log (\log (3)) \operatorname {ExpIntegralEi}(-x)+\frac {1}{3} (20-3 \log (\log (3))) \operatorname {ExpIntegralEi}(-x)-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 x}{3}-\frac {8 e^{-x}}{3}+\frac {5 e^{-x} \log (\log (3))}{3 x}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x} \]
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Rule 12
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{x^3} \, dx \\ & = \frac {1}{3} \int \left (5+\frac {e^{-x} \left (8 x^3-x (20-3 \log (\log (3)))+10 \log (\log (3))-2 x^2 (10+\log (\log (3)))\right )}{x^3}\right ) \, dx \\ & = \frac {5 x}{3}+\frac {1}{3} \int \frac {e^{-x} \left (8 x^3-x (20-3 \log (\log (3)))+10 \log (\log (3))-2 x^2 (10+\log (\log (3)))\right )}{x^3} \, dx \\ & = \frac {5 x}{3}+\frac {1}{3} \int \left (8 e^{-x}+\frac {10 e^{-x} \log (\log (3))}{x^3}-\frac {2 e^{-x} (10+\log (\log (3)))}{x}+\frac {e^{-x} (-20+3 \log (\log (3)))}{x^2}\right ) \, dx \\ & = \frac {5 x}{3}+\frac {8}{3} \int e^{-x} \, dx+\frac {1}{3} (10 \log (\log (3))) \int \frac {e^{-x}}{x^3} \, dx-\frac {1}{3} (2 (10+\log (\log (3)))) \int \frac {e^{-x}}{x} \, dx+\frac {1}{3} (-20+3 \log (\log (3))) \int \frac {e^{-x}}{x^2} \, dx \\ & = -\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}-\frac {5 e^{-x} \log (\log (3))}{3 x^2}-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3)))+\frac {1}{3} (20-3 \log (\log (3))) \int \frac {e^{-x}}{x} \, dx-\frac {1}{3} (5 \log (\log (3))) \int \frac {e^{-x}}{x^2} \, dx \\ & = -\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}+\frac {1}{3} \text {Ei}(-x) (20-3 \log (\log (3)))-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 e^{-x} \log (\log (3))}{3 x}-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3)))+\frac {1}{3} (5 \log (\log (3))) \int \frac {e^{-x}}{x} \, dx \\ & = -\frac {8 e^{-x}}{3}+\frac {5 x}{3}+\frac {e^{-x} (20-3 \log (\log (3)))}{3 x}+\frac {1}{3} \text {Ei}(-x) (20-3 \log (\log (3)))-\frac {5 e^{-x} \log (\log (3))}{3 x^2}+\frac {5 e^{-x} \log (\log (3))}{3 x}+\frac {5}{3} \text {Ei}(-x) \log (\log (3))-\frac {2}{3} \text {Ei}(-x) (10+\log (\log (3))) \\ \end{align*}
Time = 0.88 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=\frac {1}{3} \left (5 x+e^{-x} \left (-8-\frac {5 \log (\log (3))}{x^2}+\frac {2 (10+\log (\log (3)))}{x}\right )\right ) \]
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Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {5 x}{3}+\frac {\left (2 \ln \left (\ln \left (3\right )\right ) x -8 x^{2}-5 \ln \left (\ln \left (3\right )\right )+20 x \right ) {\mathrm e}^{-x}}{3 x^{2}}\) | \(34\) |
norman | \(\frac {\left (\left (\frac {2 \ln \left (\ln \left (3\right )\right )}{3}+\frac {20}{3}\right ) x -\frac {8 x^{2}}{3}+\frac {5 \,{\mathrm e}^{x} x^{3}}{3}-\frac {5 \ln \left (\ln \left (3\right )\right )}{3}\right ) {\mathrm e}^{-x}}{x^{2}}\) | \(36\) |
parallelrisch | \(\frac {\left (5 \,{\mathrm e}^{x} x^{3}+2 \ln \left (\ln \left (3\right )\right ) x -8 x^{2}-5 \ln \left (\ln \left (3\right )\right )+20 x \right ) {\mathrm e}^{-x}}{3 x^{2}}\) | \(37\) |
parts | \(\frac {5 x}{3}-\frac {8 \,{\mathrm e}^{-x}}{3}+\frac {20 \,{\mathrm e}^{-x}}{3 x}-\frac {5 \,{\mathrm e}^{-x} \ln \left (\ln \left (3\right )\right )}{3 x^{2}}+\frac {2 \,{\mathrm e}^{-x} \ln \left (\ln \left (3\right )\right )}{3 x}\) | \(44\) |
default | \(\frac {5 x}{3}-\frac {8 \,{\mathrm e}^{-x}}{3}+\frac {20 \,{\mathrm e}^{-x}}{3 x}+\frac {10 \ln \left (\ln \left (3\right )\right ) \left (-\frac {{\mathrm e}^{-x}}{2 x^{2}}+\frac {{\mathrm e}^{-x}}{2 x}-\frac {\operatorname {Ei}_{1}\left (x \right )}{2}\right )}{3}+\ln \left (\ln \left (3\right )\right ) \left (-\frac {{\mathrm e}^{-x}}{x}+\operatorname {Ei}_{1}\left (x \right )\right )+\frac {2 \ln \left (\ln \left (3\right )\right ) \operatorname {Ei}_{1}\left (x \right )}{3}\) | \(74\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=\frac {{\left (5 \, x^{3} e^{x} - 8 \, x^{2} + {\left (2 \, x - 5\right )} \log \left (\log \left (3\right )\right ) + 20 \, x\right )} e^{\left (-x\right )}}{3 \, x^{2}} \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=\frac {5 x}{3} + \frac {\left (- 8 x^{2} + 2 x \log {\left (\log {\left (3 \right )} \right )} + 20 x - 5 \log {\left (\log {\left (3 \right )} \right )}\right ) e^{- x}}{3 x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=-\frac {2}{3} \, {\rm Ei}\left (-x\right ) \log \left (\log \left (3\right )\right ) - \Gamma \left (-1, x\right ) \log \left (\log \left (3\right )\right ) - \frac {10}{3} \, \Gamma \left (-2, x\right ) \log \left (\log \left (3\right )\right ) + \frac {5}{3} \, x - \frac {20}{3} \, {\rm Ei}\left (-x\right ) - \frac {8}{3} \, e^{\left (-x\right )} + \frac {20}{3} \, \Gamma \left (-1, x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=\frac {5 \, x^{3} - 8 \, x^{2} e^{\left (-x\right )} + 2 \, x e^{\left (-x\right )} \log \left (\log \left (3\right )\right ) + 20 \, x e^{\left (-x\right )} - 5 \, e^{\left (-x\right )} \log \left (\log \left (3\right )\right )}{3 \, x^{2}} \]
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Time = 10.76 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-x} \left (-20 x-20 x^2+8 x^3+5 e^x x^3+\left (10+3 x-2 x^2\right ) \log (\log (3))\right )}{3 x^3} \, dx=\frac {5\,x}{3}-\frac {8\,{\mathrm {e}}^{-x}}{3}-\frac {\frac {5\,{\mathrm {e}}^{-x}\,\ln \left (\ln \left (3\right )\right )}{3}-\frac {x\,{\mathrm {e}}^{-x}\,\left (2\,\ln \left (\ln \left (3\right )\right )+20\right )}{3}}{x^2} \]
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