Integrand size = 48, antiderivative size = 30 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {e^{-2+x^2} \left (x+\frac {1}{5} \left (4-x^2\right )\right )^2 \log ^2(8)}{x^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(86\) vs. \(2(30)=60\).
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.87, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6873, 6874, 2235, 2245, 2241, 2240, 2243} \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {1}{25} e^{x^2-2} x^2 \log ^2(8)-\frac {2}{5} e^{x^2-2} x \log ^2(8)+\frac {17}{25} e^{x^2-2} \log ^2(8)+\frac {8 e^{x^2-2} \log ^2(8)}{5 x}+\frac {16 e^{x^2-2} \log ^2(8)}{25 x^2} \]
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Rule 12
Rule 2235
Rule 2240
Rule 2241
Rule 2243
Rule 2245
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \log ^2(8) \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right )}{x^3} \, dx \\ & = \frac {1}{25} \log ^2(8) \int \frac {2 e^{-2+x^2} \left (-16-20 x+16 x^2+35 x^3+18 x^4-10 x^5+x^6\right )}{x^3} \, dx \\ & = \frac {1}{25} \left (2 \log ^2(8)\right ) \int \frac {e^{-2+x^2} \left (-16-20 x+16 x^2+35 x^3+18 x^4-10 x^5+x^6\right )}{x^3} \, dx \\ & = \frac {1}{25} \left (2 \log ^2(8)\right ) \int \left (35 e^{-2+x^2}-\frac {16 e^{-2+x^2}}{x^3}-\frac {20 e^{-2+x^2}}{x^2}+\frac {16 e^{-2+x^2}}{x}+18 e^{-2+x^2} x-10 e^{-2+x^2} x^2+e^{-2+x^2} x^3\right ) \, dx \\ & = \frac {1}{25} \left (2 \log ^2(8)\right ) \int e^{-2+x^2} x^3 \, dx-\frac {1}{5} \left (4 \log ^2(8)\right ) \int e^{-2+x^2} x^2 \, dx-\frac {1}{25} \left (32 \log ^2(8)\right ) \int \frac {e^{-2+x^2}}{x^3} \, dx+\frac {1}{25} \left (32 \log ^2(8)\right ) \int \frac {e^{-2+x^2}}{x} \, dx+\frac {1}{25} \left (36 \log ^2(8)\right ) \int e^{-2+x^2} x \, dx-\frac {1}{5} \left (8 \log ^2(8)\right ) \int \frac {e^{-2+x^2}}{x^2} \, dx+\frac {1}{5} \left (14 \log ^2(8)\right ) \int e^{-2+x^2} \, dx \\ & = \frac {18}{25} e^{-2+x^2} \log ^2(8)+\frac {16 e^{-2+x^2} \log ^2(8)}{25 x^2}+\frac {8 e^{-2+x^2} \log ^2(8)}{5 x}-\frac {2}{5} e^{-2+x^2} x \log ^2(8)+\frac {1}{25} e^{-2+x^2} x^2 \log ^2(8)+\frac {7 \sqrt {\pi } \text {erfi}(x) \log ^2(8)}{5 e^2}+\frac {16 \text {Ei}\left (x^2\right ) \log ^2(8)}{25 e^2}-\frac {1}{25} \left (2 \log ^2(8)\right ) \int e^{-2+x^2} x \, dx+\frac {1}{5} \left (2 \log ^2(8)\right ) \int e^{-2+x^2} \, dx-\frac {1}{25} \left (32 \log ^2(8)\right ) \int \frac {e^{-2+x^2}}{x} \, dx-\frac {1}{5} \left (16 \log ^2(8)\right ) \int e^{-2+x^2} \, dx \\ & = \frac {17}{25} e^{-2+x^2} \log ^2(8)+\frac {16 e^{-2+x^2} \log ^2(8)}{25 x^2}+\frac {8 e^{-2+x^2} \log ^2(8)}{5 x}-\frac {2}{5} e^{-2+x^2} x \log ^2(8)+\frac {1}{25} e^{-2+x^2} x^2 \log ^2(8) \\ \end{align*}
Time = 0.92 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {e^{-2+x^2} \left (-4-5 x+x^2\right )^2 \log ^2(8)}{25 x^2} \]
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Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
method | result | size |
gosper | \(\frac {9 \left (x^{2}-5 x -4\right )^{2} \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{25 x^{2}}\) | \(26\) |
risch | \(\frac {9 \ln \left (2\right )^{2} \left (x^{4}-10 x^{3}+17 x^{2}+40 x +16\right ) {\mathrm e}^{x^{2}-2}}{25 x^{2}}\) | \(34\) |
parallelrisch | \(\frac {9 \ln \left (2\right )^{2} \left (x^{4} {\mathrm e}^{x^{2}-2}-10 x^{3} {\mathrm e}^{x^{2}-2}+17 \,{\mathrm e}^{x^{2}-2} x^{2}+40 x \,{\mathrm e}^{x^{2}-2}+16 \,{\mathrm e}^{x^{2}-2}\right )}{25 x^{2}}\) | \(60\) |
norman | \(\frac {\frac {144 \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{25}+\frac {72 x \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{5}+\frac {153 x^{2} \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{25}-\frac {18 x^{3} \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{5}+\frac {9 x^{4} \ln \left (2\right )^{2} {\mathrm e}^{x^{2}-2}}{25}}{x^{2}}\) | \(76\) |
default | \(\frac {18 \ln \left (2\right )^{2} \left ({\mathrm e}^{-2} \left (\frac {x^{2} {\mathrm e}^{x^{2}}}{2}-\frac {{\mathrm e}^{x^{2}}}{2}\right )+\frac {35 \,{\mathrm e}^{-2} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2}-16 \,{\mathrm e}^{-2} \left (-\frac {{\mathrm e}^{x^{2}}}{2 x^{2}}-\frac {\operatorname {Ei}_{1}\left (-x^{2}\right )}{2}\right )-20 \,{\mathrm e}^{-2} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \operatorname {erfi}\left (x \right )\right )-8 \,{\mathrm e}^{-2} \operatorname {Ei}_{1}\left (-x^{2}\right )+9 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-2}-10 \,{\mathrm e}^{-2} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{4}\right )\right )}{25}\) | \(117\) |
meijerg | \(\frac {144 \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (\frac {1}{x^{2}}+1-2 \ln \left (x \right )-i \pi -\frac {2 x^{2}+2}{2 x^{2}}+\frac {{\mathrm e}^{x^{2}}}{x^{2}}+\ln \left (-x^{2}\right )+\operatorname {Ei}_{1}\left (-x^{2}\right )\right )}{25}+\frac {9 \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (1-\frac {\left (-2 x^{2}+2\right ) {\mathrm e}^{x^{2}}}{2}\right )}{25}-\frac {18 i \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (-i {\mathrm e}^{x^{2}} x +\frac {i \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2}\right )}{5}-\frac {162 \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (1-{\mathrm e}^{x^{2}}\right )}{25}+\frac {63 \ln \left (2\right )^{2} {\mathrm e}^{-2} \sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{5}+\frac {144 \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (2 \ln \left (x \right )+i \pi -\ln \left (-x^{2}\right )-\operatorname {Ei}_{1}\left (-x^{2}\right )\right )}{25}-\frac {36 i \ln \left (2\right )^{2} {\mathrm e}^{-2} \left (\frac {2 i {\mathrm e}^{x^{2}}}{x}-2 i \sqrt {\pi }\, \operatorname {erfi}\left (x \right )\right )}{5}\) | \(196\) |
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {9 \, {\left (x^{4} - 10 \, x^{3} + 17 \, x^{2} + 40 \, x + 16\right )} e^{\left (x^{2} - 2\right )} \log \left (2\right )^{2}}{25 \, x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {\left (9 x^{4} \log {\left (2 \right )}^{2} - 90 x^{3} \log {\left (2 \right )}^{2} + 153 x^{2} \log {\left (2 \right )}^{2} + 360 x \log {\left (2 \right )}^{2} + 144 \log {\left (2 \right )}^{2}\right ) e^{x^{2} - 2}}{25 x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.93 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=-\frac {9}{25} \, {\left (40 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{\left (-2\right )} - 16 \, {\rm Ei}\left (x^{2}\right ) e^{\left (-2\right )} - {\left (x^{2} - 1\right )} e^{\left (x^{2} - 2\right )} + 10 \, x e^{\left (x^{2} - 2\right )} + 16 \, e^{\left (-2\right )} \Gamma \left (-1, -x^{2}\right ) - \frac {20 \, \sqrt {-x^{2}} e^{\left (-2\right )} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{x} - 18 \, e^{\left (x^{2} - 2\right )}\right )} \log \left (2\right )^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {9 \, {\left (x^{4} e^{\left (x^{2}\right )} - 10 \, x^{3} e^{\left (x^{2}\right )} + 17 \, x^{2} e^{\left (x^{2}\right )} + 40 \, x e^{\left (x^{2}\right )} + 16 \, e^{\left (x^{2}\right )}\right )} e^{\left (-2\right )} \log \left (2\right )^{2}}{25 \, x^{2}} \]
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Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-2+x^2} \left (-32-40 x+32 x^2+70 x^3+36 x^4-20 x^5+2 x^6\right ) \log ^2(8)}{25 x^3} \, dx=\frac {9\,{\mathrm {e}}^{x^2-2}\,{\ln \left (2\right )}^2\,{\left (-x^2+5\,x+4\right )}^2}{25\,x^2} \]
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