Optimal. Leaf size=22 \[ \frac {1}{2} x \left (x+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}\right ) \]
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Rubi [F] time = 0.61, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{x^2} \, dx\\ &=\frac {1}{2} \int \left (2 x+\frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2 (1+x)} x^4\right )\right )}{x^2}\right ) \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2 (1+x)} x^4\right )\right )}{x^2} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \left (\frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2\right )}{x^2}-\frac {2 \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \log \left (e^{-2-2 x} x^4\right )}{x^2}\right ) \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2\right )}{x^2} \, dx-\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \log \left (e^{-2-2 x} x^4\right )}{x^2} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \left (\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}+\frac {4 \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2}-\frac {2 \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x}\right ) \, dx-\log \left (e^{-2-2 x} x^4\right ) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx+\int \frac {2 (2-x) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx}{x} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \, dx+2 \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx+2 \int \frac {(2-x) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx}{x} \, dx-\log \left (e^{-2-2 x} x^4\right ) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx-\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \, dx+2 \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx+2 \int \left (-\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx+\frac {2 \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx}{x}\right ) \, dx-\log \left (e^{-2-2 x} x^4\right ) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx-\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} \int \left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \, dx+2 \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx-2 \int \left (\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx\right ) \, dx+4 \int \frac {\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx}{x} \, dx-\log \left (e^{-2-2 x} x^4\right ) \int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x^2} \, dx-\int \frac {\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.16, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \left (x+\left (e^{-2 (1+x)} x^4\right )^{\frac {1}{x^2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 23, normalized size = 1.05 \begin {gather*} \frac {1}{2} \, \left (x^{4} e^{\left (-2 \, x - 2\right )}\right )^{\left (\frac {1}{x^{2}}\right )} x + \frac {1}{2} \, x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{3} + {\left (x^{2} - 2 \, x - 2 \, \log \left (x^{4} e^{\left (-2 \, x - 2\right )}\right ) + 4\right )} \left (x^{4} e^{\left (-2 \, x - 2\right )}\right )^{\left (\frac {1}{x^{2}}\right )}}{2 \, x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 26, normalized size = 1.18
method | result | size |
default | \(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2 x -2}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\) | \(26\) |
risch | \(\frac {x^{2}}{2}+\frac {x \,x^{\frac {4}{x^{2}}} \left ({\mathrm e}^{x +1}\right )^{-\frac {2}{x^{2}}} {\mathrm e}^{-\frac {i \pi \left (\mathrm {csgn}\left (i x^{2}\right )^{3}-2 \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )-\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+\mathrm {csgn}\left (i x^{4} {\mathrm e}^{-2 x -2}\right )^{3}-\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x^{4} {\mathrm e}^{-2 x -2}\right )^{2}-\mathrm {csgn}\left (i x^{4} {\mathrm e}^{-2 x -2}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-2 x -2}\right )+\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x^{4} {\mathrm e}^{-2 x -2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x -2}\right )+\mathrm {csgn}\left (i x^{3}\right )^{3}-\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+\mathrm {csgn}\left (i x^{4}\right )^{3}-\mathrm {csgn}\left (i {\mathrm e}^{2 x +2}\right )^{3}+2 \mathrm {csgn}\left (i {\mathrm e}^{2 x +2}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x +1}\right )-\mathrm {csgn}\left (i {\mathrm e}^{2 x +2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x +1}\right )^{2}\right )}{2 x^{2}}}}{2}\) | \(357\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 28, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{\left (-\frac {2}{x} + \frac {4 \, \log \relax (x)}{x^{2}} - \frac {2}{x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.68, size = 28, normalized size = 1.27 \begin {gather*} \frac {x^2}{2}+\frac {x\,{\mathrm {e}}^{-\frac {2}{x}}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\left (x^4\right )}^{\frac {1}{x^2}}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int 2 x\, dx + \int e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}\, dx + \int \frac {8 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}}{x^{2}}\, dx + \int \left (- \frac {2 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}}}{x}\right )\, dx + \int \left (- \frac {2 e^{- \frac {2}{x^{2}}} e^{\frac {\log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}} \log {\left (x^{4} e^{- 2 x} \right )}}{x^{2}}\right )\, dx}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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