Optimal. Leaf size=24 \[ e^{-3+5 x \left (-e^{x/2}+4 x^4\right ) \log (x)} \]
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Rubi [F] time = 3.35, 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 {1}{2} e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} \left (-10 e^{x/2}+40 x^4+\left (e^{x/2} (-10-5 x)+200 x^4\right ) \log (x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} \left (-10 e^{x/2}+40 x^4+\left (e^{x/2} (-10-5 x)+200 x^4\right ) \log (x)\right ) \, dx\\ &=\frac {1}{2} \int \left (-10 e^{-3+\frac {x}{2}+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)}+40 e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4-5 e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} \left (2 e^{x/2}+e^{x/2} x-40 x^4\right ) \log (x)\right ) \, dx\\ &=-\left (\frac {5}{2} \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} \left (2 e^{x/2}+e^{x/2} x-40 x^4\right ) \log (x) \, dx\right )-5 \int e^{-3+\frac {x}{2}+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx\\ &=-\left (\frac {5}{2} \int \left (-40 e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \log (x)+e^{-3+\frac {x}{2}+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} (2+x) \log (x)\right ) \, dx\right )-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx\\ &=-\left (\frac {5}{2} \int e^{-3+\frac {x}{2}+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} (2+x) \log (x) \, dx\right )-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx+100 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \log (x) \, dx\\ &=-\left (\frac {5}{2} \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) (2+x) \log (x) \, dx\right )-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx+100 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \log (x) \, dx\\ &=-\left (\frac {5}{2} \int \left (2 \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \log (x)+\exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) x \log (x)\right ) \, dx\right )-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx+100 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \log (x) \, dx\\ &=-\left (\frac {5}{2} \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) x \log (x) \, dx\right )-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \, dx-5 \int \exp \left (\frac {1}{2} \left (-6+x-10 e^{x/2} x \log (x)+40 x^5 \log (x)\right )\right ) \log (x) \, dx+20 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \, dx+100 \int e^{-3+\left (-5 e^{x/2} x+20 x^5\right ) \log (x)} x^4 \log (x) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 24, normalized size = 1.00 \begin {gather*} \frac {x^{5 x \left (-e^{x/2}+4 x^4\right )}}{e^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 20, normalized size = 0.83 \begin {gather*} e^{\left (5 \, {\left (4 \, x^{5} - x e^{\left (\frac {1}{2} \, x\right )}\right )} \log \relax (x) - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 19, normalized size = 0.79 \begin {gather*} e^{\left (20 \, x^{5} \log \relax (x) - 5 \, x e^{\left (\frac {1}{2} \, x\right )} \log \relax (x) - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 19, normalized size = 0.79
method | result | size |
risch | \(x^{-5 x \left (-4 x^{4}+{\mathrm e}^{\frac {x}{2}}\right )} {\mathrm e}^{-3}\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.82, size = 19, normalized size = 0.79 \begin {gather*} e^{\left (20 \, x^{5} \log \relax (x) - 5 \, x e^{\left (\frac {1}{2} \, x\right )} \log \relax (x) - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.72, size = 18, normalized size = 0.75 \begin {gather*} x^{20\,x^5-5\,x\,{\mathrm {e}}^{x/2}}\,{\mathrm {e}}^{-3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.53, size = 19, normalized size = 0.79 \begin {gather*} e^{\left (20 x^{5} - 5 x e^{\frac {x}{2}}\right ) \log {\relax (x )} - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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