Optimal. Leaf size=31 \[ 5+\frac {1}{4} \left (-4-\frac {1}{5} e^{-x} x \left (-25+\frac {1}{4} x \log \left (x^2\right )\right )^2\right ) \]
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Rubi [F] time = 0.58, 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}{320} e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{320} \int e^{-x} \left (-10000+10400 x+\left (400 x-204 x^2\right ) \log \left (x^2\right )+\left (-3 x^2+x^3\right ) \log ^2\left (x^2\right )\right ) \, dx\\ &=\frac {1}{320} \int \left (-10000 e^{-x}+10400 e^{-x} x-4 e^{-x} x (-100+51 x) \log \left (x^2\right )+e^{-x} (-3+x) x^2 \log ^2\left (x^2\right )\right ) \, dx\\ &=\frac {1}{320} \int e^{-x} (-3+x) x^2 \log ^2\left (x^2\right ) \, dx-\frac {1}{80} \int e^{-x} x (-100+51 x) \log \left (x^2\right ) \, dx-\frac {125}{4} \int e^{-x} \, dx+\frac {65}{2} \int e^{-x} x \, dx\\ &=\frac {125 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int \left (-3 e^{-x} x^2 \log ^2\left (x^2\right )+e^{-x} x^3 \log ^2\left (x^2\right )\right ) \, dx+\frac {1}{80} \int \frac {2 e^{-x} \left (-2-2 x-51 x^2\right )}{x} \, dx+\frac {65}{2} \int e^{-x} \, dx\\ &=-\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx+\frac {1}{40} \int \frac {e^{-x} \left (-2-2 x-51 x^2\right )}{x} \, dx\\ &=-\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx+\frac {1}{40} \int \left (-2 e^{-x}-\frac {2 e^{-x}}{x}-51 e^{-x} x\right ) \, dx\\ &=-\frac {5 e^{-x}}{4}-\frac {65 e^{-x} x}{2}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx-\frac {1}{20} \int e^{-x} \, dx-\frac {1}{20} \int \frac {e^{-x}}{x} \, dx-\frac {51}{40} \int e^{-x} x \, dx\\ &=-\frac {6 e^{-x}}{5}-\frac {1249 e^{-x} x}{40}-\frac {\text {Ei}(-x)}{20}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx-\frac {51}{40} \int e^{-x} \, dx\\ &=\frac {3 e^{-x}}{40}-\frac {1249 e^{-x} x}{40}-\frac {\text {Ei}(-x)}{20}+\frac {1}{40} e^{-x} \log \left (x^2\right )+\frac {1}{40} e^{-x} x \log \left (x^2\right )+\frac {51}{80} e^{-x} x^2 \log \left (x^2\right )+\frac {1}{320} \int e^{-x} x^3 \log ^2\left (x^2\right ) \, dx-\frac {3}{320} \int e^{-x} x^2 \log ^2\left (x^2\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 20, normalized size = 0.65 \begin {gather*} -\frac {1}{320} e^{-x} x \left (-100+x \log \left (x^2\right )\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.67, size = 36, normalized size = 1.16 \begin {gather*} -\frac {1}{320} \, x^{3} e^{\left (-x\right )} \log \left (x^{2}\right )^{2} + \frac {5}{8} \, x^{2} e^{\left (-x\right )} \log \left (x^{2}\right ) - \frac {125}{4} \, x e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.64, size = 36, normalized size = 1.16 \begin {gather*} -\frac {1}{320} \, x^{3} e^{\left (-x\right )} \log \left (x^{2}\right )^{2} + \frac {5}{8} \, x^{2} e^{\left (-x\right )} \log \left (x^{2}\right ) - \frac {125}{4} \, x e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.08, size = 254, normalized size = 8.19
method | result | size |
risch | \(-\frac {x^{3} {\mathrm e}^{-x} \ln \relax (x )^{2}}{80}+\frac {i x^{2} \left (x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-200 i\right ) {\mathrm e}^{-x} \ln \relax (x )}{160}+\frac {\left (-40000 x -400 i \pi \,x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+800 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-400 i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi ^{2} x^{3} \mathrm {csgn}\left (i x \right )^{4} \mathrm {csgn}\left (i x^{2}\right )^{2}-4 \pi ^{2} x^{3} \mathrm {csgn}\left (i x \right )^{3} \mathrm {csgn}\left (i x^{2}\right )^{3}+6 \pi ^{2} x^{3} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )^{4}-4 \pi ^{2} x^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{5}+\pi ^{2} x^{3} \mathrm {csgn}\left (i x^{2}\right )^{6}\right ) {\mathrm e}^{-x}}{1280}\) | \(254\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 42, normalized size = 1.35 \begin {gather*} -\frac {1}{80} \, {\left (x^{3} \log \relax (x)^{2} - 100 \, x^{2} \log \relax (x) - 100 \, x - 100\right )} e^{\left (-x\right )} - \frac {65}{2} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {125}{4} \, e^{\left (-x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.28, size = 17, normalized size = 0.55 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-x}\,{\left (x\,\ln \left (x^2\right )-100\right )}^2}{320} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 27, normalized size = 0.87 \begin {gather*} \frac {\left (- x^{3} \log {\left (x^{2} \right )}^{2} + 200 x^{2} \log {\left (x^{2} \right )} - 10000 x\right ) e^{- x}}{320} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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