Optimal. Leaf size=23 \[ x^{3 x \left (\frac {2}{5} e^{-x} \left (4+\frac {1}{x}\right )+5 x\right )} \]
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Rubi [F] time = 2.14, 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}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\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}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx\\ &=\frac {1}{5} \int \left (24 e^{-x} x^{\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )}+6 e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )}+75 x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )}+6 e^{-x} x^{\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (3-4 x+25 e^x x\right ) \log (x)\right ) \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+\frac {6}{5} \int e^{-x} x^{\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (3-4 x+25 e^x x\right ) \log (x) \, dx+\frac {24}{5} \int e^{-x} x^{\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-\frac {6}{5} \int \frac {3 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+25 \int x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx-4 \int e^{-x} x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+\frac {24}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+\frac {1}{5} (18 \log (x)) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx-\frac {1}{5} (24 \log (x)) \int e^{-x} x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+(30 \log (x)) \int x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-\frac {6}{5} \int \left (\frac {3 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+25 \int x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x}-\frac {4 \int e^{-x} x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x}\right ) \, dx+\frac {24}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+\frac {1}{5} (18 \log (x)) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx-\frac {1}{5} (24 \log (x)) \int e^{-x} x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+(30 \log (x)) \int x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-\frac {6}{5} \int \frac {3 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+25 \int x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+\frac {24}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+\frac {24}{5} \int \frac {\int e^{-x} x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+\frac {1}{5} (18 \log (x)) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx-\frac {1}{5} (24 \log (x)) \int e^{-x} x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+(30 \log (x)) \int x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-\frac {6}{5} \int \left (\frac {3 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx}{x}+\frac {25 \int x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x}\right ) \, dx+\frac {24}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+\frac {24}{5} \int \frac {\int e^{-x} x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx+\frac {1}{5} (18 \log (x)) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx-\frac {1}{5} (24 \log (x)) \int e^{-x} x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+(30 \log (x)) \int x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx\\ &=\frac {6}{5} \int e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-\frac {18}{5} \int \frac {\int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx}{x} \, dx+\frac {24}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+\frac {24}{5} \int \frac {\int e^{-x} x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+15 \int x^{1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \, dx-30 \int \frac {\int x^{1+15 x^2+\frac {6}{5} e^{-x} (1+4 x)} \, dx}{x} \, dx+\frac {1}{5} (18 \log (x)) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx-\frac {1}{5} (24 \log (x)) \int e^{-x} x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx+(30 \log (x)) \int x^{1+\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.45, size = 24, normalized size = 1.04 \begin {gather*} x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 20, normalized size = 0.87 \begin {gather*} x^{\frac {3}{5} \, {\left (25 \, x^{2} e^{x} + 8 \, x + 2\right )} e^{\left (-x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.07, size = 26, normalized size = 1.13 \begin {gather*} e^{\left (15 \, x^{2} \log \relax (x) + \frac {24}{5} \, x e^{\left (-x\right )} \log \relax (x) + \frac {6}{5} \, e^{\left (-x\right )} \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 22, normalized size = 0.96
method | result | size |
risch | \(x^{15 x^{2}+\frac {24 x \,{\mathrm e}^{-x}}{5}+\frac {6 \,{\mathrm e}^{-x}}{5}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.57, size = 26, normalized size = 1.13 \begin {gather*} e^{\left (15 \, x^{2} \log \relax (x) + \frac {24}{5} \, x e^{\left (-x\right )} \log \relax (x) + \frac {6}{5} \, e^{\left (-x\right )} \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.92, size = 20, normalized size = 0.87 \begin {gather*} x^{\frac {3\,{\mathrm {e}}^{-x}\,\left (8\,x+25\,x^2\,{\mathrm {e}}^x+2\right )}{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.61, size = 24, normalized size = 1.04 \begin {gather*} e^{\left (15 x^{2} e^{x} + \frac {24 x}{5} + \frac {6}{5}\right ) e^{- x} \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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