Optimal. Leaf size=23 \[ \left (-\frac {2}{5} e^{-x} \left (-2-e+\frac {3 x}{10}\right )+x\right )^2 \]
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Rubi [B] time = 0.91, antiderivative size = 150, normalized size of antiderivative = 6.52, number of steps used = 20, number of rules used = 6, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 6741, 6742, 2194, 2176, 2196} \begin {gather*} \frac {9}{625} e^{-2 x} x^2-\frac {6}{25} e^{-x} x^2+x^2-\frac {24}{125} e^{-2 x} x-\frac {12 e^{-x} x}{25}+\frac {4}{25} (13+5 e) e^{-x} x-\frac {6}{125} e^{1-2 x}-\frac {12 e^{-2 x}}{125}-\frac {12 e^{-x}}{25}+\frac {2}{125} e^{1-2 x} (43-6 x)+\frac {4}{125} \left (23+5 e^2\right ) e^{-2 x}+\frac {4}{25} (13+5 e) e^{-x}-\frac {4}{5} (2+e) e^{-x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{625} \int e^{-2 x} \left (-920-200 e^2+258 x+1250 e^{2 x} x-18 x^2+e (-860+120 x)+e^x \left (1000+e (500-500 x)-1300 x+150 x^2\right )\right ) \, dx\\ &=\frac {1}{625} \int e^{-2 x} \left (-920 \left (1+\frac {5 e^2}{23}\right )+258 x+1250 e^{2 x} x-18 x^2+e (-860+120 x)+e^x \left (1000+e (500-500 x)-1300 x+150 x^2\right )\right ) \, dx\\ &=\frac {1}{625} \int \left (-40 e^{-2 x} \left (23+5 e^2\right )+1250 x+258 e^{-2 x} x-18 e^{-2 x} x^2+20 e^{1-2 x} (-43+6 x)+50 e^{-x} \left (10 (2+e)-2 (13+5 e) x+3 x^2\right )\right ) \, dx\\ &=x^2-\frac {18}{625} \int e^{-2 x} x^2 \, dx+\frac {4}{125} \int e^{1-2 x} (-43+6 x) \, dx+\frac {2}{25} \int e^{-x} \left (10 (2+e)-2 (13+5 e) x+3 x^2\right ) \, dx+\frac {258}{625} \int e^{-2 x} x \, dx-\frac {1}{125} \left (8 \left (23+5 e^2\right )\right ) \int e^{-2 x} \, dx\\ &=\frac {4}{125} e^{-2 x} \left (23+5 e^2\right )+\frac {2}{125} e^{1-2 x} (43-6 x)-\frac {129}{625} e^{-2 x} x+x^2+\frac {9}{625} e^{-2 x} x^2-\frac {18}{625} \int e^{-2 x} x \, dx+\frac {2}{25} \int \left (10 e^{-x} (2+e)-2 e^{-x} (13+5 e) x+3 e^{-x} x^2\right ) \, dx+\frac {12}{125} \int e^{1-2 x} \, dx+\frac {129}{625} \int e^{-2 x} \, dx\\ &=-\frac {6}{125} e^{1-2 x}-\frac {129 e^{-2 x}}{1250}+\frac {4}{125} e^{-2 x} \left (23+5 e^2\right )+\frac {2}{125} e^{1-2 x} (43-6 x)-\frac {24}{125} e^{-2 x} x+x^2+\frac {9}{625} e^{-2 x} x^2-\frac {9}{625} \int e^{-2 x} \, dx+\frac {6}{25} \int e^{-x} x^2 \, dx+\frac {1}{5} (4 (2+e)) \int e^{-x} \, dx-\frac {1}{25} (4 (13+5 e)) \int e^{-x} x \, dx\\ &=-\frac {6}{125} e^{1-2 x}-\frac {12 e^{-2 x}}{125}-\frac {4}{5} e^{-x} (2+e)+\frac {4}{125} e^{-2 x} \left (23+5 e^2\right )+\frac {2}{125} e^{1-2 x} (43-6 x)-\frac {24}{125} e^{-2 x} x+\frac {4}{25} e^{-x} (13+5 e) x+x^2+\frac {9}{625} e^{-2 x} x^2-\frac {6}{25} e^{-x} x^2+\frac {12}{25} \int e^{-x} x \, dx-\frac {1}{25} (4 (13+5 e)) \int e^{-x} \, dx\\ &=-\frac {6}{125} e^{1-2 x}-\frac {12 e^{-2 x}}{125}-\frac {4}{5} e^{-x} (2+e)+\frac {4}{25} e^{-x} (13+5 e)+\frac {4}{125} e^{-2 x} \left (23+5 e^2\right )+\frac {2}{125} e^{1-2 x} (43-6 x)-\frac {24}{125} e^{-2 x} x-\frac {12 e^{-x} x}{25}+\frac {4}{25} e^{-x} (13+5 e) x+x^2+\frac {9}{625} e^{-2 x} x^2-\frac {6}{25} e^{-x} x^2+\frac {12}{25} \int e^{-x} \, dx\\ &=-\frac {6}{125} e^{1-2 x}-\frac {12 e^{-2 x}}{125}-\frac {12 e^{-x}}{25}-\frac {4}{5} e^{-x} (2+e)+\frac {4}{25} e^{-x} (13+5 e)+\frac {4}{125} e^{-2 x} \left (23+5 e^2\right )+\frac {2}{125} e^{1-2 x} (43-6 x)-\frac {24}{125} e^{-2 x} x-\frac {12 e^{-x} x}{25}+\frac {4}{25} e^{-x} (13+5 e) x+x^2+\frac {9}{625} e^{-2 x} x^2-\frac {6}{25} e^{-x} x^2\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.56, size = 25, normalized size = 1.09 \begin {gather*} \frac {1}{625} e^{-2 x} \left (20+10 e-3 x+25 e^x x\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 56, normalized size = 2.43 \begin {gather*} \frac {1}{625} \, {\left (625 \, x^{2} e^{\left (2 \, x\right )} + 9 \, x^{2} - 20 \, {\left (3 \, x - 20\right )} e - 50 \, {\left (3 \, x^{2} - 10 \, x e - 20 \, x\right )} e^{x} - 120 \, x + 100 \, e^{2} + 400\right )} e^{\left (-2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 76, normalized size = 3.30 \begin {gather*} -\frac {6}{25} \, x^{2} e^{\left (-x\right )} + \frac {9}{625} \, x^{2} e^{\left (-2 \, x\right )} + x^{2} + \frac {8}{5} \, x e^{\left (-x\right )} - \frac {24}{125} \, x e^{\left (-2 \, x\right )} + \frac {4}{5} \, x e^{\left (-x + 1\right )} - \frac {12}{125} \, x e^{\left (-2 \, x + 1\right )} + \frac {16}{25} \, e^{\left (-2 \, x\right )} + \frac {4}{25} \, e^{\left (-2 \, x + 2\right )} + \frac {16}{25} \, e^{\left (-2 \, x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 54, normalized size = 2.35
method | result | size |
risch | \(x^{2}+\frac {\left (500 x \,{\mathrm e}-150 x^{2}+1000 x \right ) {\mathrm e}^{-x}}{625}+\frac {\left (100 \,{\mathrm e}^{2}-60 x \,{\mathrm e}+9 x^{2}+400 \,{\mathrm e}-120 x +400\right ) {\mathrm e}^{-2 x}}{625}\) | \(54\) |
norman | \(\left (\left (-\frac {24}{125}-\frac {12 \,{\mathrm e}}{125}\right ) x +{\mathrm e}^{2 x} x^{2}+\left (\frac {4 \,{\mathrm e}}{5}+\frac {8}{5}\right ) x \,{\mathrm e}^{x}+\frac {9 x^{2}}{625}-\frac {6 \,{\mathrm e}^{x} x^{2}}{25}+\frac {16}{25}+\frac {16 \,{\mathrm e}}{25}+\frac {4 \,{\mathrm e}^{2}}{25}\right ) {\mathrm e}^{-2 x}\) | \(56\) |
default | \(x^{2}+\frac {16 \,{\mathrm e}^{-2 x}}{25}-\frac {24 x \,{\mathrm e}^{-2 x}}{125}+\frac {8 x \,{\mathrm e}^{-x}}{5}+\frac {9 x^{2} {\mathrm e}^{-2 x}}{625}-\frac {6 x^{2} {\mathrm e}^{-x}}{25}-\frac {4 \,{\mathrm e} \,{\mathrm e}^{-x}}{5}+\frac {86 \,{\mathrm e}^{-2 x} {\mathrm e}}{125}+\frac {4 \,{\mathrm e}^{-2 x} {\mathrm e}^{2}}{25}+\frac {24 \,{\mathrm e} \left (-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{125}-\frac {4 \,{\mathrm e} \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )}{5}\) | \(105\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 117, normalized size = 5.09 \begin {gather*} x^{2} - \frac {6}{25} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + \frac {4}{5} \, {\left (x e + e\right )} e^{\left (-x\right )} + \frac {52}{25} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {9}{1250} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {6}{125} \, {\left (2 \, x e + e\right )} e^{\left (-2 \, x\right )} - \frac {129}{1250} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {8}{5} \, e^{\left (-x\right )} + \frac {92}{125} \, e^{\left (-2 \, x\right )} - \frac {4}{5} \, e^{\left (-x + 1\right )} + \frac {4}{25} \, e^{\left (-2 \, x + 2\right )} + \frac {86}{125} \, e^{\left (-2 \, x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.24, size = 22, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}\,{\left (10\,\mathrm {e}-3\,x+25\,x\,{\mathrm {e}}^x+20\right )}^2}{625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.21, size = 56, normalized size = 2.43 \begin {gather*} x^{2} + \frac {\left (- 3750 x^{2} + 25000 x + 12500 e x\right ) e^{- x}}{15625} + \frac {\left (225 x^{2} - 1500 e x - 3000 x + 10000 + 2500 e^{2} + 10000 e\right ) e^{- 2 x}}{15625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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