Optimal. Leaf size=22 \[ \frac {8}{9} (4+x)^2 \left (4+e^{2-x} x\right )^2 \]
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Rubi [B] time = 0.46, antiderivative size = 92, normalized size of antiderivative = 4.18, number of steps used = 29, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 6742, 2196, 2194, 2176} \begin {gather*} \frac {8}{9} e^{4-2 x} x^4+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {1024}{9} e^{2-x} x+\frac {128}{9} (x+4)^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2176
Rule 2194
Rule 2196
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int e^{-2 x} \left (e^{2 x} (1024+256 x)+e^{2+x} \left (1024-320 x^2-64 x^3\right )+e^4 \left (256 x-64 x^2-96 x^3-16 x^4\right )\right ) \, dx\\ &=\frac {1}{9} \int \left (256 (4+x)-64 e^{2-x} \left (-16+5 x^2+x^3\right )-16 e^{4-2 x} x \left (-16+4 x+6 x^2+x^3\right )\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int e^{4-2 x} x \left (-16+4 x+6 x^2+x^3\right ) \, dx-\frac {64}{9} \int e^{2-x} \left (-16+5 x^2+x^3\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int \left (-16 e^{4-2 x} x+4 e^{4-2 x} x^2+6 e^{4-2 x} x^3+e^{4-2 x} x^4\right ) \, dx-\frac {64}{9} \int \left (-16 e^{2-x}+5 e^{2-x} x^2+e^{2-x} x^3\right ) \, dx\\ &=\frac {128}{9} (4+x)^2-\frac {16}{9} \int e^{4-2 x} x^4 \, dx-\frac {64}{9} \int e^{4-2 x} x^2 \, dx-\frac {64}{9} \int e^{2-x} x^3 \, dx-\frac {32}{3} \int e^{4-2 x} x^3 \, dx+\frac {256}{9} \int e^{4-2 x} x \, dx-\frac {320}{9} \int e^{2-x} x^2 \, dx+\frac {1024}{9} \int e^{2-x} \, dx\\ &=-\frac {1024 e^{2-x}}{9}-\frac {128}{9} e^{4-2 x} x+\frac {32}{9} e^{4-2 x} x^2+\frac {320}{9} e^{2-x} x^2+\frac {16}{3} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {32}{9} \int e^{4-2 x} x^3 \, dx-\frac {64}{9} \int e^{4-2 x} x \, dx+\frac {128}{9} \int e^{4-2 x} \, dx-16 \int e^{4-2 x} x^2 \, dx-\frac {64}{3} \int e^{2-x} x^2 \, dx-\frac {640}{9} \int e^{2-x} x \, dx\\ &=-\frac {64}{9} e^{4-2 x}-\frac {1024 e^{2-x}}{9}-\frac {32}{3} e^{4-2 x} x+\frac {640}{9} e^{2-x} x+\frac {104}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {32}{9} \int e^{4-2 x} \, dx-\frac {16}{3} \int e^{4-2 x} x^2 \, dx-16 \int e^{4-2 x} x \, dx-\frac {128}{3} \int e^{2-x} x \, dx-\frac {640}{9} \int e^{2-x} \, dx\\ &=-\frac {16}{3} e^{4-2 x}-\frac {128 e^{2-x}}{3}-\frac {8}{3} e^{4-2 x} x+\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {16}{3} \int e^{4-2 x} x \, dx-8 \int e^{4-2 x} \, dx-\frac {128}{3} \int e^{2-x} \, dx\\ &=-\frac {4}{3} e^{4-2 x}+\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2-\frac {8}{3} \int e^{4-2 x} \, dx\\ &=\frac {1024}{9} e^{2-x} x+\frac {128}{9} e^{4-2 x} x^2+\frac {512}{9} e^{2-x} x^2+\frac {64}{9} e^{4-2 x} x^3+\frac {64}{9} e^{2-x} x^3+\frac {8}{9} e^{4-2 x} x^4+\frac {128}{9} (4+x)^2\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 43, normalized size = 1.95 \begin {gather*} \frac {8}{9} e^{-2 x} x \left (8 e^{2+x} (4+x)^2+e^4 x (4+x)^2+16 e^{2 x} (8+x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.73, size = 59, normalized size = 2.68 \begin {gather*} \frac {8}{9} \, {\left ({\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} e^{8} + 16 \, {\left (x^{2} + 8 \, x\right )} e^{\left (2 \, x + 4\right )} + 8 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} e^{\left (x + 6\right )}\right )} e^{\left (-2 \, x - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 73, normalized size = 3.32 \begin {gather*} \frac {8}{9} \, x^{4} e^{\left (-2 \, x + 4\right )} + \frac {64}{9} \, x^{3} e^{\left (-x + 2\right )} + \frac {64}{9} \, x^{3} e^{\left (-2 \, x + 4\right )} + \frac {512}{9} \, x^{2} e^{\left (-x + 2\right )} + \frac {128}{9} \, x^{2} e^{\left (-2 \, x + 4\right )} + \frac {128}{9} \, x^{2} + \frac {1024}{9} \, x e^{\left (-x + 2\right )} + \frac {1024}{9} \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.12, size = 64, normalized size = 2.91
method | result | size |
risch | \(\frac {128 x^{2}}{9}+\frac {1024 x}{9}+\frac {\left (64 x^{3} {\mathrm e}^{2}+512 x^{2} {\mathrm e}^{2}+1024 \,{\mathrm e}^{2} x \right ) {\mathrm e}^{-x}}{9}+\frac {\left (8 x^{4} {\mathrm e}^{4}+64 x^{3} {\mathrm e}^{4}+128 x^{2} {\mathrm e}^{4}\right ) {\mathrm e}^{-2 x}}{9}\) | \(64\) |
norman | \(\left (\frac {1024 x \,{\mathrm e}^{2 x}}{9}+\frac {128 x^{2} {\mathrm e}^{4}}{9}+\frac {64 x^{3} {\mathrm e}^{4}}{9}+\frac {8 x^{4} {\mathrm e}^{4}}{9}+\frac {128 \,{\mathrm e}^{2 x} x^{2}}{9}+\frac {1024 x \,{\mathrm e}^{2} {\mathrm e}^{x}}{9}+\frac {512 x^{2} {\mathrm e}^{2} {\mathrm e}^{x}}{9}+\frac {64 x^{3} {\mathrm e}^{2} {\mathrm e}^{x}}{9}\right ) {\mathrm e}^{-2 x}\) | \(75\) |
default | \(\frac {128 x^{2}}{9}+\frac {1024 x}{9}-\frac {1024 \,{\mathrm e}^{2} {\mathrm e}^{-x}}{9}+\frac {256 \,{\mathrm e}^{4} \left (-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {64 \,{\mathrm e}^{4} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {32 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-2 x} x^{3}}{2}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{4}-\frac {3 x \,{\mathrm e}^{-2 x}}{4}-\frac {3 \,{\mathrm e}^{-2 x}}{8}\right )}{3}-\frac {16 \,{\mathrm e}^{4} \left (-\frac {x^{4} {\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-2 x} x^{3}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{2}-\frac {3 x \,{\mathrm e}^{-2 x}}{2}-\frac {3 \,{\mathrm e}^{-2 x}}{4}\right )}{9}-\frac {320 \,{\mathrm e}^{2} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )}{9}-\frac {64 \,{\mathrm e}^{2} \left (-x^{3} {\mathrm e}^{-x}-3 x^{2} {\mathrm e}^{-x}-6 x \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-x}\right )}{9}\) | \(215\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 170, normalized size = 7.73 \begin {gather*} \frac {128}{9} \, x^{2} + \frac {64}{9} \, {\left (x^{3} e^{2} + 3 \, x^{2} e^{2} + 6 \, x e^{2} + 6 \, e^{2}\right )} e^{\left (-x\right )} + \frac {320}{9} \, {\left (x^{2} e^{2} + 2 \, x e^{2} + 2 \, e^{2}\right )} e^{\left (-x\right )} + \frac {4}{9} \, {\left (2 \, x^{4} e^{4} + 4 \, x^{3} e^{4} + 6 \, x^{2} e^{4} + 6 \, x e^{4} + 3 \, e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {4}{3} \, {\left (4 \, x^{3} e^{4} + 6 \, x^{2} e^{4} + 6 \, x e^{4} + 3 \, e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {16}{9} \, {\left (2 \, x^{2} e^{4} + 2 \, x e^{4} + e^{4}\right )} e^{\left (-2 \, x\right )} - \frac {64}{9} \, {\left (2 \, x e^{4} + e^{4}\right )} e^{\left (-2 \, x\right )} + \frac {1024}{9} \, x - \frac {1024}{9} \, e^{\left (-x + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.44, size = 61, normalized size = 2.77 \begin {gather*} \frac {1024\,x}{9}+{\mathrm {e}}^{-x}\,\left (\frac {64\,{\mathrm {e}}^2\,x^3}{9}+\frac {512\,{\mathrm {e}}^2\,x^2}{9}+\frac {1024\,{\mathrm {e}}^2\,x}{9}\right )+{\mathrm {e}}^{-2\,x}\,\left (\frac {8\,{\mathrm {e}}^4\,x^4}{9}+\frac {64\,{\mathrm {e}}^4\,x^3}{9}+\frac {128\,{\mathrm {e}}^4\,x^2}{9}\right )+\frac {128\,x^2}{9} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.20, size = 71, normalized size = 3.23 \begin {gather*} \frac {128 x^{2}}{9} + \frac {1024 x}{9} + \frac {\left (576 x^{3} e^{2} + 4608 x^{2} e^{2} + 9216 x e^{2}\right ) e^{- x}}{81} + \frac {\left (72 x^{4} e^{4} + 576 x^{3} e^{4} + 1152 x^{2} e^{4}\right ) e^{- 2 x}}{81} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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