Optimal. Leaf size=24 \[ -e^{-6-2 x}+e^{\frac {x}{5 (-2-x)}} \]
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Rubi [A] time = 0.27, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {27, 12, 6742, 2194, 2230, 2209} \begin {gather*} e^{\frac {2}{5 (x+2)}-\frac {1}{5}}-e^{-2 x-6} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 2194
Rule 2209
Rule 2230
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 e^{-\frac {x}{10+5 x}}+e^{-6-2 x} \left (40+40 x+10 x^2\right )}{5 (2+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-2 e^{-\frac {x}{10+5 x}}+e^{-6-2 x} \left (40+40 x+10 x^2\right )}{(2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (10 e^{-6-2 x}-\frac {2 e^{-\frac {x}{5 (2+x)}}}{(2+x)^2}\right ) \, dx\\ &=-\left (\frac {2}{5} \int \frac {e^{-\frac {x}{5 (2+x)}}}{(2+x)^2} \, dx\right )+2 \int e^{-6-2 x} \, dx\\ &=-e^{-6-2 x}-\frac {2}{5} \int \frac {e^{-\frac {1}{5}+\frac {2}{5 (2+x)}}}{(2+x)^2} \, dx\\ &=-e^{-6-2 x}+e^{-\frac {1}{5}+\frac {2}{5 (2+x)}}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{5} \left (5 e^{-\frac {x}{5 (2+x)}}-5 e^{-2 (3+x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 18, normalized size = 0.75 \begin {gather*} -e^{\left (-2 \, x - 6\right )} + e^{\left (-\frac {x}{5 \, {\left (x + 2\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 18, normalized size = 0.75 \begin {gather*} -e^{\left (-2 \, x - 6\right )} + e^{\left (-\frac {x}{5 \, {\left (x + 2\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 19, normalized size = 0.79
method | result | size |
risch | \(-{\mathrm e}^{-2 x -6}+{\mathrm e}^{-\frac {x}{5 \left (2+x \right )}}\) | \(19\) |
norman | \(\frac {x \,{\mathrm e}^{-\frac {x}{5 x +10}}-{\mathrm e}^{-2 x -6} x +2 \,{\mathrm e}^{-\frac {x}{5 x +10}}-2 \,{\mathrm e}^{-2 x -6}}{2+x}\) | \(51\) |
default | \({\mathrm e}^{-\frac {1}{5}+\frac {2}{5 \left (2+x \right )}}+8 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{-2 x}}{2+x}+2 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )+8 \,{\mathrm e}^{-6} \left (\frac {2 \,{\mathrm e}^{-2 x}}{2+x}-5 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )+2 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{-2 x}}{2}-\frac {4 \,{\mathrm e}^{-2 x}}{2+x}+12 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )\) | \(101\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {8 \, e^{\left (-2\right )} E_{2}\left (2 \, x + 4\right )}{x + 2} - \frac {{\left (x^{2} e^{\frac {1}{5}} + 4 \, x e^{\frac {1}{5}}\right )} e^{\left (-2 \, x\right )} - {\left (x^{2} e^{6} + 4 \, x e^{6} + 4 \, e^{6}\right )} e^{\left (\frac {2}{5 \, {\left (x + 2\right )}}\right )}}{x^{2} e^{\frac {31}{5}} + 4 \, x e^{\frac {31}{5}} + 4 \, e^{\frac {31}{5}}} + 8 \, \int \frac {e^{\left (-2 \, x\right )}}{x^{3} e^{6} + 6 \, x^{2} e^{6} + 12 \, x e^{6} + 8 \, e^{6}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.08, size = 20, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{-\frac {x}{5\,x+10}}-{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.36, size = 17, normalized size = 0.71 \begin {gather*} - e^{- 2 x - 6} + e^{- \frac {x}{5 x + 10}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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