Optimal. Leaf size=29 \[ -e^4+\frac {2}{4+x-\frac {x^2}{e^{2+x}-x}} \]
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Rubi [F] time = 3.62, 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 {-2 e^{4+2 x}-4 x^2+e^{2+x} \left (8 x-2 x^2\right )}{16 x^2+16 x^3+4 x^4+e^{4+2 x} \left (16+8 x+x^2\right )+e^{2+x} \left (-32 x-24 x^2-4 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-e^{4+2 x}-e^{2+x} (-4+x) x-2 x^2\right )}{\left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx\\ &=2 \int \frac {-e^{4+2 x}-e^{2+x} (-4+x) x-2 x^2}{\left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx\\ &=2 \int \left (-\frac {1}{(4+x)^2}+\frac {x \left (-8+4 x+x^2\right )}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )}-\frac {2 x^2 \left (-8+5 x^2+x^3\right )}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}\right ) \, dx\\ &=\frac {2}{4+x}+2 \int \frac {x \left (-8+4 x+x^2\right )}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )} \, dx-4 \int \frac {x^2 \left (-8+5 x^2+x^3\right )}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx\\ &=\frac {2}{4+x}+2 \int \frac {x \left (-8+4 x+x^2\right )}{(4+x)^2 \left (2 x (2+x)-e^{2+x} (4+x)\right )} \, dx-4 \int \frac {x^2 \left (-8+5 x^2+x^3\right )}{(4+x)^2 \left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx\\ &=\frac {2}{4+x}+2 \int \left (\frac {4}{4 e^{2+x}-4 x+e^{2+x} x-2 x^2}+\frac {x}{-4 e^{2+x}+4 x-e^{2+x} x+2 x^2}+\frac {32}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )}+\frac {8}{(4+x) \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )}\right ) \, dx-4 \int \left (-\frac {24}{\left (4 e^{2+x}-4 x+e^{2+x} x-2 x^2\right )^2}+\frac {8 x}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}-\frac {3 x^2}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}+\frac {x^3}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}+\frac {128}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}+\frac {64}{(4+x) \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2}\right ) \, dx\\ &=\frac {2}{4+x}+2 \int \frac {x}{-4 e^{2+x}+4 x-e^{2+x} x+2 x^2} \, dx-4 \int \frac {x^3}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx+8 \int \frac {1}{4 e^{2+x}-4 x+e^{2+x} x-2 x^2} \, dx+12 \int \frac {x^2}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx+16 \int \frac {1}{(4+x) \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )} \, dx-32 \int \frac {x}{\left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx+64 \int \frac {1}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )} \, dx+96 \int \frac {1}{\left (4 e^{2+x}-4 x+e^{2+x} x-2 x^2\right )^2} \, dx-256 \int \frac {1}{(4+x) \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx-512 \int \frac {1}{(4+x)^2 \left (-4 e^{2+x}+4 x-e^{2+x} x+2 x^2\right )^2} \, dx\\ &=\frac {2}{4+x}+2 \int \frac {x}{2 x (2+x)-e^{2+x} (4+x)} \, dx-4 \int \frac {x^3}{\left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx+8 \int \frac {1}{-2 x (2+x)+e^{2+x} (4+x)} \, dx+12 \int \frac {x^2}{\left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx+16 \int \frac {1}{(4+x) \left (2 x (2+x)-e^{2+x} (4+x)\right )} \, dx-32 \int \frac {x}{\left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx+64 \int \frac {1}{(4+x)^2 \left (2 x (2+x)-e^{2+x} (4+x)\right )} \, dx+96 \int \frac {1}{\left (4 e^{2+x}-4 x+e^{2+x} x-2 x^2\right )^2} \, dx-256 \int \frac {1}{(4+x) \left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx-512 \int \frac {1}{(4+x)^2 \left (2 x (2+x)-e^{2+x} (4+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.57, size = 29, normalized size = 1.00 \begin {gather*} -\frac {2 \left (-e^{2+x}+x\right )}{-2 x (2+x)+e^{2+x} (4+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 30, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (x - e^{\left (x + 2\right )}\right )}}{2 \, x^{2} - {\left (x + 4\right )} e^{\left (x + 2\right )} + 4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 34, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (x - e^{\left (x + 2\right )}\right )}}{2 \, x^{2} - x e^{\left (x + 2\right )} + 4 \, x - 4 \, e^{\left (x + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 36, normalized size = 1.24
method | result | size |
norman | \(\frac {-2 \,{\mathrm e}^{2+x}+2 x}{2 x^{2}-x \,{\mathrm e}^{2+x}+4 x -4 \,{\mathrm e}^{2+x}}\) | \(36\) |
risch | \(\frac {2}{4+x}-\frac {2 x^{2}}{\left (4+x \right ) \left (2 x^{2}-x \,{\mathrm e}^{2+x}+4 x -4 \,{\mathrm e}^{2+x}\right )}\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 34, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (x - e^{\left (x + 2\right )}\right )}}{2 \, x^{2} - {\left (x e^{2} + 4 \, e^{2}\right )} e^{x} + 4 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.41, size = 35, normalized size = 1.21 \begin {gather*} \frac {2\,x-2\,{\mathrm {e}}^{x+2}}{4\,x-4\,{\mathrm {e}}^{x+2}-x\,{\mathrm {e}}^{x+2}+2\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.23, size = 36, normalized size = 1.24 \begin {gather*} \frac {2 x^{2}}{- 2 x^{3} - 12 x^{2} - 16 x + \left (x^{2} + 8 x + 16\right ) e^{x + 2}} + \frac {2}{x + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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