Optimal. Leaf size=24 \[ 2+x-\frac {x}{12+e^{-5+2 x} x-x^2} \]
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Rubi [F]
time = 0.64, 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 {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\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 {e^{10} \left (132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx\\ &=e^{10} \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx\\ &=e^{10} \int \left (\frac {1}{e^{10}}-\frac {2 x}{e^5 \left (-12 e^5-e^{2 x} x+e^5 x^2\right )}+\frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \left (-\frac {12}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {24 x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}+\frac {2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx-e^{10} \int \frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx+\left (2 e^{10}\right ) \int \frac {x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (12 e^{10}\right ) \int \frac {1}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (24 e^{10}\right ) \int \frac {x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 3.26, size = 27, normalized size = 1.12 \begin {gather*} x+\frac {e^5 x}{-e^{2 x} x+e^5 \left (-12+x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 21, normalized size = 0.88
method | result | size |
risch | \(x +\frac {x}{x^{2}-x \,{\mathrm e}^{2 x -5}-12}\) | \(21\) |
norman | \(\frac {x^{3}-11 x -x \,{\mathrm e}^{2 x +\ln \left (x \right )-5}}{x^{2}-{\mathrm e}^{2 x +\ln \left (x \right )-5}-12}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 42, normalized size = 1.75 \begin {gather*} \frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 36, normalized size = 1.50 \begin {gather*} \frac {x^{3} - x e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 11 \, x}{x^{2} - e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 12} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 15, normalized size = 0.62 \begin {gather*} x - \frac {x}{- x^{2} + x e^{2 x - 5} + 12} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 42, normalized size = 1.75 \begin {gather*} \frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.04, size = 22, normalized size = 0.92 \begin {gather*} x-\frac {x}{x\,{\mathrm {e}}^{2\,x-5}-x^2+12} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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