Optimal. Leaf size=33 \[ \frac {1}{3} \left (x^2+\frac {3 x^2 (4+x)}{16-e^x-x-5 x^2}\right ) \]
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Rubi [F] time = 1.34, 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 {896 x+2 e^{2 x} x+68 x^2-324 x^3+5 x^4+50 x^5+e^x \left (-88 x+7 x^2+23 x^3\right )}{768+3 e^{2 x}-96 x-477 x^2+30 x^3+75 x^4+e^x \left (-96+6 x+30 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 {x \left (896+2 e^{2 x}+68 x-324 x^2+5 x^3+50 x^4+e^x \left (-88+7 x+23 x^2\right )\right )}{3 \left (16-e^x-x-5 x^2\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {x \left (896+2 e^{2 x}+68 x-324 x^2+5 x^3+50 x^4+e^x \left (-88+7 x+23 x^2\right )\right )}{\left (16-e^x-x-5 x^2\right )^2} \, dx\\ &=\frac {1}{3} \int \left (2 x+\frac {3 x \left (-8+x+x^2\right )}{-16+e^x+x+5 x^2}-\frac {3 x^2 \left (-68-53 x+11 x^2+5 x^3\right )}{\left (-16+e^x+x+5 x^2\right )^2}\right ) \, dx\\ &=\frac {x^2}{3}+\int \frac {x \left (-8+x+x^2\right )}{-16+e^x+x+5 x^2} \, dx-\int \frac {x^2 \left (-68-53 x+11 x^2+5 x^3\right )}{\left (-16+e^x+x+5 x^2\right )^2} \, dx\\ &=\frac {x^2}{3}-\int \left (-\frac {68 x^2}{\left (-16+e^x+x+5 x^2\right )^2}-\frac {53 x^3}{\left (-16+e^x+x+5 x^2\right )^2}+\frac {11 x^4}{\left (-16+e^x+x+5 x^2\right )^2}+\frac {5 x^5}{\left (-16+e^x+x+5 x^2\right )^2}\right ) \, dx+\int \left (-\frac {8 x}{-16+e^x+x+5 x^2}+\frac {x^2}{-16+e^x+x+5 x^2}+\frac {x^3}{-16+e^x+x+5 x^2}\right ) \, dx\\ &=\frac {x^2}{3}-5 \int \frac {x^5}{\left (-16+e^x+x+5 x^2\right )^2} \, dx-8 \int \frac {x}{-16+e^x+x+5 x^2} \, dx-11 \int \frac {x^4}{\left (-16+e^x+x+5 x^2\right )^2} \, dx+53 \int \frac {x^3}{\left (-16+e^x+x+5 x^2\right )^2} \, dx+68 \int \frac {x^2}{\left (-16+e^x+x+5 x^2\right )^2} \, dx+\int \frac {x^2}{-16+e^x+x+5 x^2} \, dx+\int \frac {x^3}{-16+e^x+x+5 x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.27, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{3} x^2 \left (1-\frac {3 (4+x)}{-16+e^x+x+5 x^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 36, normalized size = 1.09 \begin {gather*} \frac {5 \, x^{4} - 2 \, x^{3} + x^{2} e^{x} - 28 \, x^{2}}{3 \, {\left (5 \, x^{2} + x + e^{x} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.29, size = 36, normalized size = 1.09 \begin {gather*} \frac {5 \, x^{4} - 2 \, x^{3} + x^{2} e^{x} - 28 \, x^{2}}{3 \, {\left (5 \, x^{2} + x + e^{x} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 27, normalized size = 0.82
method | result | size |
risch | \(\frac {x^{2}}{3}-\frac {x^{2} \left (4+x \right )}{5 x^{2}+{\mathrm e}^{x}+x -16}\) | \(27\) |
norman | \(\frac {\frac {28 x}{15}+\frac {28 \,{\mathrm e}^{x}}{15}-\frac {2 x^{3}}{3}+\frac {5 x^{4}}{3}+\frac {{\mathrm e}^{x} x^{2}}{3}-\frac {448}{15}}{5 x^{2}+{\mathrm e}^{x}+x -16}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.52, size = 36, normalized size = 1.09 \begin {gather*} \frac {5 \, x^{4} - 2 \, x^{3} + x^{2} e^{x} - 28 \, x^{2}}{3 \, {\left (5 \, x^{2} + x + e^{x} - 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.69, size = 35, normalized size = 1.06 \begin {gather*} -\frac {x^2\,\left (2\,x-{\mathrm {e}}^x-5\,x^2+28\right )}{3\,\left (x+{\mathrm {e}}^x+5\,x^2-16\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 26, normalized size = 0.79 \begin {gather*} \frac {x^{2}}{3} + \frac {- x^{3} - 4 x^{2}}{5 x^{2} + x + e^{x} - 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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