Optimal. Leaf size=29 \[ e^{-1+\frac {x \left (-2+9 e^{-x} x^2\right )}{-3+\frac {x^3}{5}}} \]
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Rubi [F] time = 10.55, 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 {\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) \left (-2025 x^2+675 x^3-45 x^6+e^x \left (150+20 x^3\right )\right )}{225-30 x^3+x^6} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) \left (-2025 x^2+675 x^3-45 x^6+e^x \left (150+20 x^3\right )\right )}{\left (-15+x^3\right )^2} \, dx\\ &=\int \left (\frac {10 \exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) \left (15+2 x^3\right )}{\left (-15+x^3\right )^2}-\frac {45 \exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) x^2 \left (45-15 x+x^4\right )}{\left (-15+x^3\right )^2}\right ) \, dx\\ &=10 \int \frac {\exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) \left (15+2 x^3\right )}{\left (-15+x^3\right )^2} \, dx-45 \int \frac {\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) x^2 \left (45-15 x+x^4\right )}{\left (-15+x^3\right )^2} \, dx\\ &=10 \int \left (\frac {45 \exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{\left (-15+x^3\right )^2}+\frac {2 \exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{-15+x^3}\right ) \, dx-45 \int \left (\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )+\frac {45 \exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) x^2}{\left (-15+x^3\right )^2}+\frac {15 \exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{-15+x^3}\right ) \, dx\\ &=20 \int \frac {\exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{-15+x^3} \, dx-45 \int \exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) \, dx+450 \int \frac {\exp \left (\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{\left (-15+x^3\right )^2} \, dx-675 \int \frac {\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right )}{-15+x^3} \, dx-2025 \int \frac {\exp \left (-x+\frac {e^{-x} \left (45 x^3+e^x \left (15-10 x-x^3\right )\right )}{-15+x^3}\right ) x^2}{\left (-15+x^3\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 1.78, size = 31, normalized size = 1.07 \begin {gather*} e^{-1-\frac {10 x}{-15+x^3}+\frac {45 e^{-x} x^3}{-15+x^3}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 36, normalized size = 1.24 \begin {gather*} e^{\left (x + \frac {{\left (45 \, x^{3} - {\left (x^{4} + x^{3} - 5 \, x - 15\right )} e^{x}\right )} e^{\left (-x\right )}}{x^{3} - 15}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5 \, {\left (9 \, x^{6} - 135 \, x^{3} + 405 \, x^{2} - 2 \, {\left (2 \, x^{3} + 15\right )} e^{x}\right )} e^{\left (-x + \frac {{\left (45 \, x^{3} - {\left (x^{3} + 10 \, x - 15\right )} e^{x}\right )} e^{\left (-x\right )}}{x^{3} - 15}\right )}}{x^{6} - 30 \, x^{3} + 225}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 36, normalized size = 1.24
method | result | size |
risch | \({\mathrm e}^{-\frac {\left ({\mathrm e}^{x} x^{3}-45 x^{3}+10 \,{\mathrm e}^{x} x -15 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{3}-15}}\) | \(36\) |
norman | \(\frac {\left ({\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {\left (\left (-x^{3}-10 x +15\right ) {\mathrm e}^{x}+45 x^{3}\right ) {\mathrm e}^{-x}}{x^{3}-15}}-15 \,{\mathrm e}^{x} {\mathrm e}^{\frac {\left (\left (-x^{3}-10 x +15\right ) {\mathrm e}^{x}+45 x^{3}\right ) {\mathrm e}^{-x}}{x^{3}-15}}\right ) {\mathrm e}^{-x}}{x^{3}-15}\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 32, normalized size = 1.10 \begin {gather*} e^{\left (-\frac {10 \, x}{x^{3} - 15} + \frac {675 \, e^{\left (-x\right )}}{x^{3} - 15} + 45 \, e^{\left (-x\right )} - 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.40, size = 52, normalized size = 1.79 \begin {gather*} {\mathrm {e}}^{-\frac {x^3}{x^3-15}}\,{\mathrm {e}}^{\frac {15}{x^3-15}}\,{\mathrm {e}}^{\frac {45\,x^3\,{\mathrm {e}}^{-x}}{x^3-15}}\,{\mathrm {e}}^{-\frac {10\,x}{x^3-15}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 26, normalized size = 0.90 \begin {gather*} e^{\frac {\left (45 x^{3} + \left (- x^{3} - 10 x + 15\right ) e^{x}\right ) e^{- x}}{x^{3} - 15}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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