Optimal. Leaf size=29 \[ \frac {e^{4-\frac {1}{x}-\frac {16 x^2}{9}+\frac {x}{3+x^2}}}{x} \]
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Rubi [F] time = 3.13, 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 {e^{\frac {-27+108 x-12 x^3-16 x^5}{27 x+9 x^3}} \left (81-81 x+81 x^2-342 x^3-201 x^5-32 x^7\right )}{81 x^3+54 x^5+9 x^7} \, 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^{\frac {-27+108 x-12 x^3-16 x^5}{27 x+9 x^3}} \left (81-81 x+81 x^2-342 x^3-201 x^5-32 x^7\right )}{x^3 \left (81+54 x^2+9 x^4\right )} \, dx\\ &=9 \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{27 x+9 x^3}} \left (81-81 x+81 x^2-342 x^3-201 x^5-32 x^7\right )}{x^3 \left (27+9 x^2\right )^2} \, dx\\ &=9 \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} \left (81-81 x+81 x^2-342 x^3-201 x^5-32 x^7\right )}{x^3 \left (27+9 x^2\right )^2} \, dx\\ &=9 \int \left (-\frac {32}{81} e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}+\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{9 x^3}-\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{9 x^2}+\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{27 x}-\frac {2 e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{9 \left (3+x^2\right )^2}-\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{27 \left (3+x^2\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x} \, dx-\frac {1}{3} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{3+x^2} \, dx-2 \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{\left (3+x^2\right )^2} \, dx-\frac {32}{9} \int e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} \, dx+\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^3} \, dx-\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x} \, dx-\frac {1}{3} \int \left (-\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{2 \left (i \sqrt {3}-x\right )}+\frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{2 \left (i \sqrt {3}+x\right )}\right ) \, dx-2 \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{\left (3+x^2\right )^2} \, dx-\frac {32}{9} \int e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} \, dx+\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^3} \, dx-\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^2} \, dx\\ &=\frac {1}{6} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{i \sqrt {3}-x} \, dx-\frac {1}{6} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{i \sqrt {3}+x} \, dx+\frac {1}{3} \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x} \, dx-2 \int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} x}{\left (3+x^2\right )^2} \, dx-\frac {32}{9} \int e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}} \, dx+\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^3} \, dx-\int \frac {e^{\frac {-27+108 x-12 x^3-16 x^5}{x \left (27+9 x^2\right )}}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.81, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{4-\frac {1}{x}-\frac {16 x^2}{9}+\frac {x}{3+x^2}}}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 31, normalized size = 1.07 \begin {gather*} \frac {e^{\left (-\frac {16 \, x^{5} + 12 \, x^{3} - 108 \, x + 27}{9 \, {\left (x^{3} + 3 \, x\right )}}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 31, normalized size = 1.07 \begin {gather*} \frac {e^{\left (-\frac {16 \, x^{5} + 12 \, x^{3} - 108 \, x + 27}{9 \, {\left (x^{3} + 3 \, x\right )}}\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 33, normalized size = 1.14
method | result | size |
gosper | \(\frac {{\mathrm e}^{-\frac {16 x^{5}+12 x^{3}-108 x +27}{9 \left (x^{2}+3\right ) x}}}{x}\) | \(33\) |
risch | \(\frac {{\mathrm e}^{-\frac {16 x^{5}+12 x^{3}-108 x +27}{9 \left (x^{2}+3\right ) x}}}{x}\) | \(33\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {-16 x^{5}-12 x^{3}+108 x -27}{9 x^{3}+27 x}}+3 x \,{\mathrm e}^{\frac {-16 x^{5}-12 x^{3}+108 x -27}{9 x^{3}+27 x}}}{x^{2} \left (x^{2}+3\right )}\) | \(76\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 26, normalized size = 0.90 \begin {gather*} \frac {e^{\left (-\frac {16}{9} \, x^{2} + \frac {x}{x^{2} + 3} - \frac {1}{x} + 4\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.47, size = 56, normalized size = 1.93 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {3}{x^3+3\,x}}\,{\mathrm {e}}^{-\frac {4\,x^2}{3\,\left (x^2+3\right )}}\,{\mathrm {e}}^{-\frac {16\,x^4}{9\,\left (x^2+3\right )}}\,{\mathrm {e}}^{\frac {12}{x^2+3}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 26, normalized size = 0.90 \begin {gather*} \frac {e^{\frac {- 16 x^{5} - 12 x^{3} + 108 x - 27}{9 x^{3} + 27 x}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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