Optimal. Leaf size=28 \[ \left (e^{4+x}+2 e^{-\frac {3 e^{-\frac {e^{16}}{x}}}{x}} x\right )^2 \]
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Rubi [F] time = 7.32, 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 {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (-24 e^{16} x+24 x^2+2 e^{8+\frac {e^{16}}{x}+\frac {6 e^{-\frac {e^{16}}{x}}}{x}+2 x} x^2+8 e^{\frac {e^{16}}{x}} x^3+e^{\frac {3 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{4+x} \left (-12 e^{16}+12 x\right )+e^{4+\frac {e^{16}}{x}+x} \left (4 x^2+4 x^3\right )\right )\right )}{x^2} \, 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 e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{4+\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}+2 x\right ) \left (-6 e^{16}+6 x+2 e^{\frac {e^{16}}{x}} x^2+e^{\frac {e^{16}+3 e^{-\frac {e^{16}}{x}}+x (4+x)}{x}} x^2\right )}{x^2} \, dx\\ &=2 \int \frac {e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{4+\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}+2 x\right ) \left (-6 e^{16}+6 x+2 e^{\frac {e^{16}}{x}} x^2+e^{\frac {e^{16}+3 e^{-\frac {e^{16}}{x}}+x (4+x)}{x}} x^2\right )}{x^2} \, dx\\ &=2 \int \left (e^{8+2 x}+\frac {4 e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (-3 e^{16}+3 x+e^{\frac {e^{16}}{x}} x^2\right )}{x}-\frac {2 e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \left (3 e^{16}-3 x-e^{\frac {e^{16}}{x}} x^2-e^{\frac {e^{16}}{x}} x^3\right )}{x^2}\right ) \, dx\\ &=2 \int e^{8+2 x} \, dx-4 \int \frac {e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \left (3 e^{16}-3 x-e^{\frac {e^{16}}{x}} x^2-e^{\frac {e^{16}}{x}} x^3\right )}{x^2} \, dx+8 \int \frac {e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (-3 e^{16}+3 x+e^{\frac {e^{16}}{x}} x^2\right )}{x} \, dx\\ &=e^{8+2 x}-4 \int \frac {e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \left (3 e^{16}-3 x-e^{\frac {e^{16}}{x}} x^2 (1+x)\right )}{x^2} \, dx+8 \int \left (-\frac {3 e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{16}-x\right )}{x}+e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} x\right ) \, dx\\ &=e^{8+2 x}-4 \int \left (\frac {3 e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \left (e^{16}-x\right )}{x^2}-e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} (1+x)\right ) \, dx+8 \int e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} x \, dx-24 \int \frac {e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{16}-x\right )}{x} \, dx\\ &=e^{8+2 x}+4 \int e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} (1+x) \, dx+8 \int e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} x \, dx-12 \int \frac {e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \left (e^{16}-x\right )}{x^2} \, dx-24 \int \left (-e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}}+\frac {e^{16-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}}}{x}\right ) \, dx\\ &=e^{8+2 x}+4 \int \left (e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}+e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} x\right ) \, dx+8 \int e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} x \, dx-12 \int \left (\frac {e^{20-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}}{x^2}-\frac {e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}}{x}\right ) \, dx+24 \int e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \, dx-24 \int \frac {e^{16-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}}}{x} \, dx\\ &=e^{8+2 x}+4 \int e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} \, dx+4 \int e^{4-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x} x \, dx+8 \int e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} x \, dx-12 \int \frac {e^{20-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}}{x^2} \, dx+12 \int \frac {e^{4-\frac {e^{16}}{x}-\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}}{x} \, dx+24 \int e^{-\frac {e^{16}}{x}-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \, dx+24 \operatorname {Subst}\left (\int \frac {e^{16-e^{16} x-6 e^{-e^{16} x} x}}{x} \, dx,x,\frac {1}{x}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 44, normalized size = 1.57 \begin {gather*} e^{-\frac {6 e^{-\frac {e^{16}}{x}}}{x}} \left (e^{4+\frac {3 e^{-\frac {e^{16}}{x}}}{x}+x}+2 x\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 103, normalized size = 3.68 \begin {gather*} {\left (4 \, x^{2} e^{\left (\frac {2 \, e^{16}}{x}\right )} + 4 \, x e^{\left (\frac {x^{2} + 4 \, x + e^{16}}{x} + \frac {e^{16}}{x} + \frac {3 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\right )} + e^{\left (\frac {2 \, {\left (x^{2} + 4 \, x + e^{16}\right )}}{x} + \frac {6 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\right )}\right )} e^{\left (-\frac {2 \, e^{16}}{x} - \frac {6 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\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 {2 \, {\left (4 \, x^{3} e^{\left (\frac {e^{16}}{x}\right )} + x^{2} e^{\left (2 \, x + \frac {e^{16}}{x} + \frac {6 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x} + 8\right )} + 12 \, x^{2} - 12 \, x e^{16} + 2 \, {\left ({\left (x^{3} + x^{2}\right )} e^{\left (x + \frac {e^{16}}{x} + 4\right )} + 3 \, {\left (x - e^{16}\right )} e^{\left (x + 4\right )}\right )} e^{\left (\frac {3 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\right )}\right )} e^{\left (-\frac {e^{16}}{x} - \frac {6 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\right )}}{x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 52, normalized size = 1.86
method | result | size |
risch | \({\mathrm e}^{2 x +8}+4 x \,{\mathrm e}^{\frac {x^{2}-3 \,{\mathrm e}^{-\frac {{\mathrm e}^{16}}{x}}+4 x}{x}}+4 x^{2} {\mathrm e}^{-\frac {6 \,{\mathrm e}^{-\frac {{\mathrm e}^{16}}{x}}}{x}}\) | \(52\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, x e^{\left (x - \frac {3 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x} + 4\right )} + e^{\left (2 \, x + 8\right )} + 2 \, \int \frac {4 \, {\left (x^{2} e^{\left (\frac {e^{16}}{x}\right )} + 3 \, x - 3 \, e^{16}\right )} e^{\left (-\frac {e^{16}}{x} - \frac {6 \, e^{\left (-\frac {e^{16}}{x}\right )}}{x}\right )}}{x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}}{x}}\,{\mathrm {e}}^{-\frac {6\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}}{x}}}{x}}\,\left (8\,x^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{16}}{x}}-24\,x\,{\mathrm {e}}^{16}+{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}}{x}}}{x}}\,\left ({\mathrm {e}}^{x+4}\,\left (12\,x-12\,{\mathrm {e}}^{16}\right )+{\mathrm {e}}^{\frac {{\mathrm {e}}^{16}}{x}}\,{\mathrm {e}}^{x+4}\,\left (4\,x^3+4\,x^2\right )\right )+24\,x^2+2\,x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{16}}{x}}\,{\mathrm {e}}^{2\,x+8}\,{\mathrm {e}}^{\frac {6\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}}{x}}}{x}}\right )}{x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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