Optimal. Leaf size=21 \[ \frac {10}{e^3 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \]
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Rubi [F] time = 4.66, 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 {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\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 {10 e^{-3+\frac {4}{x}} \left (-2 e^{4/x} x^2-e^x \left (4+x^2\right )\right )}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx\\ &=10 \int \frac {e^{-3+\frac {4}{x}} \left (-2 e^{4/x} x^2-e^x \left (4+x^2\right )\right )}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx\\ &=10 \int \left (-\frac {e^{-3+\frac {4}{x}} \left (4+x^2\right )}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )}+\frac {2 e^{-3+\frac {8}{x}} \left (8+4 x+x^2+x^3\right )}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-3+\frac {4}{x}} \left (4+x^2\right )}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )} \, dx\right )+20 \int \frac {e^{-3+\frac {8}{x}} \left (8+4 x+x^2+x^3\right )}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2} \, dx\\ &=-\left (10 \int \frac {4+x^2}{e^3 x^2 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \, dx\right )+20 \int \frac {e^{-3+\frac {8}{x}} \left (8+4 x+x^2+x^3\right )}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx\\ &=20 \int \left (\frac {e^{-3+\frac {8}{x}}}{\left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2}+\frac {8 e^{-3+\frac {8}{x}}}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2}+\frac {4 e^{-3+\frac {8}{x}}}{x \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2}+\frac {e^{-3+\frac {8}{x}} x}{\left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2}\right ) \, dx-\frac {10 \int \frac {4+x^2}{x^2 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \, dx}{e^3}\\ &=20 \int \frac {e^{-3+\frac {8}{x}}}{\left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2} \, dx+20 \int \frac {e^{-3+\frac {8}{x}} x}{\left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2} \, dx+80 \int \frac {e^{-3+\frac {8}{x}}}{x \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2} \, dx+160 \int \frac {e^{-3+\frac {8}{x}}}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )^2} \, dx-\frac {10 \int \left (\frac {e^{4/x}}{4 e^{4/x}+e^x+2 e^{4/x} x}+\frac {4 e^{4/x}}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )}\right ) \, dx}{e^3}\\ &=20 \int \frac {e^{-3+\frac {8}{x}}}{\left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+20 \int \frac {e^{-3+\frac {8}{x}} x}{\left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+80 \int \frac {e^{-3+\frac {8}{x}}}{x \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+160 \int \frac {e^{-3+\frac {8}{x}}}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx-\frac {10 \int \frac {e^{4/x}}{4 e^{4/x}+e^x+2 e^{4/x} x} \, dx}{e^3}-\frac {40 \int \frac {e^{4/x}}{x^2 \left (4 e^{4/x}+e^x+2 e^{4/x} x\right )} \, dx}{e^3}\\ &=20 \int \frac {e^{-3+\frac {8}{x}}}{\left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+20 \int \frac {e^{-3+\frac {8}{x}} x}{\left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+80 \int \frac {e^{-3+\frac {8}{x}}}{x \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx+160 \int \frac {e^{-3+\frac {8}{x}}}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )^2} \, dx-\frac {10 \int \frac {e^{4/x}}{e^x+2 e^{4/x} (2+x)} \, dx}{e^3}-\frac {40 \int \frac {e^{4/x}}{x^2 \left (e^x+2 e^{4/x} (2+x)\right )} \, dx}{e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.79, size = 21, normalized size = 1.00 \begin {gather*} \frac {10}{e^3 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 25, normalized size = 1.19 \begin {gather*} \frac {10}{2 \, {\left (x + 2\right )} e^{3} + e^{\left (\frac {x^{2} + 3 \, x - 4}{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 555, normalized size = 26.43 \begin {gather*} \frac {10 \, {\left (2 \, x^{4} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + x^{3} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 6 \, x^{3} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} - x^{2} e^{\left (x + 6\right )} + 2 \, x^{2} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 12 \, x^{2} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + 4 \, x e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 32 \, x e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + 8 \, e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 32 \, e^{\left (\frac {3 \, x + 4}{x} + 3\right )}\right )}}{4 \, x^{5} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 2 \, x^{4} e^{\left (x + 9\right )} + 2 \, x^{4} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 20 \, x^{4} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + x^{3} e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 6 \, x^{3} e^{\left (x + 9\right )} + 6 \, x^{3} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 48 \, x^{3} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + x^{2} e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 12 \, x^{2} e^{\left (x + 9\right )} + 12 \, x^{2} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 112 \, x^{2} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 4 \, x e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 32 \, x e^{\left (x + 9\right )} + 32 \, x e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 192 \, x e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 8 \, e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 32 \, e^{\left (x + 9\right )} + 32 \, e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 128 \, e^{\left (\frac {3 \, x + 4}{x} + 6\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 23, normalized size = 1.10
method | result | size |
risch | \(\frac {10 \,{\mathrm e}^{-3}}{4+{\mathrm e}^{\frac {\left (x -2\right ) \left (2+x \right )}{x}}+2 x}\) | \(23\) |
norman | \(\frac {10 \,{\mathrm e}^{-3}}{4+{\mathrm e}^{\frac {x^{2}-4}{x}}+2 x}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.87, size = 32, normalized size = 1.52 \begin {gather*} \frac {10 \, e^{\frac {4}{x}}}{2 \, {\left (x e^{3} + 2 \, e^{3}\right )} e^{\frac {4}{x}} + e^{\left (x + 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.81, size = 25, normalized size = 1.19 \begin {gather*} \frac {10}{4\,{\mathrm {e}}^3+2\,x\,{\mathrm {e}}^3+{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {4}{x}}\,{\mathrm {e}}^x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 24, normalized size = 1.14 \begin {gather*} \frac {10}{2 x e^{3} + e^{3} e^{\frac {x^{2} - 4}{x}} + 4 e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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