Optimal. Leaf size=24 \[ e^{\frac {4 x}{5 \left (x+e^{2 x} x^4\right )}}-2 x \]
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Rubi [F] time = 2.61, 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 {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} 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 {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5 \left (1+e^{2 x} x^3\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-10-\frac {4 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx\\ &=-2 x-\frac {4}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ &=-2 x-\frac {4}{5} \int \left (\frac {3 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2}+\frac {2 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx\\ &=-2 x-\frac {8}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2} \, dx-\frac {12}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.46, size = 31, normalized size = 1.29 \begin {gather*} -\frac {2}{5} \left (-\frac {5}{2} e^{\frac {4}{5 \left (1+e^{2 x} x^3\right )}}+5 x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 45, normalized size = 1.88 \begin {gather*} -{\left (2 \, x e^{\left (2 \, x\right )} - e^{\left (\frac {2 \, {\left (5 \, x^{4} e^{\left (2 \, x\right )} + 5 \, x + 2\right )}}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )}\right )} e^{\left (-2 \, 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 (5 \, x^{6} e^{\left (4 \, x\right )} + 10 \, x^{3} e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} + 5\right )}}{5 \, {\left (x^{6} e^{\left (4 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 20, normalized size = 0.83
method | result | size |
risch | \({\mathrm e}^{\frac {4}{5 \left ({\mathrm e}^{2 x} x^{3}+1\right )}}-2 x\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.39, size = 19, normalized size = 0.79 \begin {gather*} -2 \, x + e^{\left (\frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.21, size = 20, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{\frac {4}{5\,x^3\,{\mathrm {e}}^{2\,x}+5}}-2\,x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.29, size = 17, normalized size = 0.71 \begin {gather*} - 2 x + e^{\frac {4}{5 x^{3} e^{2 x} + 5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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