Optimal. Leaf size=26 \[ 15 (16-x) x+\frac {x}{-e^{e^{x/3}}+x} \]
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Rubi [F] time = 2.10, 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^{2 e^{x/3}} (720-90 x)+720 x^2-90 x^3+e^{e^{x/3}} \left (-3-1440 x+e^{x/3} x+180 x^2\right )}{3 e^{2 e^{x/3}}-6 e^{e^{x/3}} x+3 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 {e^{2 e^{x/3}} (720-90 x)+720 x^2-90 x^3+e^{e^{x/3}} \left (-3-1440 x+e^{x/3} x+180 x^2\right )}{3 \left (e^{e^{x/3}}-x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{2 e^{x/3}} (720-90 x)+720 x^2-90 x^3+e^{e^{x/3}} \left (-3-1440 x+e^{x/3} x+180 x^2\right )}{\left (e^{e^{x/3}}-x\right )^2} \, dx\\ &=\frac {1}{3} \int \left (-\frac {3 e^{e^{x/3}}}{\left (e^{e^{x/3}}-x\right )^2}-\frac {90 e^{2 e^{x/3}} (-8+x)}{\left (e^{e^{x/3}}-x\right )^2}-\frac {1440 e^{e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2}+\frac {e^{\frac {1}{3} \left (3 e^{x/3}+x\right )} x}{\left (e^{e^{x/3}}-x\right )^2}+\frac {720 x^2}{\left (e^{e^{x/3}}-x\right )^2}+\frac {180 e^{e^{x/3}} x^2}{\left (e^{e^{x/3}}-x\right )^2}-\frac {90 x^3}{\left (e^{e^{x/3}}-x\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {1}{3} \left (3 e^{x/3}+x\right )} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-30 \int \frac {e^{2 e^{x/3}} (-8+x)}{\left (e^{e^{x/3}}-x\right )^2} \, dx-30 \int \frac {x^3}{\left (e^{e^{x/3}}-x\right )^2} \, dx+60 \int \frac {e^{e^{x/3}} x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx+240 \int \frac {x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx-480 \int \frac {e^{e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-\int \frac {e^{e^{x/3}}}{\left (e^{e^{x/3}}-x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {1}{3} \left (3 e^{x/3}+x\right )} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-3 \operatorname {Subst}\left (\int \frac {e^{e^x}}{\left (e^{e^x}-3 x\right )^2} \, dx,x,\frac {x}{3}\right )-30 \int \frac {x^3}{\left (e^{e^{x/3}}-x\right )^2} \, dx-30 \int \left (-\frac {8 e^{2 e^{x/3}}}{\left (e^{e^{x/3}}-x\right )^2}+\frac {e^{2 e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2}\right ) \, dx+60 \int \frac {e^{e^{x/3}} x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx+240 \int \frac {x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx-480 \int \frac {e^{e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {1}{3} \left (3 e^{x/3}+x\right )} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-3 \operatorname {Subst}\left (\int \frac {e^{e^x}}{\left (e^{e^x}-3 x\right )^2} \, dx,x,\frac {x}{3}\right )-30 \int \frac {e^{2 e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-30 \int \frac {x^3}{\left (e^{e^{x/3}}-x\right )^2} \, dx+60 \int \frac {e^{e^{x/3}} x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx+240 \int \frac {e^{2 e^{x/3}}}{\left (e^{e^{x/3}}-x\right )^2} \, dx+240 \int \frac {x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx-480 \int \frac {e^{e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {1}{3} \left (3 e^{x/3}+x\right )} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-3 \operatorname {Subst}\left (\int \frac {e^{e^x}}{\left (e^{e^x}-3 x\right )^2} \, dx,x,\frac {x}{3}\right )-30 \int \frac {e^{2 e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx-30 \int \frac {x^3}{\left (e^{e^{x/3}}-x\right )^2} \, dx+60 \int \frac {e^{e^{x/3}} x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx+240 \int \frac {x^2}{\left (e^{e^{x/3}}-x\right )^2} \, dx-480 \int \frac {e^{e^{x/3}} x}{\left (e^{e^{x/3}}-x\right )^2} \, dx+720 \operatorname {Subst}\left (\int \frac {e^{2 e^x}}{\left (e^{e^x}-3 x\right )^2} \, dx,x,\frac {x}{3}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{3} \left (720 x-\frac {3 x}{e^{e^{x/3}}-x}-45 x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 41, normalized size = 1.58 \begin {gather*} -\frac {15 \, x^{3} - 240 \, x^{2} - 15 \, {\left (x^{2} - 16 \, x\right )} e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )} - x}{x - e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.96, size = 45, normalized size = 1.73 \begin {gather*} -\frac {15 \, x^{3} - 15 \, x^{2} e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )} - 240 \, x^{2} + 240 \, x e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )} - x}{x - e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 23, normalized size = 0.88
method | result | size |
risch | \(-15 x^{2}+240 x +\frac {x}{-{\mathrm e}^{{\mathrm e}^{\frac {x}{3}}}+x}\) | \(23\) |
norman | \(\frac {{\mathrm e}^{{\mathrm e}^{\frac {x}{3}}}+240 x^{2}-15 x^{3}-240 x \,{\mathrm e}^{{\mathrm e}^{\frac {x}{3}}}+15 \,{\mathrm e}^{{\mathrm e}^{\frac {x}{3}}} x^{2}}{-{\mathrm e}^{{\mathrm e}^{\frac {x}{3}}}+x}\) | \(47\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.54, size = 41, normalized size = 1.58 \begin {gather*} -\frac {15 \, x^{3} - 240 \, x^{2} - 15 \, {\left (x^{2} - 16 \, x\right )} e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )} - x}{x - e^{\left (e^{\left (\frac {1}{3} \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.84, size = 26, normalized size = 1.00 \begin {gather*} 240\,x+\frac {{\mathrm {e}}^{{\left ({\mathrm {e}}^x\right )}^{1/3}}}{x-{\mathrm {e}}^{{\left ({\mathrm {e}}^x\right )}^{1/3}}}-15\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 17, normalized size = 0.65 \begin {gather*} - 15 x^{2} + 240 x - \frac {x}{- x + e^{e^{\frac {x}{3}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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