Optimal. Leaf size=29 \[ \frac {20 \left (4-\frac {4}{x}\right )}{e^2+e^{e^{3+4 e^4+x}}}+x \]
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Rubi [F] time = 3.44, 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 {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx\\ &=\int \left (\frac {e^4}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2}+\frac {e^{2 e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2}+\frac {2 e^{2+e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2}+\frac {80 e^2}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2}+\frac {80 e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2}+\frac {80 e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x} (1-x)}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x}\right ) \, dx\\ &=2 \int \frac {e^{2+e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x} (1-x)}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x} \, dx+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+e^4 \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2} \, dx+\int \frac {e^{2 e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {e^{2+x}}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )+80 \int \left (-\frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2}+\frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x}\right ) \, dx+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+e^4 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )+\operatorname {Subst}\left (\int \frac {e^{2 x}}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )\\ &=-\left (80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2} \, dx\right )+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x} \, dx+\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^x}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+e^4 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )+\operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {e^4}{\left (e^2+e^x\right )^2 x}-\frac {2 e^2}{\left (e^2+e^x\right ) x}\right ) \, dx,x,e^{3+4 e^4+x}\right )\\ &=-\frac {2 e^{-1-4 e^4-x}}{e^2+e^{e^{3+4 e^4+x}}}+x+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x} \, dx-80 \operatorname {Subst}\left (\int \frac {e^x}{\left (e^2+e^x\right )^2} \, dx,x,e^{3+4 e^4+x}\right )-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x^2} \, dx,x,e^{3+4 e^4+x}\right )-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x} \, dx,x,e^{3+4 e^4+x}\right )+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+2 \left (e^4 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )\right )\\ &=-\frac {2 e^{-1-4 e^4-x}}{e^2+e^{e^{3+4 e^4+x}}}+x+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x} \, dx-80 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+x\right )^2} \, dx,x,e^{e^{3+4 e^4+x}}\right )-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x^2} \, dx,x,e^{3+4 e^4+x}\right )-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x} \, dx,x,e^{3+4 e^4+x}\right )+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+2 \left (e^4 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )\right )\\ &=\frac {80}{e^2+e^{e^{3+4 e^4+x}}}-\frac {2 e^{-1-4 e^4-x}}{e^2+e^{e^{3+4 e^4+x}}}+x+80 \int \frac {e^{e^{3+4 e^4+x}}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+80 \int \frac {e^{e^{3+4 e^4+x}+3 \left (1+\frac {4 e^4}{3}\right )+x}}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x} \, dx-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x^2} \, dx,x,e^{3+4 e^4+x}\right )-\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x} \, dx,x,e^{3+4 e^4+x}\right )+\left (80 e^2\right ) \int \frac {1}{\left (e^2+e^{e^{3+4 e^4+x}}\right )^2 x^2} \, dx+2 \left (e^4 \operatorname {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x} \, dx,x,e^{3+4 e^4+x}\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 28, normalized size = 0.97 \begin {gather*} \frac {80 (-1+x)}{\left (e^2+e^{e^{3+4 e^4+x}}\right ) x}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 50, normalized size = 1.72 \begin {gather*} \frac {x^{2} e^{4} + x^{2} e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )} + 2\right )} + 80 \, {\left (x - 1\right )} e^{2}}{x e^{4} + x e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )} + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 86, normalized size = 2.97 \begin {gather*} \frac {x^{2} e^{\left (x + 4 \, e^{4} + e^{\left (x + 4 \, e^{4} + 3\right )} + 3\right )} + x^{2} e^{\left (x + 4 \, e^{4} + 5\right )} + 80 \, x e^{\left (x + 4 \, e^{4} + 3\right )} - 80 \, e^{\left (x + 4 \, e^{4} + 3\right )}}{x e^{\left (x + 4 \, e^{4} + e^{\left (x + 4 \, e^{4} + 3\right )} + 3\right )} + x e^{\left (x + 4 \, e^{4} + 5\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 25, normalized size = 0.86
method | result | size |
risch | \(x +\frac {80 x -80}{x \left ({\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}\right )}\) | \(25\) |
norman | \(\frac {-80+x^{2} {\mathrm e}^{2}+x^{2} {\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}+80 x}{x \left ({\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}\right )}\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.83, size = 43, normalized size = 1.48 \begin {gather*} \frac {x^{2} e^{2} + x^{2} e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )}\right )} + 80 \, x - 80}{x e^{2} + x e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.10, size = 27, normalized size = 0.93 \begin {gather*} x+\frac {80\,x-80}{x\,\left ({\mathrm {e}}^{{\mathrm {e}}^{4\,{\mathrm {e}}^4}\,{\mathrm {e}}^3\,{\mathrm {e}}^x}+{\mathrm {e}}^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 24, normalized size = 0.83 \begin {gather*} x + \frac {80 x - 80}{x e^{e^{x + 3 + 4 e^{4}}} + x e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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