Optimal. Leaf size=36 \[ \frac {-3+\frac {x}{2-\frac {\log (3)}{4 e^3 \left (-e^{2 e^3}+e^x\right )}}}{x} \]
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Rubi [B] time = 1.93, antiderivative size = 164, normalized size of antiderivative = 4.56, number of steps used = 11, number of rules used = 8, integrand size = 144, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6, 6741, 6742, 2282, 44, 36, 29, 31} \begin {gather*} -\frac {3}{x}+\frac {x \log (3)}{16 e^{3+2 e^3}+\log (9)}-\frac {x \log (3)}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}+\frac {\log (3) \log \left (-8 e^{x+3}+8 e^{3+2 e^3}+\log (3)\right )}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3) \log \left (-16 e^{x+3}+16 e^{3+2 e^3}+\log (9)\right )}{16 e^{3+2 e^3}+\log (9)}-\frac {\log (3)}{2 \left (-8 e^{x+3}+8 e^{3+2 e^3}+\log (3)\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 29
Rule 31
Rule 36
Rule 44
Rule 2282
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {192 e^{6+4 e^3}+192 e^{6+2 x}+e^{3+x} \left (-48-4 x^2\right ) \log (3)+3 \log ^2(3)+e^{2 e^3} \left (-384 e^{6+x}+48 e^3 \log (3)\right )}{64 e^{6+2 x} x^2-16 e^{3+x} x^2 \log (3)+e^{2 e^3} \left (-128 e^{6+x} x^2+16 e^3 x^2 \log (3)\right )+x^2 \left (64 e^{6+4 e^3}+\log ^2(3)\right )} \, dx\\ &=\int \frac {192 e^{6+2 x}+e^{3+x} \left (-48-4 x^2\right ) \log (3)+e^{2 e^3} \left (-384 e^{6+x}+48 e^3 \log (3)\right )+192 e^{6+4 e^3} \left (1+\frac {1}{64} e^{-6-4 e^3} \log ^2(3)\right )}{x^2 \left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2} \, dx\\ &=\int \left (\frac {3}{x^2}+\frac {\left (-8 e^{3+2 e^3}-\log (3)\right ) \log (3)}{2 \left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2}+\frac {\log (3)}{-16 e^{3+x}+16 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )}\right ) \, dx\\ &=-\frac {3}{x}+\log (3) \int \frac {1}{-16 e^{3+x}+16 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )} \, dx-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \int \frac {1}{\left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2} \, dx\\ &=-\frac {3}{x}+\log (3) \operatorname {Subst}\left (\int \frac {1}{x \left (16 e^{3+2 e^3}-16 x+\log (9)\right )} \, dx,x,e^{3+x}\right )-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (8 e^{3+2 e^3}-8 x+\log (3)\right )^2} \, dx,x,e^{3+x}\right )\\ &=-\frac {3}{x}-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x \left (8 e^{3+2 e^3}+\log (3)\right )^2}+\frac {8}{\left (8 e^{3+2 e^3}+\log (3)\right ) \left (8 e^{3+2 e^3}-8 x+\log (3)\right )^2}+\frac {8}{\left (8 e^{3+2 e^3}+\log (3)\right )^2 \left (8 e^{3+2 e^3}-8 x+\log (3)\right )}\right ) \, dx,x,e^{3+x}\right )+\frac {\log (3) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{3+x}\right )}{16 e^{3+2 e^3}+\log (9)}+\frac {(16 \log (3)) \operatorname {Subst}\left (\int \frac {1}{16 e^{3+2 e^3}-16 x+\log (9)} \, dx,x,e^{3+x}\right )}{16 e^{3+2 e^3}+\log (9)}\\ &=-\frac {3}{x}-\frac {x \log (3)}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3)}{2 \left (8 e^{3+2 e^3}-8 e^{3+x}+\log (3)\right )}+\frac {x \log (3)}{16 e^{3+2 e^3}+\log (9)}+\frac {\log (3) \log \left (8 e^{3+2 e^3}-8 e^{3+x}+\log (3)\right )}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3) \log \left (16 e^{3+2 e^3}-16 e^{3+x}+\log (9)\right )}{16 e^{3+2 e^3}+\log (9)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 33, normalized size = 0.92 \begin {gather*} -\frac {3}{x}-\frac {\log (3)}{16 e^{3+2 e^3}-16 e^{3+x}+\log (9)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 52, normalized size = 1.44 \begin {gather*} -\frac {{\left (x + 6\right )} e^{3} \log \relax (3) - 48 \, e^{\left (x + 6\right )} + 48 \, e^{\left (2 \, e^{3} + 6\right )}}{2 \, {\left (x e^{3} \log \relax (3) - 8 \, x e^{\left (x + 6\right )} + 8 \, x e^{\left (2 \, e^{3} + 6\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 51, normalized size = 1.42 \begin {gather*} \frac {x \log \relax (3) - 48 \, e^{\left (x + 3\right )} + 48 \, e^{\left (2 \, e^{3} + 3\right )} + 6 \, \log \relax (3)}{2 \, {\left (8 \, x e^{\left (x + 3\right )} - 8 \, x e^{\left (2 \, e^{3} + 3\right )} - x \log \relax (3)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 31, normalized size = 0.86
method | result | size |
risch | \(-\frac {3}{x}-\frac {\ln \relax (3)}{2 \left (8 \,{\mathrm e}^{3+2 \,{\mathrm e}^{3}}-8 \,{\mathrm e}^{3+x}+\ln \relax (3)\right )}\) | \(31\) |
norman | \(\frac {-\frac {x \ln \relax (3)}{2}+24 \,{\mathrm e}^{x} {\mathrm e}^{3}-24 \,{\mathrm e}^{3} {\mathrm e}^{2 \,{\mathrm e}^{3}}-3 \ln \relax (3)}{x \left (8 \,{\mathrm e}^{3} {\mathrm e}^{2 \,{\mathrm e}^{3}}-8 \,{\mathrm e}^{x} {\mathrm e}^{3}+\ln \relax (3)\right )}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 50, normalized size = 1.39 \begin {gather*} -\frac {x \log \relax (3) - 48 \, e^{\left (x + 3\right )} + 48 \, e^{\left (2 \, e^{3} + 3\right )} + 6 \, \log \relax (3)}{2 \, {\left (x {\left (8 \, e^{\left (2 \, e^{3} + 3\right )} + \log \relax (3)\right )} - 8 \, x e^{\left (x + 3\right )}\right )}} \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 = -0.03 \begin {gather*} \int \frac {192\,{\mathrm {e}}^{4\,{\mathrm {e}}^3+6}+192\,{\mathrm {e}}^{2\,x+6}-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (384\,{\mathrm {e}}^{x+6}-48\,{\mathrm {e}}^3\,\ln \relax (3)\right )+3\,{\ln \relax (3)}^2-{\mathrm {e}}^{x+3}\,\ln \relax (3)\,\left (4\,x^2+48\right )}{x^2\,{\ln \relax (3)}^2-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (128\,x^2\,{\mathrm {e}}^{x+6}-16\,x^2\,{\mathrm {e}}^3\,\ln \relax (3)\right )+64\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^3+6}+64\,x^2\,{\mathrm {e}}^{2\,x+6}-16\,x^2\,{\mathrm {e}}^{x+3}\,\ln \relax (3)} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 31, normalized size = 0.86 \begin {gather*} \frac {\log {\relax (3 )}}{16 e^{3} e^{x} - 16 e^{3} e^{2 e^{3}} - 2 \log {\relax (3 )}} - \frac {3}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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