Optimal. Leaf size=26 \[ 2 \left (-1+\left (1+e^{\log ^2\left (\frac {16}{x^2}\right )}-\frac {3}{x}\right )^2-x\right ) \]
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Rubi [B] time = 0.21, antiderivative size = 69, normalized size of antiderivative = 2.65, number of steps used = 7, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2276, 2204, 2209, 2288} \begin {gather*} \frac {18}{x^2}+2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 x-\frac {12}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2204
Rule 2209
Rule 2276
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 \left (18-6 x+x^3\right )}{x^3}-\frac {16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {18-6 x+x^3}{x^3} \, dx\right )-4 \int \frac {e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2} \, dx-16 \int \frac {e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x} \, dx\\ &=-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 \int \left (1+\frac {18}{x^3}-\frac {6}{x^2}\right ) \, dx+8 \operatorname {Subst}\left (\int e^{2 x^2} x \, dx,x,\log \left (\frac {16}{x^2}\right )\right )\\ &=2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}+\frac {18}{x^2}-\frac {12}{x}-2 x-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.19, size = 46, normalized size = 1.77 \begin {gather*} -2 \left (-e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {9}{x^2}+\frac {6}{x}-\frac {2 e^{\log ^2\left (\frac {16}{x^2}\right )} (-3+x)}{x}+x\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 47, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, {\left (x^{2} - 3 \, x\right )} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.72, size = 55, normalized size = 2.12 \begin {gather*} -\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, x^{2} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 44, normalized size = 1.69
method | result | size |
risch | \(-2 x +\frac {-12 x +18}{x^{2}}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}+\frac {4 \left (x -3\right ) {\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}\) | \(44\) |
default | \(\frac {4 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x -12 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}-2 x +\frac {18}{x^{2}}-\frac {12}{x}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}\) | \(56\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.63, size = 77, normalized size = 2.96 \begin {gather*} -2 \, x + \frac {2 \, {\left (x e^{\left (32 \, \log \relax (2)^{2} - 32 \, \log \relax (2) \log \relax (x) + 8 \, \log \relax (x)^{2}\right )} + 2 \, {\left (x e^{\left (16 \, \log \relax (2)^{2}\right )} - 3 \, e^{\left (16 \, \log \relax (2)^{2}\right )}\right )} e^{\left (-16 \, \log \relax (2) \log \relax (x) + 4 \, \log \relax (x)^{2}\right )}\right )}}{x} - \frac {12}{x} + \frac {18}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.64, size = 93, normalized size = 3.58 \begin {gather*} 2\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x^2}\right )}^2+32\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{16\,\ln \relax (2)}-2\,x+4\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \relax (2)}+\frac {18\,x-x^2\,\left (12\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \relax (2)}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \relax (2)}+12\right )}{x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.34, size = 42, normalized size = 1.62 \begin {gather*} - 2 x + \frac {2 x e^{2 \log {\left (\frac {16}{x^{2}} \right )}^{2}} + \left (4 x - 12\right ) e^{\log {\left (\frac {16}{x^{2}} \right )}^{2}}}{x} - \frac {12 x - 18}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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