Optimal. Leaf size=22 \[ \left (e^3+x\right ) \log \left (\frac {-\frac {7}{2}+x+x^2}{10 x^2}\right ) \]
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Rubi [B] time = 0.33, antiderivative size = 134, normalized size of antiderivative = 6.09, number of steps used = 13, number of rules used = 7, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1594, 6688, 1628, 632, 31, 2523, 12} \begin {gather*} x \log \left (-\frac {-2 x^2-2 x+7}{20 x^2}\right )-2 e^3 \log (x)-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (2 x-\sqrt {15}+1\right )+\frac {1}{2} \left (1-\sqrt {15}\right ) \log \left (2 x-\sqrt {15}+1\right )-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (2 x+\sqrt {15}+1\right )+\frac {1}{2} \left (1+\sqrt {15}\right ) \log \left (2 x+\sqrt {15}+1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 632
Rule 1594
Rule 1628
Rule 2523
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{x \left (-7+2 x+2 x^2\right )} \, dx\\ &=\int \left (-\frac {2 (-7+x) \left (e^3+x\right )}{x \left (-7+2 x+2 x^2\right )}+\log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )\right ) \, dx\\ &=-\left (2 \int \frac {(-7+x) \left (e^3+x\right )}{x \left (-7+2 x+2 x^2\right )} \, dx\right )+\int \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right ) \, dx\\ &=x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-2 \int \left (\frac {e^3}{x}+\frac {7+e^3-\left (1-2 e^3\right ) x}{7-2 x-2 x^2}\right ) \, dx-\int \frac {2 (-7+x)}{7-2 x-2 x^2} \, dx\\ &=-2 e^3 \log (x)+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-2 \int \frac {-7+x}{7-2 x-2 x^2} \, dx-2 \int \frac {7+e^3-\left (1-2 e^3\right ) x}{7-2 x-2 x^2} \, dx\\ &=-2 e^3 \log (x)+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-\left (1-\sqrt {15}\right ) \int \frac {1}{-1+\sqrt {15}-2 x} \, dx-\left (1+\sqrt {15}\right ) \int \frac {1}{-1-\sqrt {15}-2 x} \, dx-\left (-1-\sqrt {15}+2 e^3\right ) \int \frac {1}{-1-\sqrt {15}-2 x} \, dx-\left (-1+\sqrt {15}+2 e^3\right ) \int \frac {1}{-1+\sqrt {15}-2 x} \, dx\\ &=-2 e^3 \log (x)+\frac {1}{2} \left (1-\sqrt {15}\right ) \log \left (1-\sqrt {15}+2 x\right )-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (1-\sqrt {15}+2 x\right )+\frac {1}{2} \left (1+\sqrt {15}\right ) \log \left (1+\sqrt {15}+2 x\right )-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (1+\sqrt {15}+2 x\right )+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.09, size = 135, normalized size = 6.14 \begin {gather*} -\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (-1+\sqrt {15}-2 x\right )-2 e^3 \log (x)-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (1+\sqrt {15}+2 x\right )-2 \left (-\frac {1}{4} \left (1-\sqrt {15}\right ) \log \left (1-\sqrt {15}+2 x\right )-\frac {1}{4} \left (1+\sqrt {15}\right ) \log \left (1+\sqrt {15}+2 x\right )\right )+x \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 21, normalized size = 0.95 \begin {gather*} {\left (x + e^{3}\right )} \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 39, normalized size = 1.77 \begin {gather*} e^{3} \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, e^{3} \log \relax (x) + x \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 38, normalized size = 1.73
method | result | size |
derivativedivides | \(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )-x \ln \left (20\right )+\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right ) {\mathrm e}^{3}\) | \(38\) |
default | \(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )-x \ln \left (20\right )+\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right ) {\mathrm e}^{3}\) | \(38\) |
norman | \({\mathrm e}^{3} \ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right )+\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x\) | \(39\) |
risch | \(\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x -2 \,{\mathrm e}^{3} \ln \left (-x \right )+{\mathrm e}^{3} \ln \left (2 x^{2}+2 x -7\right )\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.54, size = 106, normalized size = 4.82 \begin {gather*} -\frac {1}{15} \, \sqrt {15} e^{3} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) - x {\left (\log \relax (5) + 2 \, \log \relax (2)\right )} + \frac {1}{15} \, {\left (\sqrt {15} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) + 15 \, \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 30 \, \log \relax (x)\right )} e^{3} + x \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, x \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.33, size = 41, normalized size = 1.86 \begin {gather*} x\,\ln \left (\frac {1}{x^2}\right )-x\,\ln \left (20\right )-2\,{\mathrm {e}}^3\,\ln \relax (x)+x\,\ln \left (2\,x^2+2\,x-7\right )+\ln \left (x^2+x-\frac {7}{2}\right )\,{\mathrm {e}}^3 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.31, size = 41, normalized size = 1.86 \begin {gather*} x \log {\left (\frac {\frac {x^{2}}{10} + \frac {x}{10} - \frac {7}{20}}{x^{2}} \right )} - 2 e^{3} \log {\relax (x )} + e^{3} \log {\left (2 x^{2} + 2 x - 7 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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