Integrand size = 68, antiderivative size = 22 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=\left (e^3+x\right ) \log \left (\frac {-\frac {7}{2}+x+x^2}{10 x^2}\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(134\) vs. \(2(22)=44\).
Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 6.09, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1608, 6820, 1642, 646, 31, 2603, 12} \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=x \log \left (-\frac {-2 x^2-2 x+7}{20 x^2}\right )-2 e^3 \log (x)-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (2 x-\sqrt {15}+1\right )+\frac {1}{2} \left (1-\sqrt {15}\right ) \log \left (2 x-\sqrt {15}+1\right )-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (2 x+\sqrt {15}+1\right )+\frac {1}{2} \left (1+\sqrt {15}\right ) \log \left (2 x+\sqrt {15}+1\right ) \]
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Rule 12
Rule 31
Rule 646
Rule 1608
Rule 1642
Rule 2603
Rule 6820
Rubi steps \begin{align*} \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{align*}
Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(22)=44\).
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 6.14 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (-1+\sqrt {15}-2 x\right )-2 e^3 \log (x)-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (1+\sqrt {15}+2 x\right )-2 \left (-\frac {1}{4} \left (1-\sqrt {15}\right ) \log \left (1-\sqrt {15}+2 x\right )-\frac {1}{4} \left (1+\sqrt {15}\right ) \log \left (1+\sqrt {15}+2 x\right )\right )+x \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(38\) vs. \(2(17)=34\).
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77
method | result | size |
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 | \(\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x -2 \,{\mathrm e}^{3} \ln \left (-x \right )+{\mathrm e}^{3} \ln \left (2 x^{2}+2 x -7\right )\) | \(42\) |
derivativedivides | \(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )+\frac {\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right )}{2}-x \ln \left (20\right )+\frac {\left (14 \,{\mathrm e}^{3}-7\right ) \ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right )}{14}\) | \(58\) |
default | \(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )+\frac {\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right )}{2}-x \ln \left (20\right )+\frac {\left (14 \,{\mathrm e}^{3}-7\right ) \ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right )}{14}\) | \(58\) |
parallelrisch | \(-2 \ln \left (x \right ) {\mathrm e}^{3}+{\mathrm e}^{3} \ln \left (x^{2}-\frac {7}{2}+x \right )+\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x -4 \ln \left (x \right )+2 \ln \left (x^{2}-\frac {7}{2}+x \right )-2 \ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right )\) | \(67\) |
parts | \(-\frac {\left (-2 \,{\mathrm e}^{3}+1\right ) \ln \left (2 x^{2}+2 x -7\right )}{2}-\sqrt {15}\, \operatorname {arctanh}\left (\frac {\left (4 x +2\right ) \sqrt {15}}{30}\right )-2 \ln \left (x \right ) {\mathrm e}^{3}+x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )-\ln \left (\frac {1}{x}\right )+\frac {\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right )}{2}+\sqrt {15}\, \operatorname {arctanh}\left (\frac {\left (\frac {14}{x}-2\right ) \sqrt {15}}{30}\right )-x \ln \left (20\right )\) | \(101\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx={\left (x + e^{3}\right )} \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=x \log {\left (\frac {\frac {x^{2}}{10} + \frac {x}{10} - \frac {7}{20}}{x^{2}} \right )} - 2 e^{3} \log {\left (x \right )} + e^{3} \log {\left (2 x^{2} + 2 x - 7 \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (21) = 42\).
Time = 0.36 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.82 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=-\frac {1}{15} \, \sqrt {15} e^{3} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) - x {\left (\log \left (5\right ) + 2 \, \log \left (2\right )\right )} + \frac {1}{15} \, {\left (\sqrt {15} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) + 15 \, \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 30 \, \log \left (x\right )\right )} e^{3} + x \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, x \log \left (x\right ) \]
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Time = 0.35 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=e^{3} \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, e^{3} \log \left (x\right ) + x \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \]
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Time = 14.40 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{-7 x+2 x^2+2 x^3} \, dx=x\,\ln \left (\frac {1}{x^2}\right )-x\,\ln \left (20\right )-2\,{\mathrm {e}}^3\,\ln \left (x\right )+x\,\ln \left (2\,x^2+2\,x-7\right )+\ln \left (x^2+x-\frac {7}{2}\right )\,{\mathrm {e}}^3 \]
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