Integrand size = 74, antiderivative size = 27 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log \left (\frac {5 e^{-\frac {-4+x^2}{x}} (-3+x)}{6-3 x}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 26, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {1608, 6860, 630, 31, 719, 29, 646, 717, 715, 2628, 1642} \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right ) \]
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Rule 29
Rule 31
Rule 630
Rule 646
Rule 715
Rule 717
Rule 719
Rule 1608
Rule 1642
Rule 2628
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{x \left (6-5 x+x^2\right )} \, dx \\ & = \int \left (\frac {20}{6-5 x+x^2}-\frac {24}{x \left (6-5 x+x^2\right )}-\frac {9 x}{6-5 x+x^2}+\frac {5 x^2}{6-5 x+x^2}-\frac {x^3}{6-5 x+x^2}+\log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right )\right ) \, dx \\ & = 5 \int \frac {x^2}{6-5 x+x^2} \, dx-9 \int \frac {x}{6-5 x+x^2} \, dx+20 \int \frac {1}{6-5 x+x^2} \, dx-24 \int \frac {1}{x \left (6-5 x+x^2\right )} \, dx-\int \frac {x^3}{6-5 x+x^2} \, dx+\int \log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right ) \, dx \\ & = 5 x+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )-4 \int \frac {1}{x} \, dx-4 \int \frac {5-x}{6-5 x+x^2} \, dx+5 \int \frac {-6+5 x}{6-5 x+x^2} \, dx+18 \int \frac {1}{-2+x} \, dx+20 \int \frac {1}{-3+x} \, dx-20 \int \frac {1}{-2+x} \, dx-27 \int \frac {1}{-3+x} \, dx-\int \frac {-24+20 x-9 x^2+5 x^3-x^4}{x \left (6-5 x+x^2\right )} \, dx-\int \left (5+x-\frac {30-19 x}{6-5 x+x^2}\right ) \, dx \\ & = -\frac {x^2}{2}-2 \log (2-x)-7 \log (3-x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )-4 \log (x)-8 \int \frac {1}{-3+x} \, dx+12 \int \frac {1}{-2+x} \, dx-20 \int \frac {1}{-2+x} \, dx+45 \int \frac {1}{-3+x} \, dx-\int \left (\frac {3}{-3+x}-\frac {2}{-2+x}-\frac {4}{x}-x\right ) \, dx+\int \frac {30-19 x}{6-5 x+x^2} \, dx \\ & = -8 \log (2-x)+27 \log (3-x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )+8 \int \frac {1}{-2+x} \, dx-27 \int \frac {1}{-3+x} \, dx \\ & = x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(27)=54\).
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=-4-2 \log (2-x)+3 \log (3-x)-3 \log (-3+x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right )+2 \log (-2+x) \]
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Time = 0.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(x \ln \left (-\frac {5 \,{\mathrm e}^{-\frac {x^{2}-4}{x}} \left (-3+x \right )}{-6+3 x}\right )\) | \(28\) |
default | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
norman | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
parts | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
risch | \(-x \ln \left ({\mathrm e}^{\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right )-x \ln \left (-2+x \right )+\ln \left (-3+x \right ) x -i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}-\frac {i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i}{-2+x}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )}{2}-\frac {i \pi x \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{3}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{-2+x}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}}{2}+\frac {i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{3}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}}{2}+i \pi x +x \ln \left (5\right )-x \ln \left (3\right )\) | \(359\) |
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log \left (-\frac {5 \, {\left (x - 3\right )} e^{\left (-\frac {x^{2} - 4}{x}\right )}}{3 \, {\left (x - 2\right )}}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log {\left (\frac {\left (15 - 5 x\right ) e^{- \frac {x^{2} - 4}{x}}}{3 x - 6} \right )} \]
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Result contains complex when optimal does not.
Time = 0.37 (sec) , antiderivative size = 200, normalized size of antiderivative = 7.41 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx={\left (i \, \pi + \log \left (5\right ) - \log \left (3\right )\right )} x - x^{2} + {\left (6 i \, \pi - x + 6 \, \log \left (5\right ) - 6 \, \log \left (3\right ) + 6 \, \log \left (x - 3\right ) + 2\right )} \log \left (x - 2\right ) - 3 \, \log \left (x - 2\right )^{2} + {\left (-6 i \, \pi + x - 6 \, \log \left (5\right ) + 6 \, \log \left (3\right ) + 7\right )} \log \left (x - 3\right ) - 3 \, \log \left (x - 3\right )^{2} - 6 \, {\left (\log \left (x - 2\right ) - \log \left (x - 3\right )\right )} \log \left (-\frac {5 \, x e^{\left (-x + \frac {4}{x}\right )}}{3 \, {\left (x - 2\right )}} + \frac {5 \, e^{\left (-x + \frac {4}{x}\right )}}{x - 2}\right ) - \frac {3 \, x \log \left (x - 2\right )^{2} + 3 \, x \log \left (x - 3\right )^{2} + 6 \, {\left (x^{2} - x \log \left (x - 3\right ) - 4\right )} \log \left (x - 2\right ) - 2 \, {\left (3 \, x^{2} - 5 \, x - 12\right )} \log \left (x - 3\right )}{x} - 2 \, \log \left (x - 2\right ) + 3 \, \log \left (x - 3\right ) \]
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Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=-x^{2} + x \log \left (-\frac {5 \, {\left (x - 3\right )}}{3 \, {\left (x - 2\right )}}\right ) \]
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Time = 12.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x\,\ln \left (-\frac {5\,x-15}{3\,x-6}\right )-x^2 \]
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