Integrand size = 75, antiderivative size = 27 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x} \]
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Time = 0.93 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.173, Rules used = {1607, 6820, 12, 14, 2209, 6874, 45, 2463, 2437, 2339, 29, 30, 2635} \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{2 x} \]
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Rule 12
Rule 14
Rule 29
Rule 30
Rule 45
Rule 1607
Rule 2209
Rule 2339
Rule 2437
Rule 2463
Rule 2635
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2 (-6+2 x) \log (-3+x)} \, dx \\ & = \int \frac {1+\left (2+e^x\right ) x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x^2} \, dx \\ & = \frac {1}{2} \int \frac {1+\left (2+e^x\right ) x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^x}{x}+\frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)-3 \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )+x \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{(-3+x) x^2 \log (-3+x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x}{x} \, dx+\frac {1}{2} \int \frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)-3 \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )+x \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{(-3+x) x^2 \log (-3+x)} \, dx \\ & = \frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \frac {1+2 x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx \\ & = \frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \left (\frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)}{(-3+x) x^2 \log (-3+x)}+\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2}\right ) \, dx \\ & = \frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)}{(-3+x) x^2 \log (-3+x)} \, dx+\frac {1}{2} \int \frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx \\ & = \frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \left (\frac {1+2 x}{x^2}-\frac {1}{(-3+x) x \log (-3+x)}\right ) \, dx-\frac {1}{2} \int \frac {1+e^x x-\frac {x}{(-3+x) \log (-3+x)}}{x^2} \, dx \\ & = \frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \frac {1+2 x}{x^2} \, dx-\frac {1}{2} \int \frac {1}{(-3+x) x \log (-3+x)} \, dx-\frac {1}{2} \int \left (\frac {e^x}{x}+\frac {-x-3 \log (-3+x)+x \log (-3+x)}{(-3+x) x^2 \log (-3+x)}\right ) \, dx \\ & = \frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \left (\frac {1}{x^2}+\frac {2}{x}\right ) \, dx-\frac {1}{2} \int \frac {e^x}{x} \, dx-\frac {1}{2} \int \left (\frac {1}{3 (-3+x) \log (-3+x)}-\frac {1}{3 x \log (-3+x)}\right ) \, dx-\frac {1}{2} \int \frac {-x-3 \log (-3+x)+x \log (-3+x)}{(-3+x) x^2 \log (-3+x)} \, dx \\ & = -\frac {1}{2 x}+\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}-\frac {1}{6} \int \frac {1}{(-3+x) \log (-3+x)} \, dx+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{2} \int \frac {1-\frac {x}{(-3+x) \log (-3+x)}}{x^2} \, dx \\ & = -\frac {1}{2 x}+\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{6} \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3+x\right )-\frac {1}{2} \int \left (\frac {1}{x^2}-\frac {1}{(-3+x) x \log (-3+x)}\right ) \, dx \\ & = \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{6} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3+x)\right )+\frac {1}{2} \int \frac {1}{(-3+x) x \log (-3+x)} \, dx \\ & = \log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx+\frac {1}{2} \int \left (\frac {1}{3 (-3+x) \log (-3+x)}-\frac {1}{3 x \log (-3+x)}\right ) \, dx \\ & = \log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{(-3+x) \log (-3+x)} \, dx \\ & = \log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3+x\right ) \\ & = \log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3+x)\right ) \\ & = \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {1}{2} \left (2 \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x}\right ) \]
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Time = 7.93 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {-12 x \ln \left (x \right )+6 \ln \left (\frac {4 \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{x}\right )}{12 x}\) | \(29\) |
risch | \(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )}{2 x}+\frac {-i \pi \,\operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{2}-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \ln \left (-3+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}+i \pi \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{3}+4 x \ln \left (x \right )-4 \ln \left (2\right )+2 \ln \left (x \right )-2 \ln \left (\ln \left (-3+x \right )\right )}{4 x}\) | \(271\) |
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {2 \, x \log \left (x\right ) - \log \left (\frac {4 \, e^{\left (-e^{x}\right )} \log \left (x - 3\right )}{x}\right )}{2 \, x} \]
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Time = 3.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\log {\left (x \right )} - \frac {\log {\left (\frac {4 e^{- e^{x}} \log {\left (x - 3 \right )}}{x} \right )}}{2 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {x \log \left (x - 3\right ) + {\left (5 \, x + 3\right )} \log \left (x\right ) + 3 \, e^{x} - 6 \, \log \left (2\right ) - 3 \, \log \left (\log \left (x - 3\right )\right ) + 3}{6 \, x} - \frac {1}{2 \, x} - \frac {1}{6} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x\right ) \]
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Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {2 \, x \log \left (x\right ) + e^{x} + \log \left (x\right ) - \log \left (4 \, \log \left (x - 3\right )\right )}{2 \, x} \]
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Time = 12.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\ln \left (x\right )+\frac {{\mathrm {e}}^x}{2\,x}-\frac {\ln \left (\frac {\ln \left (x-3\right )}{x}\right )}{2\,x}-\frac {\ln \left (2\right )}{x} \]
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