Integrand size = 89, antiderivative size = 33 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {-x^2+\log \left (\frac {e^x}{3}\right )}{(-2+x) x}-\frac {-1+x}{\log (x)} \]
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Time = 0.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24, number of steps used = 19, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.169, Rules used = {1608, 27, 6820, 6874, 78, 75, 2634, 36, 31, 29, 2395, 2334, 2335, 2339, 30} \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {2}{2-x}-\frac {x}{\log (x)}+\frac {1}{\log (x)}-\frac {\log \left (\frac {e^x}{3}\right )}{(2-x) x} \]
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Rule 27
Rule 29
Rule 30
Rule 31
Rule 36
Rule 75
Rule 78
Rule 1608
Rule 2334
Rule 2335
Rule 2339
Rule 2395
Rule 2634
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{x^2 \left (4-4 x+x^2\right ) \log ^2(x)} \, dx \\ & = \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{(-2+x)^2 x^2 \log ^2(x)} \, dx \\ & = \int \left (\frac {x (-2+3 x)-2 (-1+x) \log \left (\frac {e^x}{3}\right )}{(-2+x)^2 x^2}+\frac {-1+x}{x \log ^2(x)}-\frac {1}{\log (x)}\right ) \, dx \\ & = \int \frac {x (-2+3 x)-2 (-1+x) \log \left (\frac {e^x}{3}\right )}{(-2+x)^2 x^2} \, dx+\int \frac {-1+x}{x \log ^2(x)} \, dx-\int \frac {1}{\log (x)} \, dx \\ & = -\operatorname {LogIntegral}(x)+\int \left (\frac {-2+3 x}{(-2+x)^2 x}-\frac {2 (-1+x) \log \left (\frac {e^x}{3}\right )}{(-2+x)^2 x^2}\right ) \, dx+\int \left (\frac {1}{\log ^2(x)}-\frac {1}{x \log ^2(x)}\right ) \, dx \\ & = -\operatorname {LogIntegral}(x)-2 \int \frac {(-1+x) \log \left (\frac {e^x}{3}\right )}{(-2+x)^2 x^2} \, dx+\int \frac {-2+3 x}{(-2+x)^2 x} \, dx+\int \frac {1}{\log ^2(x)} \, dx-\int \frac {1}{x \log ^2(x)} \, dx \\ & = -\frac {\log \left (\frac {e^x}{3}\right )}{(2-x) x}-\frac {x}{\log (x)}-\operatorname {LogIntegral}(x)+2 \int \frac {1}{(4-2 x) x} \, dx+\int \left (\frac {2}{(-2+x)^2}+\frac {1}{2 (-2+x)}-\frac {1}{2 x}\right ) \, dx+\int \frac {1}{\log (x)} \, dx-\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = \frac {2}{2-x}-\frac {\log \left (\frac {e^x}{3}\right )}{(2-x) x}+\frac {1}{2} \log (2-x)+\frac {1}{\log (x)}-\frac {x}{\log (x)}-\frac {\log (x)}{2}+\frac {1}{2} \int \frac {1}{x} \, dx+\int \frac {1}{4-2 x} \, dx \\ & = \frac {2}{2-x}-\frac {\log \left (\frac {e^x}{3}\right )}{(2-x) x}+\frac {1}{\log (x)}-\frac {x}{\log (x)} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {4+\log (3)}{2 (2-x)}-\frac {\log \left (\frac {e^x}{3}\right )}{2 x}-\frac {\log \left (e^x\right )}{2 (2-x)}+\frac {1-x}{\log (x)} \]
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Time = 0.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27
method | result | size |
default | \(-\frac {2}{-2+x}+\frac {1}{\ln \left (x \right )}-\frac {x}{\ln \left (x \right )}-\frac {\ln \left (\frac {{\mathrm e}^{x}}{3}\right )}{2 x}+\frac {\ln \left (\frac {{\mathrm e}^{x}}{3}\right )}{2 x -4}\) | \(42\) |
parts | \(-\frac {2}{-2+x}+\frac {1}{\ln \left (x \right )}-\frac {x}{\ln \left (x \right )}-\frac {\ln \left (\frac {{\mathrm e}^{x}}{3}\right )}{2 x}+\frac {\ln \left (\frac {{\mathrm e}^{x}}{3}\right )}{2 x -4}\) | \(42\) |
parallelrisch | \(\frac {-4 x -4 x \ln \left (x \right )-2 x^{3}+6 x^{2}+2 \ln \left (\frac {{\mathrm e}^{x}}{3}\right ) \ln \left (x \right )}{2 x \ln \left (x \right ) \left (-2+x \right )}\) | \(43\) |
risch | \(\frac {\ln \left ({\mathrm e}^{x}\right )}{\left (-2+x \right ) x}-\frac {2 x^{3}+2 \ln \left (3\right ) \ln \left (x \right )-6 x^{2}+4 x \ln \left (x \right )+4 x}{2 \ln \left (x \right ) \left (-2+x \right ) x}\) | \(53\) |
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Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{3} - 3 \, x^{2} + {\left (x + \log \left (3\right )\right )} \log \left (x\right ) + 2 \, x}{{\left (x^{2} - 2 \, x\right )} \log \left (x\right )} \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.58 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {1 - x}{\log {\left (x \right )}} + \frac {- x - \log {\left (3 \right )}}{x^{2} - 2 x} \]
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Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{3} - 3 \, x^{2} + {\left (x + \log \left (3\right )\right )} \log \left (x\right ) + 2 \, x}{{\left (x^{2} - 2 \, x\right )} \log \left (x\right )} \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {\log \left (3\right ) + 2}{2 \, {\left (x - 2\right )}} + \frac {\log \left (3\right )}{2 \, x} - \frac {x - 1}{\log \left (x\right )} \]
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Time = 8.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-4 x+8 x^2-5 x^3+x^4+\left (-4 x^2+4 x^3-x^4\right ) \log (x)+\left (-2 x+3 x^2\right ) \log ^2(x)+(2-2 x) \log \left (\frac {e^x}{3}\right ) \log ^2(x)}{\left (4 x^2-4 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {x+\ln \left (3\right )}{2\,x-x^2}-x+\frac {x\,\ln \left (x\right )-x+1}{\ln \left (x\right )} \]
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