Integrand size = 61, antiderivative size = 25 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=1+e^{\frac {-\log ^2(2)+\frac {x}{9 \log ^4(x)}}{x}}+x \]
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Time = 0.64 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6874, 6838} \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x+e^{\frac {1}{9 \log ^4(x)}-\frac {\log ^2(2)}{x}} \]
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Rule 12
Rule 6838
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)} \, dx \\ & = \frac {1}{9} \int \left (9-\frac {e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}} \left (4 x-9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)}\right ) \, dx \\ & = x-\frac {1}{9} \int \frac {e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}} \left (4 x-9 \log ^2(2) \log ^5(x)\right )}{x^2 \log ^5(x)} \, dx \\ & = e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}}+x \\ \end{align*}
Time = 0.47 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=e^{-\frac {\log ^2(2)}{x}+\frac {1}{9 \log ^4(x)}}+x \]
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Time = 9.83 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {9 \ln \left (2\right )^{2} \ln \left (x \right )^{4}-x}{9 x \ln \left (x \right )^{4}}}\) | \(27\) |
parallelrisch | \(x +{\mathrm e}^{-\frac {9 \ln \left (2\right )^{2} \ln \left (x \right )^{4}-x}{9 x \ln \left (x \right )^{4}}}\) | \(27\) |
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none
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\left (-\frac {9 \, \log \left (2\right )^{2} \log \left (x\right )^{4} - x}{9 \, x \log \left (x\right )^{4}}\right )} \]
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Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\frac {\frac {x}{9} - \log {\left (2 \right )}^{2} \log {\left (x \right )}^{4}}{x \log {\left (x \right )}^{4}}} \]
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Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x + e^{\left (-\frac {\log \left (2\right )^{2}}{x} + \frac {1}{9 \, \log \left (x\right )^{4}}\right )} \]
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\[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=\int { \frac {9 \, x^{2} \log \left (x\right )^{5} + {\left (9 \, \log \left (2\right )^{2} \log \left (x\right )^{5} - 4 \, x\right )} e^{\left (-\frac {9 \, \log \left (2\right )^{2} \log \left (x\right )^{4} - x}{9 \, x \log \left (x\right )^{4}}\right )}}{9 \, x^{2} \log \left (x\right )^{5}} \,d x } \]
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Time = 12.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {9 x^2 \log ^5(x)+e^{\frac {x-9 \log ^2(2) \log ^4(x)}{9 x \log ^4(x)}} \left (-4 x+9 \log ^2(2) \log ^5(x)\right )}{9 x^2 \log ^5(x)} \, dx=x+{\mathrm {e}}^{\frac {1}{9\,{\ln \left (x\right )}^4}}\,{\mathrm {e}}^{-\frac {{\ln \left (2\right )}^2}{x}} \]
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