Integrand size = 69, antiderivative size = 20 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=-4+e^x+e^{\frac {1}{4 (-1+\log (x))}}+2 x \]
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Time = 0.96 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 32, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6873, 12, 6874, 2240, 2334, 2336, 2209, 2407, 2225} \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2 x+e^x+e^{-\frac {1}{4 (1-\log (x))}} \]
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Rule 12
Rule 2209
Rule 2225
Rule 2240
Rule 2334
Rule 2336
Rule 2407
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x (1-\log (x))^2} \, dx \\ & = \frac {1}{4} \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{x (1-\log (x))^2} \, dx \\ & = \frac {1}{4} \int \left (-\frac {e^{\frac {1}{4 (-1+\log (x))}}}{x (1-\log (x))^2}+\frac {8}{(-1+\log (x))^2}+\frac {4 e^x}{(-1+\log (x))^2}-\frac {8 \left (2+e^x\right ) \log (x)}{(-1+\log (x))^2}+\frac {4 \left (2+e^x\right ) \log ^2(x)}{(-1+\log (x))^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{\frac {1}{4 (-1+\log (x))}}}{x (1-\log (x))^2} \, dx\right )+2 \int \frac {1}{(-1+\log (x))^2} \, dx-2 \int \frac {\left (2+e^x\right ) \log (x)}{(-1+\log (x))^2} \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \frac {\left (2+e^x\right ) \log ^2(x)}{(-1+\log (x))^2} \, dx \\ & = \frac {2 x}{1-\log (x)}-\frac {1}{4} \text {Subst}\left (\int \frac {e^{\frac {1}{4 (-1+x)}}}{(-1+x)^2} \, dx,x,\log (x)\right )+2 \int \frac {1}{-1+\log (x)} \, dx-2 \int \left (\frac {2 \log (x)}{(-1+\log (x))^2}+\frac {e^x \log (x)}{(-1+\log (x))^2}\right ) \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \left (\frac {2 \log ^2(x)}{(-1+\log (x))^2}+\frac {e^x \log ^2(x)}{(-1+\log (x))^2}\right ) \, dx \\ & = e^{-\frac {1}{4 (1-\log (x))}}+\frac {2 x}{1-\log (x)}-2 \int \frac {e^x \log (x)}{(-1+\log (x))^2} \, dx+2 \int \frac {\log ^2(x)}{(-1+\log (x))^2} \, dx+2 \text {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )-4 \int \frac {\log (x)}{(-1+\log (x))^2} \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \frac {e^x \log ^2(x)}{(-1+\log (x))^2} \, dx \\ & = e^{-\frac {1}{4 (1-\log (x))}}+2 e \operatorname {ExpIntegralEi}(-1+\log (x))+\frac {2 x}{1-\log (x)}+2 \int \left (1+\frac {1}{(-1+\log (x))^2}+\frac {2}{-1+\log (x)}\right ) \, dx-2 \int \left (\frac {e^x}{(-1+\log (x))^2}+\frac {e^x}{-1+\log (x)}\right ) \, dx-4 \int \left (\frac {1}{(-1+\log (x))^2}+\frac {1}{-1+\log (x)}\right ) \, dx+\int \left (e^x+\frac {e^x}{(-1+\log (x))^2}+\frac {2 e^x}{-1+\log (x)}\right ) \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx \\ & = e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \operatorname {ExpIntegralEi}(-1+\log (x))+\frac {2 x}{1-\log (x)}+2 \int \frac {1}{(-1+\log (x))^2} \, dx-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx-4 \int \frac {1}{(-1+\log (x))^2} \, dx+\int e^x \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx \\ & = e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \operatorname {ExpIntegralEi}(-1+\log (x))-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \int \frac {1}{-1+\log (x)} \, dx-4 \int \frac {1}{-1+\log (x)} \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx \\ & = e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \operatorname {ExpIntegralEi}(-1+\log (x))-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \text {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )-4 \text {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx \\ & = e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=\frac {1}{4} \left (4 e^x+4 e^{\frac {1}{4 (-1+\log (x))}}+8 x\right ) \]
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Time = 0.56 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80
method | result | size |
risch | \(2 x +{\mathrm e}^{x}+{\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}\) | \(16\) |
parallelrisch | \(2 x +{\mathrm e}^{x}+{\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}\) | \(16\) |
parts | \(2 x +\frac {\ln \left (x \right ) {\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}-{\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}}{\ln \left (x \right )-1}+{\mathrm e}^{x}\) | \(38\) |
default | \(2 x +{\mathrm e}^{x}+\frac {4 \ln \left (x \right ) {\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}-4 \,{\mathrm e}^{\frac {1}{4 \ln \left (x \right )-4}}}{4 \ln \left (x \right )-4}\) | \(40\) |
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Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \left (x\right ) - 1\right )}}\right )} \]
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Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2 x + e^{x} + e^{\frac {1}{4 \log {\left (x \right )} - 4}} \]
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Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \left (x\right ) - 1\right )}}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \left (x\right ) - 1\right )}}\right )} \]
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Time = 11.88 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx=2\,x+{\mathrm {e}}^{\frac {1}{4\,\ln \left (x\right )-4}}+{\mathrm {e}}^x \]
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