Integrand size = 47, antiderivative size = 29 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=-x+2 x^2+\frac {1}{4} \left (-e^x-3 \log ^2(x)\right )+\log (2+x) \]
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Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {1607, 6820, 2225, 712, 2338} \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=2 x^2-x-\frac {e^x}{4}-\frac {3 \log ^2(x)}{4}+\log (x+2) \]
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Rule 712
Rule 1607
Rule 2225
Rule 2338
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{x (8+4 x)} \, dx \\ & = \int \left (-\frac {e^x}{4}+\frac {-1+7 x+4 x^2}{2+x}-\frac {3 \log (x)}{2 x}\right ) \, dx \\ & = -\frac {\int e^x \, dx}{4}-\frac {3}{2} \int \frac {\log (x)}{x} \, dx+\int \frac {-1+7 x+4 x^2}{2+x} \, dx \\ & = -\frac {e^x}{4}-\frac {3 \log ^2(x)}{4}+\int \left (-1+4 x+\frac {1}{2+x}\right ) \, dx \\ & = -\frac {e^x}{4}-x+2 x^2-\frac {3 \log ^2(x)}{4}+\log (2+x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=-\frac {e^x}{4}-9 (2+x)+2 (2+x)^2-\frac {3 \log ^2(x)}{4}+\log (2+x) \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
default | \(\ln \left (2+x \right )-x -\frac {{\mathrm e}^{x}}{4}-\frac {3 \ln \left (x \right )^{2}}{4}+2 x^{2}\) | \(24\) |
norman | \(\ln \left (2+x \right )-x -\frac {{\mathrm e}^{x}}{4}-\frac {3 \ln \left (x \right )^{2}}{4}+2 x^{2}\) | \(24\) |
risch | \(\ln \left (2+x \right )-x -\frac {{\mathrm e}^{x}}{4}-\frac {3 \ln \left (x \right )^{2}}{4}+2 x^{2}\) | \(24\) |
parallelrisch | \(\ln \left (2+x \right )-x -\frac {{\mathrm e}^{x}}{4}-\frac {3 \ln \left (x \right )^{2}}{4}+2 x^{2}\) | \(24\) |
parts | \(\ln \left (2+x \right )-x -\frac {{\mathrm e}^{x}}{4}-\frac {3 \ln \left (x \right )^{2}}{4}+2 x^{2}\) | \(24\) |
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Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=2 \, x^{2} - \frac {3}{4} \, \log \left (x\right )^{2} - x - \frac {1}{4} \, e^{x} + \log \left (x + 2\right ) \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=2 x^{2} - x - \frac {e^{x}}{4} - \frac {3 \log {\left (x \right )}^{2}}{4} + \log {\left (x + 2 \right )} \]
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\[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=\int { \frac {16 \, x^{3} + 28 \, x^{2} - {\left (x^{2} + 2 \, x\right )} e^{x} - 6 \, {\left (x + 2\right )} \log \left (x\right ) - 4 \, x}{4 \, {\left (x^{2} + 2 \, x\right )}} \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=2 \, x^{2} - \frac {3}{4} \, \log \left (x\right )^{2} - x - \frac {1}{4} \, e^{x} + \log \left (x + 2\right ) \]
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Time = 13.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x+28 x^2+16 x^3+e^x \left (-2 x-x^2\right )+(-12-6 x) \log (x)}{8 x+4 x^2} \, dx=\ln \left (x+2\right )-x-\frac {{\mathrm {e}}^x}{4}-\frac {3\,{\ln \left (x\right )}^2}{4}+2\,x^2 \]
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