Integrand size = 104, antiderivative size = 30 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=e^3-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 (4+x))}{x} \]
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Time = 2.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 17, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6820, 6874, 14, 36, 29, 31, 2442, 2631, 30, 2637, 2634} \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 x+16)}{x} \]
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Rule 14
Rule 29
Rule 30
Rule 31
Rule 36
Rule 2442
Rule 2631
Rule 2634
Rule 2637
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (4+5 x+x^2\right ) \log (4 (4+x))-\left (-1+2 e^x x\right ) \log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+(4+x) \log (4 (4+x)))}{x^2 (4+x) \left (1-2 e^x x\right )} \, dx \\ & = \int \left (\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+4 \log (4 (4+x))+x \log (4 (4+x)))}{x^2 (4+x)}+\frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )}\right ) \, dx \\ & = \int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+4 \log (4 (4+x))+x \log (4 (4+x)))}{x^2 (4+x)} \, dx+\int \frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )} \, dx \\ & = -\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx+\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+(4+x) \log (4 (4+x)))}{x^2 (4+x)} \, dx-\int \frac {-\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx \\ & = -\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx+\int \left (-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x (4+x)}+\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x^2}\right ) \, dx-\int \left (-\frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x}-\frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x}\right ) \, dx \\ & = -\left (\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx\right )-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx-\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x (4+x)} \, dx+\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x^2} \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx \\ & = -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}-\log (16+4 x) \int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\log (16+4 x) \int \frac {1}{x \left (-1+2 e^x x\right )} \, dx-\int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{(-4-x) x} \, dx-\int \left (\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 x}-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 (4+x)}\right ) \, dx-\int \frac {(1+x) \log (16+4 x)}{x^2 \left (1-2 e^x x\right )} \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx \\ & = -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}-\frac {1}{4} \int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{x} \, dx+\frac {1}{4} \int \frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4+x} \, dx-\int \left (-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 x}+\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right )}{4 (4+x)}\right ) \, dx+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {-\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx-\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx \\ & = -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x}+\int \frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x} \, dx+\int \left (-\frac {\int \frac {1}{x^2 \left (-1+2 e^x x\right )} \, dx}{4+x}-\frac {\int \frac {1}{x \left (-1+2 e^x x\right )} \, dx}{4+x}\right ) \, dx \\ & = -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (16+4 x)}{x} \\ \end{align*}
Time = 0.84 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\log (4+x)-\frac {\left (x+\log \left (5-\frac {5 e^{-x}}{2 x}\right )\right ) \log (4 (4+x))}{x} \]
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Time = 7.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {\ln \left (4 x +16\right ) \ln \left (\frac {5 \left (2 \,{\mathrm e}^{x} x -1\right ) {\mathrm e}^{-x}}{2 x}\right )}{x}\) | \(29\) |
risch | \(\frac {\ln \left (4 x +16\right ) \ln \left ({\mathrm e}^{x}\right )}{x}+\frac {\ln \left (4 x +16\right ) \left (i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{3}-2 \ln \left (5\right )+2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )}{2 x}\) | \(290\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {\log \left (4 \, x + 16\right ) \log \left (\frac {5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}}{2 \, x}\right )}{x} \]
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Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=- \frac {\log {\left (\frac {\left (5 x e^{x} - \frac {5}{2}\right ) e^{- x}}{x} \right )} \log {\left (4 x + 16 \right )}}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {2 \, \log \left (5\right ) \log \left (2\right ) - 2 \, \log \left (2\right )^{2} + {\left (2 \, \log \left (2\right ) + \log \left (x + 4\right )\right )} \log \left (2 \, x e^{x} - 1\right ) - {\left (x - \log \left (5\right ) + \log \left (2\right ) + \log \left (x\right )\right )} \log \left (x + 4\right ) - 2 \, \log \left (2\right ) \log \left (x\right )}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (26) = 52\).
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\frac {2 \, \log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) + \log \left (2\right ) \log \left (x + 4\right ) - \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) \log \left (x + 4\right ) + 2 \, \log \left (2\right ) \log \left (x\right ) + \log \left (x + 4\right ) \log \left (x\right )}{x} \]
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Time = 11.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\ln \left (x+4\right )-\frac {\ln \left (4\,x+16\right )\,\ln \left (\frac {5\,\left (2\,x\,{\mathrm {e}}^x-1\right )}{2\,x}\right )}{x} \]
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