Integrand size = 192, antiderivative size = 23 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x+\log \left (5+x+\log \left (-e^4+x+\frac {x}{4+5 x}\right )\right ) \]
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Time = 0.71 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6820, 6860, 6816} \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x+\log \left (x+\log \left (\frac {5 x (x+1)}{5 x+4}-e^4\right )+5\right ) \]
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Rule 6816
Rule 6820
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4 (6+x) (4+5 x)^2-5 \left (4+32 x+63 x^2+39 x^3+5 x^4\right )+(4+5 x) \left (-5 x (1+x)+e^4 (4+5 x)\right ) \log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )} \, dx \\ & = \int \left (1+\frac {-4 \left (5-4 e^4\right )-20 \left (3-2 e^4\right ) x-5 \left (14-5 e^4\right ) x^2-25 x^3}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )}\right ) \, dx \\ & = x+\int \frac {-4 \left (5-4 e^4\right )-20 \left (3-2 e^4\right ) x-5 \left (14-5 e^4\right ) x^2-25 x^3}{(4+5 x) \left (4 e^4-5 \left (1-e^4\right ) x-5 x^2\right ) \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right )} \, dx \\ & = x+\log \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right ) \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x+\log \left (5+x+\log \left (-e^4+\frac {5 x (1+x)}{4+5 x}\right )\right ) \]
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Time = 1.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43
method | result | size |
norman | \(x +\ln \left (x +\ln \left (\frac {\left (-5 x -4\right ) {\mathrm e}^{4}+5 x^{2}+5 x}{4+5 x}\right )+5\right )\) | \(33\) |
risch | \(x +\ln \left (x +\ln \left (\frac {\left (-5 x -4\right ) {\mathrm e}^{4}+5 x^{2}+5 x}{4+5 x}\right )+5\right )\) | \(33\) |
parallelrisch | \(-\frac {\left (-256 \ln \left (x +\ln \left (\frac {\left (-5 x -4\right ) {\mathrm e}^{4}+5 x^{2}+5 x}{4+5 x}\right )+5\right ) {\mathrm e}^{8}-256 x \,{\mathrm e}^{8}\right ) {\mathrm e}^{-8}}{256}\) | \(51\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x + \log \left (x + \log \left (\frac {5 \, x^{2} - {\left (5 \, x + 4\right )} e^{4} + 5 \, x}{5 \, x + 4}\right ) + 5\right ) \]
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Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x + \log {\left (x + \log {\left (\frac {5 x^{2} + 5 x + \left (- 5 x - 4\right ) e^{4}}{5 x + 4} \right )} + 5 \right )} \]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x + \log \left (x + \log \left (5 \, x^{2} - 5 \, x {\left (e^{4} - 1\right )} - 4 \, e^{4}\right ) - \log \left (5 \, x + 4\right ) + 5\right ) \]
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Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=x + \log \left (x + \log \left (\frac {5 \, x^{2} - 5 \, x e^{4} + 5 \, x - 4 \, e^{4}}{5 \, x + 4}\right ) + 5\right ) \]
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Timed out. \[ \int \frac {-20-160 x-315 x^2-195 x^3-25 x^4+e^4 \left (96+256 x+190 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )}{-100 x-245 x^2-170 x^3-25 x^4+e^4 \left (80+216 x+165 x^2+25 x^3\right )+\left (-20 x-45 x^2-25 x^3+e^4 \left (16+40 x+25 x^2\right )\right ) \log \left (\frac {e^4 (-4-5 x)+5 x+5 x^2}{4+5 x}\right )} \, dx=\text {Hanged} \]
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