Integrand size = 139, antiderivative size = 30 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=e^x-x-\log \left (-2 x+e^4 x \left (x-\frac {1}{5} (3+x)^2\right )\right ) \]
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Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6820, 6860, 2225, 1642, 642} \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=-\log \left (e^4 x^2+e^4 x+9 e^4+10\right )-x+e^x-\log (x) \]
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Rule 642
Rule 1642
Rule 2225
Rule 6820
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^x x-10 (1+x)+e^{4+x} x \left (9+x+x^2\right )-e^4 \left (9+11 x+4 x^2+x^3\right )}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )} \, dx \\ & = \int \left (e^x+\frac {-10-9 e^4-\left (10+11 e^4\right ) x-4 e^4 x^2-e^4 x^3}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {-10-9 e^4-\left (10+11 e^4\right ) x-4 e^4 x^2-e^4 x^3}{x \left (10+9 e^4+e^4 x+e^4 x^2\right )} \, dx \\ & = e^x+\int \left (-1-\frac {1}{x}-\frac {e^4 (1+2 x)}{10+9 e^4+e^4 x+e^4 x^2}\right ) \, dx \\ & = e^x-x-\log (x)-e^4 \int \frac {1+2 x}{10+9 e^4+e^4 x+e^4 x^2} \, dx \\ & = e^x-x-\log (x)-\log \left (10+9 e^4+e^4 x+e^4 x^2\right ) \\ \end{align*}
Time = 2.54 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=e^x-x-\log (x)-\log \left (10+e^4 \left (9+x+x^2\right )\right ) \]
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Time = 3.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
method | result | size |
default | \(-x -\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )+{\mathrm e}^{x}\) | \(30\) |
parts | \(-x -\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )+{\mathrm e}^{x}\) | \(30\) |
norman | \(\frac {{\mathrm e}^{x} x -x^{2}}{x}-\ln \left (x \right )-\ln \left (x^{2} {\mathrm e}^{4}+x \,{\mathrm e}^{4}+9 \,{\mathrm e}^{4}+10\right )\) | \(39\) |
parallelrisch | \(\frac {2 x^{6} {\mathrm e}^{x}+38 \,{\mathrm e}^{x} x^{4}+162 \,{\mathrm e}^{x} x^{2}+36 \,{\mathrm e}^{x} x^{3}+4 x^{5} {\mathrm e}^{x}-2 x^{7}+15 x^{6}+325 x^{4}+180 x^{3}+1539 x^{2}-2 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{6}-4 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{5}-38 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{4}-36 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{3}-162 \ln \left (x -\frac {{\mathrm e}^{\ln \left (-\frac {1}{5} x^{3}-\frac {1}{5} x^{2}-\frac {9}{5} x \right )+4}}{2}\right ) x^{2}}{2 x^{2} \left (x^{2}+x +9\right )^{2}}\) | \(215\) |
risch | \(-x -\ln \left (x^{3}+x^{2}+9 x \right )+{\mathrm e}^{x}-\ln \left (5\right )+4+\ln \left (x \right )+\ln \left (x^{2}+x +9\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )+\operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \left (\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )-1\right )-\ln \left (2 x -\frac {x \left (x^{2}+x +9\right ) {\mathrm e}^{4} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{3}}{2}} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \operatorname {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi {\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}+x +9\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x^{2}+x +9\right )\right )}{2}}}{5}\right )\) | \(240\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=-x + e^{x} - \log \left ({\left (x^{3} + x^{2} + 9 \, x\right )} e^{4} + 10 \, x\right ) \]
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Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=- x + e^{x} - \log {\left (x^{3} e^{4} + x^{2} e^{4} + x \left (10 + 9 e^{4}\right ) \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (29) = 58\).
Time = 0.37 (sec) , antiderivative size = 703, normalized size of antiderivative = 23.43 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=\text {Too large to display} \]
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Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx=-{\left (x e^{4} + e^{4} \log \left (x^{2} e^{4} + x e^{4} + 9 \, e^{4} + 10\right ) + e^{4} \log \left (x\right ) - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )} \]
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Time = 1.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {18 x+20 x^2+4 x^3+2 x^4+e^x \left (-18 x^2-2 x^3-2 x^4\right )+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (-9-11 x-4 x^2-x^3+e^x \left (9 x+x^2+x^3\right )\right )}{-18 x^2-2 x^3-2 x^4+\frac {1}{5} e^4 \left (-9 x-x^2-x^3\right ) \left (9 x+x^2+x^3\right )} \, dx={\mathrm {e}}^x-\ln \left (9\,x+10\,x\,{\mathrm {e}}^{-4}+x^2+x^3\right )-x \]
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