Integrand size = 176, antiderivative size = 23 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} x \log \left (4+x+\frac {5}{2 x^2 \left (e^x+x\right )}\right ) \]
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Time = 1.95 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87, number of steps used = 19, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6874, 6820, 45, 2628} \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} x \log \left (\frac {2 x^4+8 x^3+2 e^x (x+4) x^2+5}{2 x^2 \left (x+e^x\right )}\right ) \]
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Rule 45
Rule 2628
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-1+x) x}{5 \left (e^x+x\right )}-\frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{5 (4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}+\frac {x+4 \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )+x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )}{5 (4+x)}\right ) \, dx \\ & = \frac {1}{5} \int \frac {(-1+x) x}{e^x+x} \, dx-\frac {1}{5} \int \frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx+\frac {1}{5} \int \frac {x+4 \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )+x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )}{4+x} \, dx \\ & = \frac {1}{5} \int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx-\frac {1}{5} \int \left (\frac {15}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {5 x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {8 x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {6 x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {2 x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {20}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx+\frac {1}{5} \int \left (\frac {x}{4+x}+\log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )\right ) \, dx \\ & = \frac {1}{5} \int \frac {x}{4+x} \, dx-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx \\ & = \frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \left (1-\frac {4}{4+x}\right ) \, dx-\frac {1}{5} \int \frac {2 e^{2 x} x^3+x \left (-15+2 x^4\right )+e^x \left (-10-5 x+4 x^4\right )}{\left (e^x+x\right ) \left (5+8 x^3+2 x^4+2 e^x x^2 (4+x)\right )} \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx \\ & = \frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx-\frac {1}{5} \int \left (\frac {x}{4+x}+\frac {(-1+x) x}{e^x+x}-\frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx \\ & = \frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{4+x} \, dx-\frac {1}{5} \int \frac {x}{e^x+x} \, dx-\frac {1}{5} \int \frac {(-1+x) x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx+\frac {1}{5} \int \frac {40+35 x+5 x^2-32 x^3+16 x^4+14 x^5+2 x^6}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx \\ & = \frac {x}{5}-\frac {4}{5} \log (4+x)+\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right )-\frac {1}{5} \int \frac {x}{e^x+x} \, dx+\frac {1}{5} \int \frac {x^2}{e^x+x} \, dx-\frac {1}{5} \int \left (1-\frac {4}{4+x}\right ) \, dx-\frac {1}{5} \int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx+\frac {1}{5} \int \left (\frac {15}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {5 x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {8 x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {6 x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}+\frac {2 x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4}-\frac {20}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )}\right ) \, dx-\frac {2}{5} \int \frac {x^5}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-\frac {6}{5} \int \frac {x^4}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+\frac {8}{5} \int \frac {x^3}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx-3 \int \frac {1}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx+4 \int \frac {1}{(4+x) \left (5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4\right )} \, dx-\int \frac {x}{5+8 e^x x^2+8 x^3+2 e^x x^3+2 x^4} \, dx \\ & = \frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \\ \end{align*}
Time = 0.73 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} x \log \left (\frac {5+8 x^3+2 x^4+2 e^x x^2 (4+x)}{2 x^2 \left (e^x+x\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(18)=36\).
Time = 5.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 5.22
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {\left (2 x^{3}+8 x^{2}\right ) {\mathrm e}^{x}+2 x^{4}+8 x^{3}+5}{2 x^{2} \left ({\mathrm e}^{x}+x \right )}\right ) x}{5}-\frac {4 \ln \left (x \right )}{5}-\frac {2 \ln \left ({\mathrm e}^{x}+x \right )}{5}+\frac {2 \ln \left (\frac {5}{2}+x^{4}+{\mathrm e}^{x} x^{3}+4 x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{5}-\frac {2 \ln \left (\frac {\left (2 x^{3}+8 x^{2}\right ) {\mathrm e}^{x}+2 x^{4}+8 x^{3}+5}{2 x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}{5}\) | \(120\) |
risch | \(\frac {x \ln \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{5}-\frac {x \ln \left ({\mathrm e}^{x}+x \right )}{5}-\frac {2 x \ln \left (x \right )}{5}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \operatorname {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}{10}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{2}}{10}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{2}}{10}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right )}^{3}}{10}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{2}}{10}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{{\mathrm e}^{x}+x}\right ) \operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{10}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{3}}{10}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {5}{2}+x^{4}+\left ({\mathrm e}^{x}+4\right ) x^{3}+4 \,{\mathrm e}^{x} x^{2}\right )}{x^{2} \left ({\mathrm e}^{x}+x \right )}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{10}+\frac {i x \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{10}-\frac {i x \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}}{5}+\frac {i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{10}\) | \(537\) |
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (18) = 36\).
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.40 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {x \log {\left (\frac {2 x^{4} + 8 x^{3} + \left (2 x^{3} + 8 x^{2}\right ) e^{x} + 5}{2 x^{3} + 2 x^{2} e^{x}} \right )}}{5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (18) = 36\).
Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=-\frac {1}{5} \, x \log \left (2\right ) + \frac {1}{5} \, x \log \left (2 \, x^{4} + 8 \, x^{3} + 2 \, {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 5\right ) - \frac {1}{5} \, x \log \left (x + e^{x}\right ) - \frac {2}{5} \, x \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (18) = 36\).
Time = 0.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {1}{5} \, x \log \left (\frac {2 \, x^{4} + 2 \, x^{3} e^{x} + 8 \, x^{3} + 8 \, x^{2} e^{x} + 5}{2 \, {\left (x^{3} + x^{2} e^{x}\right )}}\right ) \]
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Time = 12.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {-15 x+2 e^{2 x} x^3+2 x^5+e^x \left (-10-5 x+4 x^4\right )+\left (5 x+8 x^4+2 x^5+e^{2 x} \left (8 x^2+2 x^3\right )+e^x \left (5+16 x^3+4 x^4\right )\right ) \log \left (\frac {5+8 x^3+2 x^4+e^x \left (8 x^2+2 x^3\right )}{2 e^x x^2+2 x^3}\right )}{25 x+40 x^4+10 x^5+e^{2 x} \left (40 x^2+10 x^3\right )+e^x \left (25+80 x^3+20 x^4\right )} \, dx=\frac {x\,\ln \left (\frac {{\mathrm {e}}^x\,\left (2\,x^3+8\,x^2\right )+8\,x^3+2\,x^4+5}{2\,x^2\,{\mathrm {e}}^x+2\,x^3}\right )}{5} \]
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