Integrand size = 237, antiderivative size = 27 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=\log \left (2+e^{1-e^{x+\frac {x}{e (5+\log (3+x))}}+x}\right ) \]
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Timed out. \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
Time = 1.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=-e^{x+\frac {x}{e (5+\log (3+x))}}+\log \left (2 e^{e^{x+\frac {x}{e (5+\log (3+x))}}}+e^{1+x}\right ) \]
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Time = 26.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41
method | result | size |
risch | \(-1+\ln \left ({\mathrm e}^{-{\mathrm e}^{\frac {x \left ({\mathrm e} \ln \left (3+x \right )+5 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}}{\ln \left (3+x \right )+5}}+x +1}+2\right )\) | \(38\) |
parallelrisch | \(\ln \left ({\mathrm e}^{-{\mathrm e}^{\frac {x \left ({\mathrm e} \ln \left (3+x \right )+5 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}}{\ln \left (3+x \right )+5}}+x +1}+2\right )\) | \(38\) |
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Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=\log \left (e^{\left (x - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )} + 1\right )} + 2\right ) \]
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Time = 15.55 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=\log {\left (e^{x - e^{\frac {e x \log {\left (x + 3 \right )} + x + 5 e x}{e \log {\left (x + 3 \right )} + 5 e}} + 1} + 2 \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (24) = 48\).
Time = 0.33 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.63 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=-e^{\left (\frac {x \log \left (x + 3\right )}{\log \left (x + 3\right ) + 5} + \frac {x}{e \log \left (x + 3\right ) + 5 \, e} + \frac {5 \, x}{\log \left (x + 3\right ) + 5}\right )} + \log \left (\frac {1}{2} \, e^{\left (x + 1\right )} + e^{\left (e^{\left (\frac {x \log \left (x + 3\right )}{\log \left (x + 3\right ) + 5} + \frac {x}{e \log \left (x + 3\right ) + 5 \, e} + \frac {5 \, x}{\log \left (x + 3\right ) + 5}\right )}\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (24) = 48\).
Time = 11.64 (sec) , antiderivative size = 329, normalized size of antiderivative = 12.19 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=-\frac {x e \log \left (x + 3\right ) - e \log \left (x + 3\right ) \log \left (e^{\left (\frac {2 \, x e \log \left (x + 3\right ) + 10 \, x e - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} \log \left (x + 3\right ) + x - 5 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )}}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} + 2 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )}\right ) + 5 \, x e - 5 \, e \log \left (e^{\left (\frac {2 \, x e \log \left (x + 3\right ) + 10 \, x e - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} \log \left (x + 3\right ) + x - 5 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )}}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} + 2 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )}\right ) + x}{e \log \left (x + 3\right ) + 5 \, e} \]
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Time = 1.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} \left (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)\right )\right )}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} \left (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)\right )} \, dx=\ln \left (\mathrm {e}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {5\,x}{\ln \left (x+3\right )+5}}\,{\mathrm {e}}^{\frac {x}{5\,\mathrm {e}+\ln \left (x+3\right )\,\mathrm {e}}}\,{\left (x+3\right )}^{\frac {x}{\ln \left (x+3\right )+5}}}+2\right ) \]
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