Integrand size = 51, antiderivative size = 29 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=-3+5 e^{e^x} x \left (-e^x+\frac {1}{4} \left (-2+e^{\log ^2(5)}\right )+x\right ) \]
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\[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx \\ & = \frac {1}{4} \int \left (-10 e^{e^x}+40 e^{e^x} x-20 e^{e^x+2 x} x+5 e^{e^x+\log ^2(5)} \left (1+e^x x\right )+10 e^{e^x+x} \left (-2-3 x+2 x^2\right )\right ) \, dx \\ & = \frac {5}{4} \int e^{e^x+\log ^2(5)} \left (1+e^x x\right ) \, dx-\frac {5}{2} \int e^{e^x} \, dx+\frac {5}{2} \int e^{e^x+x} \left (-2-3 x+2 x^2\right ) \, dx-5 \int e^{e^x+2 x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x+\frac {5}{2} \int \left (-2 e^{e^x+x}-3 e^{e^x+x} x+2 e^{e^x+x} x^2\right ) \, dx-\frac {5}{2} \text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )-5 \int e^{e^x+2 x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+x} \, dx-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-5 \text {Subst}\left (\int e^x \, dx,x,e^x\right )-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ & = -5 e^{e^x}+\frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ \end{align*}
Time = 1.83 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} e^{e^x} x \left (-2-4 e^x+e^{\log ^2(5)}+4 x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {\left (5 \,{\mathrm e}^{\ln \left (5\right )^{2}} x +20 x^{2}-20 \,{\mathrm e}^{x} x -10 x \right ) {\mathrm e}^{{\mathrm e}^{x}}}{4}\) | \(28\) |
norman | \(\left (\frac {5 \,{\mathrm e}^{\ln \left (5\right )^{2}}}{4}-\frac {5}{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{x}}+5 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}\) | \(32\) |
parallelrisch | \(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{\ln \left (5\right )^{2}}}{4}+5 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}-\frac {5 x \,{\mathrm e}^{{\mathrm e}^{x}}}{2}\) | \(35\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} \, {\left (4 \, x^{2} + x e^{\left (\log \left (5\right )^{2}\right )} - 4 \, x e^{x} - 2 \, x\right )} e^{\left (e^{x}\right )} \]
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Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {\left (20 x^{2} - 20 x e^{x} - 10 x + 5 x e^{\log {\left (5 \right )}^{2}}\right ) e^{e^{x}}}{4} \]
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\[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\int { \frac {5}{4} \, {\left ({\left (x e^{x} + 1\right )} e^{\left (\log \left (5\right )^{2}\right )} - 4 \, x e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{2} - 3 \, x - 2\right )} e^{x} + 8 \, x - 2\right )} e^{\left (e^{x}\right )} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} \, {\left (4 \, x^{2} e^{\left (x + e^{x}\right )} + x e^{\left (\log \left (5\right )^{2} + x + e^{x}\right )} - 4 \, x e^{\left (2 \, x + e^{x}\right )} - 2 \, x e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x+{\mathrm {e}}^{{\ln \left (5\right )}^2}-4\,{\mathrm {e}}^x-2\right )}{4} \]
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