Integrand size = 97, antiderivative size = 21 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {1}{5} x \left (5+2 e^{-3+x}+x\right )^2} \]
[Out]
\[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=\int \frac {1}{5} \exp \left (-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \exp \left (-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int \left (8 \exp \left (-6+2 x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) (1+2 x)+8 \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (5+7 x+x^2\right )+2 \exp \left (\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (25+20 x+3 x^2\right )\right ) \, dx \\ & = \frac {2}{5} \int \exp \left (\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (25+20 x+3 x^2\right ) \, dx+\frac {8}{5} \int \exp \left (-6+2 x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) (1+2 x) \, dx+\frac {8}{5} \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \left (5+7 x+x^2\right ) \, dx \\ & = \frac {2}{5} \int e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} \left (25+20 x+3 x^2\right ) \, dx+\frac {8}{5} \int \exp \left (\frac {1}{5} \left (-30+35 x+4 e^{-6+2 x} x+10 x^2+x^3+4 e^{-3+x} x (5+x)\right )\right ) (1+2 x) \, dx+\frac {8}{5} \int \left (5 \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right )+7 \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x+\exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x^2\right ) \, dx \\ & = \frac {2}{5} \int \left (25 e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}}+20 e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} x+3 e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} x^2\right ) \, dx+\frac {8}{5} \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x^2 \, dx+\frac {8}{5} \int \left (\exp \left (\frac {1}{5} \left (-30+35 x+4 e^{-6+2 x} x+10 x^2+x^3+4 e^{-3+x} x (5+x)\right )\right )+2 \exp \left (\frac {1}{5} \left (-30+35 x+4 e^{-6+2 x} x+10 x^2+x^3+4 e^{-3+x} x (5+x)\right )\right ) x\right ) \, dx+8 \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \, dx+\frac {56}{5} \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x \, dx \\ & = \frac {6}{5} \int e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} x^2 \, dx+\frac {8}{5} \int \exp \left (\frac {1}{5} \left (-30+35 x+4 e^{-6+2 x} x+10 x^2+x^3+4 e^{-3+x} x (5+x)\right )\right ) \, dx+\frac {8}{5} \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x^2 \, dx+\frac {16}{5} \int \exp \left (\frac {1}{5} \left (-30+35 x+4 e^{-6+2 x} x+10 x^2+x^3+4 e^{-3+x} x (5+x)\right )\right ) x \, dx+8 \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) \, dx+8 \int e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} x \, dx+10 \int e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} \, dx+\frac {56}{5} \int \exp \left (-3+x+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}\right ) x \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(43\) vs. \(2(19)=38\).
Time = 0.41 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.10
method | result | size |
risch | \(2 \,{\mathrm e}^{\frac {x \left (x^{2} {\mathrm e}^{6}+4 \,{\mathrm e}^{3+x} x +10 x \,{\mathrm e}^{6}+4 \,{\mathrm e}^{2 x}+20 \,{\mathrm e}^{3+x}+25 \,{\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(44\) |
norman | \(2 \,{\mathrm e}^{\frac {\left (4 x \,{\mathrm e}^{2 x}+\left (4 x^{2}+20 x \right ) {\mathrm e}^{3} {\mathrm e}^{x}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(49\) |
parallelrisch | \(2 \,{\mathrm e}^{\frac {\left (4 x \,{\mathrm e}^{2 x}+\left (4 x^{2}+20 x \right ) {\mathrm e}^{3} {\mathrm e}^{x}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(49\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (17) = 34\).
Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 \, e^{\left (\frac {1}{5} \, {\left ({\left (x^{3} + 10 \, x^{2} + 25 \, x - 30\right )} e^{12} + 4 \, x e^{\left (2 \, x + 6\right )} + 4 \, {\left (x^{2} + 5 \, x\right )} e^{\left (x + 9\right )}\right )} e^{\left (-12\right )} + 6\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.33 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {\frac {4 x e^{2 x}}{5} + \frac {\left (4 x^{2} + 20 x\right ) e^{3} e^{x}}{5} + \frac {\left (x^{3} + 10 x^{2} + 25 x\right ) e^{6}}{5}}{e^{6}}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).
Time = 0.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 \, e^{\left (\frac {1}{5} \, x^{3} + \frac {4}{5} \, x^{2} e^{\left (x - 3\right )} + 2 \, x^{2} + \frac {4}{5} \, x e^{\left (2 \, x - 6\right )} + 4 \, x e^{\left (x - 3\right )} + 5 \, x\right )} \]
[In]
[Out]
\[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=\int { \frac {2}{5} \, {\left ({\left (3 \, x^{2} + 20 \, x + 25\right )} e^{6} + 4 \, {\left (2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} + 7 \, x + 5\right )} e^{\left (x + 3\right )}\right )} e^{\left (\frac {1}{5} \, {\left ({\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{6} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} + 5 \, x\right )} e^{\left (x + 3\right )}\right )} e^{\left (-6\right )} - 6\right )} \,d x } \]
[In]
[Out]
Time = 12.96 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{\frac {x^3}{5}}\,{\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-6}}{5}}\,{\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{5}} \]
[In]
[Out]