Integrand size = 156, antiderivative size = 24 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=e^{17+e^x+\frac {16}{-4+\frac {e^x}{3}+\log (2)}}+x \]
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\[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=\int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+\exp \left (\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}\right ) \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {9 \left (\frac {x^2}{9}+\frac {2}{3} x (-4+\log (2))+16 \left (1+\frac {1}{16} (-8+\log (2)) \log (2)\right )+\frac {1}{9} 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} x \left (96+x^2+6 x (-4+\log (2))-72 \log (2)+9 \log ^2(2)\right )\right )}{x (12-x-\log (8))^2} \, dx,x,e^x\right ) \\ & = 9 \text {Subst}\left (\int \frac {\frac {x^2}{9}+\frac {2}{3} x (-4+\log (2))+16 \left (1+\frac {1}{16} (-8+\log (2)) \log (2)\right )+\frac {1}{9} 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} x \left (96+x^2+6 x (-4+\log (2))-72 \log (2)+9 \log ^2(2)\right )}{x (12-x-\log (8))^2} \, dx,x,e^x\right ) \\ & = 9 \text {Subst}\left (\int \left (\frac {1}{9 x}+\frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (x^2-6 x (4-\log (2))+3 \left (32-24 \log (2)+3 \log ^2(2)\right )\right )}{9 (12-x-\log (8))^2}\right ) \, dx,x,e^x\right ) \\ & = x+\text {Subst}\left (\int \frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (x^2-6 x (4-\log (2))+3 \left (32-24 \log (2)+3 \log ^2(2)\right )\right )}{(12-x-\log (8))^2} \, dx,x,e^x\right ) \\ & = x+\text {Subst}\left (\int \left (2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}}+\frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \left (-48+9 \log ^2(2)-6 \log (2) \log (8)+\log ^2(8)\right )}{(-12+x+\log (8))^2}\right ) \, dx,x,e^x\right ) \\ & = x+\left (-48+9 \log ^2(2)-6 \log (2) \log (8)+\log ^2(8)\right ) \text {Subst}\left (\int \frac {2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}}}{(-12+x+\log (8))^2} \, dx,x,e^x\right )+\text {Subst}\left (\int 2^{\frac {51}{-12+x+\log (8)}} e^{\frac {-156+x^2+x (5+\log (8))}{-12+x+\log (8)}} \, dx,x,e^x\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(24)=48\).
Time = 2.60 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.08 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x+\frac {e^{17+e^x+\frac {48}{-12+e^x+\log (8)}} \left (96-24 e^x+e^{2 x}+9 \log ^2(2)-24 \log (8)+2 e^x \log (8)\right )}{\left (-12+e^x+\log (8)\right )^2 \left (1-\frac {48}{\left (-12+e^x+\log (8)\right )^2}\right )} \]
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Time = 2.98 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46
method | result | size |
risch | \(x +{\mathrm e}^{\frac {3 \,{\mathrm e}^{x} \ln \left (2\right )+5 \,{\mathrm e}^{x}+51 \ln \left (2\right )+{\mathrm e}^{2 x}-156}{{\mathrm e}^{x}+3 \ln \left (2\right )-12}}\) | \(35\) |
parallelrisch | \(x +{\mathrm e}^{\frac {\left ({\mathrm e}^{-\ln \left (3\right )+x}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+17 \,{\mathrm e}^{-\ln \left (3\right )+x}+17 \ln \left (2\right )-52}{{\mathrm e}^{-\ln \left (3\right )+x}+\ln \left (2\right )-4}}\) | \(47\) |
norman | \(\frac {\left (3 \ln \left (2\right )-12\right ) x +\left (3 \ln \left (2\right )-12\right ) {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \left (2\right )-52}{\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4}}+{\mathrm e}^{x} x +{\mathrm e}^{x} {\mathrm e}^{\frac {\left (\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4\right ) {\mathrm e}^{x}+\frac {17 \,{\mathrm e}^{x}}{3}+17 \ln \left (2\right )-52}{\frac {{\mathrm e}^{x}}{3}+\ln \left (2\right )-4}}}{{\mathrm e}^{x}+3 \ln \left (2\right )-12}\) | \(101\) |
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x + e^{\left (\frac {{\left (3 \, \log \left (2\right ) + 5\right )} e^{x} + e^{\left (2 \, x\right )} + 51 \, \log \left (2\right ) - 156}{e^{x} + 3 \, \log \left (2\right ) - 12}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x + e^{\frac {\left (\frac {e^{x}}{3} - 4 + \log {\left (2 \right )}\right ) e^{x} + \frac {17 e^{x}}{3} - 52 + 17 \log {\left (2 \right )}}{\frac {e^{x}}{3} - 4 + \log {\left (2 \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (22) = 44\).
Time = 0.40 (sec) , antiderivative size = 261, normalized size of antiderivative = 10.88 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx={\left (\frac {x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} - \frac {\log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48}\right )} \log \left (2\right )^{2} - 8 \, {\left (\frac {x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} - \frac {\log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48}\right )} \log \left (2\right ) + \frac {16 \, x}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {3 \, {\left (\log \left (2\right ) - 4\right )}}{e^{x} + 3 \, \log \left (2\right ) - 12} - \frac {6 \, \log \left (2\right )}{e^{x} + 3 \, \log \left (2\right ) - 12} - \frac {16 \, \log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right )}{\log \left (2\right )^{2} - 8 \, \log \left (2\right ) + 16} + \frac {48}{{\left (\log \left (2\right ) - 4\right )} e^{x} + 3 \, \log \left (2\right )^{2} - 24 \, \log \left (2\right ) + 48} + \frac {24}{e^{x} + 3 \, \log \left (2\right ) - 12} + e^{\left (\frac {48}{e^{x} + 3 \, \log \left (2\right ) - 12} + e^{x} + 17\right )} + \log \left (e^{x} + 3 \, \log \left (2\right ) - 12\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (22) = 44\).
Time = 0.72 (sec) , antiderivative size = 239, normalized size of antiderivative = 9.96 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=\frac {1}{2} \, {\left (2 \, {\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} + 2 \, e^{\left (x - \log \left (3\right )\right )} \log \left (e^{\left (x - \log \left (3\right )\right )} + \log \left (2\right ) - 4\right ) - 2 \, e^{\left (x - \log \left (3\right )\right )} \log \left (-e^{\left (x - \log \left (3\right )\right )} - \log \left (2\right ) + 4\right ) + e^{\left (\frac {{\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + {\left (x - \log \left (3\right )\right )} \log \left (2\right )^{2} + 3 \, e^{\left (x - \log \left (3\right )\right )} \log \left (2\right )^{2} - 4 \, {\left (x - \log \left (3\right )\right )} e^{\left (x - \log \left (3\right )\right )} - 8 \, {\left (x - \log \left (3\right )\right )} \log \left (2\right ) + 3 \, e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} \log \left (2\right ) - 24 \, e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + 16 \, x - 12 \, e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + 32 \, e^{\left (x - \log \left (3\right )\right )} - 16 \, \log \left (3\right )}{e^{\left (x - \log \left (3\right )\right )} \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, e^{\left (x - \log \left (3\right )\right )} - 8 \, \log \left (2\right ) + 16} + \frac {\log \left (2\right )^{2} + 13 \, \log \left (2\right ) - 52}{\log \left (2\right ) - 4}\right )}\right )} e^{\left (-x + \log \left (3\right )\right )} \]
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Time = 8.73 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.12 \[ \int \frac {16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))+e^{\frac {-52+\frac {17 e^x}{3}+17 \log (2)+e^x \left (-4+\frac {e^x}{3}+\log (2)\right )}{-4+\frac {e^x}{3}+\log (2)}} \left (-\frac {16 e^x}{3}+e^x \left (16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))\right )\right )}{16+\frac {e^{2 x}}{9}-8 \log (2)+\log ^2(2)+\frac {1}{3} e^x (-8+2 \log (2))} \, dx=x+2^{\frac {{\mathrm {e}}^x}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}}\,2^{\frac {17}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{3\,\left (\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4\right )}-\frac {52}{\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4}+\frac {5\,{\mathrm {e}}^x}{3\,\left (\ln \left (2\right )+\frac {{\mathrm {e}}^x}{3}-4\right )}} \]
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