Integrand size = 222, antiderivative size = 33 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \log \left (x \left (5+\log \left (5-x+\frac {e^{3-x} (2-x)}{-5+2 x}\right )\right )\right ) \]
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\[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=\int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )\right )}{(5-2 x) x \left (2 e^3-25 e^x-e^3 x+15 e^x x-2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ & = \int \left (\frac {2 e^x \left (-45+47 x-17 x^2+2 x^3\right )}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}-\frac {2 \left (-50+56 x-19 x^2+2 x^3-10 \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )+9 x \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )-2 x^2 \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}{(-2+x) x (-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}\right ) \, dx \\ & = 2 \int \frac {e^x \left (-45+47 x-17 x^2+2 x^3\right )}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-2 \int \frac {-50+56 x-19 x^2+2 x^3-10 \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )+9 x \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )-2 x^2 \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )}{(-2+x) x (-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ & = -\left (2 \int \frac {-50+56 x-19 x^2+2 x^3+\left (-10+9 x-2 x^2\right ) \log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )}{(5-2 x) (2-x) x \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx\right )+2 \int \left (\frac {21 e^x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}-\frac {3 e^x}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}-\frac {13 e^x x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}+\frac {2 e^x x^2}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}\right ) \, dx \\ & = -\left (2 \int \left (-\frac {1}{x}+\frac {11-9 x+2 x^2}{(-2+x) (-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}\right ) \, dx\right )+4 \int \frac {e^x x^2}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-6 \int \frac {e^x}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-26 \int \frac {e^x x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+42 \int \frac {e^x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ & = 2 \log (x)-2 \int \frac {11-9 x+2 x^2}{(-2+x) (-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+4 \int \frac {e^x x^2}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-6 \int \frac {e^x}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-26 \int \frac {e^x x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+42 \int \frac {e^x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ & = 2 \log (x)-2 \int \left (\frac {1}{5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )}-\frac {1}{(-2+x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}+\frac {2}{(-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )}\right ) \, dx+4 \int \frac {e^x x^2}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-6 \int \frac {e^x}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-26 \int \frac {e^x x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+42 \int \frac {e^x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ & = 2 \log (x)-2 \int \frac {1}{5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )} \, dx+2 \int \frac {1}{(-2+x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-4 \int \frac {1}{(-5+2 x) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+4 \int \frac {e^x x^2}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-6 \int \frac {e^x}{(-2+x) \left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx-26 \int \frac {e^x x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx+42 \int \frac {e^x}{\left (-2 e^3+25 e^x+e^3 x-15 e^x x+2 e^x x^2\right ) \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )} \, dx \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \left (\log (x)+\log \left (5+\log \left (-\frac {25+e^{3-x} (-2+x)-15 x+2 x^2}{-5+2 x}\right )\right )\right ) \]
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Time = 1.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27
method | result | size |
norman | \(2 \ln \left (x \right )+2 \ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}^{-x +3}-2 x^{2}+15 x -25}{-5+2 x}\right )+5\right )\) | \(42\) |
parallelrisch | \(2 \ln \left (x \right )+2 \ln \left (\ln \left (\frac {\left (2-x \right ) {\mathrm e}^{-x +3}-2 x^{2}+15 x -25}{-5+2 x}\right )+5\right )\) | \(42\) |
risch | \(2 \ln \left (x \right )+2 \ln \left (\ln \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )+\frac {i \left (-2 \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )}{x -\frac {5}{2}}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{x -\frac {5}{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )}{x -\frac {5}{2}}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x -\frac {5}{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )}{x -\frac {5}{2}}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )}{x -\frac {5}{2}}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{-x +3}}{2}-\frac {15}{2}\right ) x -{\mathrm e}^{-x +3}+\frac {25}{2}\right )}{x -\frac {5}{2}}\right )}^{3}+2 \pi +2 i \ln \left (x -\frac {5}{2}\right )-10 i\right )}{2}\right )\) | \(318\) |
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Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (-\frac {2 \, x^{2} + {\left (x - 2\right )} e^{\left (-x + 3\right )} - 15 \, x + 25}{2 \, x - 5}\right ) + 5\right ) \]
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Time = 0.53 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \log {\left (x \right )} + 2 \log {\left (\log {\left (\frac {- 2 x^{2} + 15 x + \left (2 - x\right ) e^{3 - x} - 25}{2 x - 5} \right )} + 5 \right )} \]
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Time = 0.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \, \log \left (x\right ) + 2 \, \log \left (-x + \log \left (-x e^{3} - {\left (2 \, x^{2} - 15 \, x + 25\right )} e^{x} + 2 \, e^{3}\right ) - \log \left (2 \, x - 5\right ) + 5\right ) \]
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Time = 0.63 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2 \, \log \left (x\right ) + 2 \, \log \left (\log \left (-\frac {2 \, x^{2} + x e^{\left (-x + 3\right )} - 15 \, x - 2 \, e^{\left (-x + 3\right )} + 25}{2 \, x - 5}\right ) + 5\right ) \]
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Time = 13.57 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-1250+1300 x-440 x^2+48 x^3+e^{3-x} \left (100-112 x+38 x^2-4 x^3\right )+\left (-250+250 x-80 x^2+8 x^3+e^{3-x} \left (20-18 x+4 x^2\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )}{-625 x+625 x^2-200 x^3+20 x^4+e^{3-x} \left (50 x-45 x^2+10 x^3\right )+\left (-125 x+125 x^2-40 x^3+4 x^4+e^{3-x} \left (10 x-9 x^2+2 x^3\right )\right ) \log \left (\frac {-25+e^{3-x} (2-x)+15 x-2 x^2}{-5+2 x}\right )} \, dx=2\,\ln \left (\ln \left (-\frac {2\,x^2-15\,x+{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,\left (x-2\right )+25}{2\,x-5}\right )+5\right )+2\,\ln \left (x\right ) \]
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