Integrand size = 102, antiderivative size = 28 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=-4+e^{\frac {-3+e^5+(3+x)^2}{5 (5+x-\log (8))}}+x \]
[Out]
\[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=\int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{5 x^2+10 x (5-\log (8))+5 (5-\log (8))^2} \, dx \\ & = \int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{5 (5+x-\log (8))^2} \, dx \\ & = \frac {1}{5} \int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{(5+x-\log (8))^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {50 x}{(5+x-\log (8))^2}+\frac {5 x^2}{(5+x-\log (8))^2}+\frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (24-e^5+x^2+2 x (5-\log (8))-6 \log (8)\right )}{(5+x-\log (8))^2}-\frac {10 (5+x) \log (8)}{(5+x-\log (8))^2}+\frac {5 \left (25+\log ^2(8)\right )}{(5+x-\log (8))^2}\right ) \, dx \\ & = -\frac {25+\log ^2(8)}{5+x-\log (8)}+\frac {1}{5} \int \frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (24-e^5+x^2+2 x (5-\log (8))-6 \log (8)\right )}{(5+x-\log (8))^2} \, dx+10 \int \frac {x}{(5+x-\log (8))^2} \, dx-(2 \log (8)) \int \frac {5+x}{(5+x-\log (8))^2} \, dx+\int \frac {x^2}{(5+x-\log (8))^2} \, dx \\ & = -\frac {25+\log ^2(8)}{5+x-\log (8)}+\frac {1}{5} \int \left (e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}}+\frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (-1-e^5+4 \log (8)-\log ^2(8)\right )}{(5+x-\log (8))^2}\right ) \, dx+10 \int \left (\frac {1}{5+x-\log (8)}+\frac {-5+\log (8)}{(5+x-\log (8))^2}\right ) \, dx-(2 \log (8)) \int \left (\frac {1}{5+x-\log (8)}+\frac {\log (8)}{(5+x-\log (8))^2}\right ) \, dx+\int \left (1+\frac {2 (-5+\log (8))}{5+x-\log (8)}+\frac {(-5+\log (8))^2}{(5+x-\log (8))^2}\right ) \, dx \\ & = x+\frac {10 (5-\log (8))}{5+x-\log (8)}-\frac {(5-\log (8))^2}{5+x-\log (8)}+\frac {2 \log ^2(8)}{5+x-\log (8)}-\frac {25+\log ^2(8)}{5+x-\log (8)}+10 \log (5+x-\log (8))-2 (5-\log (8)) \log (5+x-\log (8))-2 \log (8) \log (5+x-\log (8))+\frac {1}{5} \int e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \, dx+\frac {1}{5} \left (-1-e^5+4 \log (8)-\log ^2(8)\right ) \int \frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}}}{(5+x-\log (8))^2} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(28)=56\).
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=\frac {1}{5} \left (5\ 2^{-\frac {3 \log (8)}{5 (5+x-\log (8))}} e^{\frac {6+e^5+6 x+x^2+\log ^2(8)}{5 (5+x-\log (8))}}+5 (5+x-\log (8))\right ) \]
[In]
[Out]
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{5}+6 x +6}{5 \left (3 \ln \left (2\right )-x -5\right )}}\) | \(27\) |
parallelrisch | \(12 \ln \left (2\right )+x +{\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{5}+6 x +6}{5 \left (3 \ln \left (2\right )-x -5\right )}}-20\) | \(32\) |
parts | \(x +\frac {\left (3 \ln \left (2\right )-5\right ) {\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}-x \,{\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}}{3 \ln \left (2\right )-x -5}\) | \(80\) |
norman | \(\frac {\left (3 \ln \left (2\right )-5\right ) {\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}-x^{2}-x \,{\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}+9 \ln \left (2\right )^{2}-30 \ln \left (2\right )+25}{3 \ln \left (2\right )-x -5}\) | \(94\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\left (\frac {x^{2} + 6 \, x + e^{5} + 6}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}}\right )} \]
[In]
[Out]
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\frac {- x^{2} - 6 x - e^{5} - 6}{- 5 x - 25 + 15 \log {\left (2 \right )}}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (24) = 48\).
Time = 0.41 (sec) , antiderivative size = 200, normalized size of antiderivative = 7.14 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=6 \, {\left (\frac {3 \, \log \left (2\right ) - 5}{x - 3 \, \log \left (2\right ) + 5} - \log \left (x - 3 \, \log \left (2\right ) + 5\right )\right )} \log \left (2\right ) + 2 \, {\left (3 \, \log \left (2\right ) - 5\right )} \log \left (x - 3 \, \log \left (2\right ) + 5\right ) + 2^{\frac {3}{5}} e^{\left (\frac {1}{5} \, x + \frac {9 \, \log \left (2\right )^{2}}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {e^{5}}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} - \frac {12 \, \log \left (2\right )}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {1}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {1}{5}\right )} + x - \frac {9 \, \log \left (2\right )^{2}}{x - 3 \, \log \left (2\right ) + 5} - \frac {9 \, \log \left (2\right )^{2} - 30 \, \log \left (2\right ) + 25}{x - 3 \, \log \left (2\right ) + 5} - \frac {10 \, {\left (3 \, \log \left (2\right ) - 5\right )}}{x - 3 \, \log \left (2\right ) + 5} + \frac {30 \, \log \left (2\right )}{x - 3 \, \log \left (2\right ) + 5} - \frac {25}{x - 3 \, \log \left (2\right ) + 5} + 10 \, \log \left (x - 3 \, \log \left (2\right ) + 5\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (24) = 48\).
Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\left (\frac {3 \, x^{2} \log \left (2\right ) - 5 \, x^{2} + x e^{5} + 18 \, x \log \left (2\right ) - 24 \, x}{5 \, {\left (3 \, x \log \left (2\right ) - 9 \, \log \left (2\right )^{2} - 5 \, x + 30 \, \log \left (2\right ) - 25\right )}} - \frac {e^{5} + 6}{5 \, {\left (3 \, \log \left (2\right ) - 5\right )}}\right )} \]
[In]
[Out]
Timed out. \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=\int \frac {50\,x-3\,\ln \left (2\right )\,\left (10\,x+50\right )+45\,{\ln \left (2\right )}^2+{\mathrm {e}}^{\frac {x^2+6\,x+{\mathrm {e}}^5+6}{5\,x-15\,\ln \left (2\right )+25}}\,\left (10\,x-{\mathrm {e}}^5-3\,\ln \left (2\right )\,\left (2\,x+6\right )+x^2+24\right )+5\,x^2+125}{50\,x-3\,\ln \left (2\right )\,\left (10\,x+50\right )+45\,{\ln \left (2\right )}^2+5\,x^2+125} \,d x \]
[In]
[Out]