Integrand size = 128, antiderivative size = 28 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{-\frac {5}{5+e^x}-x+x^2+x (-1+5 \log (x))^2} \]
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\[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\int \frac {\exp \left (\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}\right ) \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \left (50 (-5+x)+2 e^{2 x} (-5+x)+5 e^x (-19+4 x)+40 \left (5+e^x\right )^2 \log (x)+25 \left (5+e^x\right )^2 \log ^2(x)\right )}{\left (5+e^x\right )^2} \, dx \\ & = \int \left (2 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x}-10 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}-\frac {25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2}+\frac {5 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x}+40 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x)+25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x)\right ) \, dx \\ & = 2 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x} \, dx+5 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x} \, dx-10 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \, dx-25 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2} \, dx+25 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x) \, dx+40 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x) \, dx \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \]
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Time = 2.40 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{x} x +125 x \right ) \ln \left (x \right )^{2}+\left (-10 \,{\mathrm e}^{x} x -50 x \right ) \ln \left (x \right )+{\mathrm e}^{x} x^{2}+5 x^{2}-5}{{\mathrm e}^{x}+5}}\) | \(48\) |
risch | \({\mathrm e}^{\frac {25 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}+125 x \ln \left (x \right )^{2}-10 x \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{x} x^{2}-50 x \ln \left (x \right )+5 x^{2}-5}{{\mathrm e}^{x}+5}}\) | \(50\) |
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Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right )^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right ) - 5}{e^{x} + 5}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.55 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\frac {x^{2} e^{x} + 5 x^{2} + \left (- 10 x e^{x} - 50 x\right ) \log {\left (x \right )} + \left (25 x e^{x} + 125 x\right ) \log {\left (x \right )}^{2} - 5}{e^{x} + 5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (26) = 52\).
Time = 0.42 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.04 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\left (\frac {25 \, x e^{x} \log \left (x\right )^{2}}{e^{x} + 5} + \frac {x^{2} e^{x}}{e^{x} + 5} - \frac {10 \, x e^{x} \log \left (x\right )}{e^{x} + 5} + \frac {125 \, x \log \left (x\right )^{2}}{e^{x} + 5} + \frac {5 \, x^{2}}{e^{x} + 5} - \frac {50 \, x \log \left (x\right )}{e^{x} + 5} - \frac {5}{e^{x} + 5}\right )} \]
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\[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\int { \frac {{\left (25 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \left (x\right )^{2} + 2 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} + 5 \, {\left (4 \, x - 19\right )} e^{x} + 40 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \left (x\right ) + 50 \, x - 250\right )} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right )^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right ) - 5}{e^{x} + 5}\right )}}{e^{\left (2 \, x\right )} + 10 \, e^{x} + 25} \,d x } \]
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Time = 14.91 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.57 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {125\,x\,{\ln \left (x\right )}^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {5\,x^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2}{{\mathrm {e}}^x+5}}}{x^{10\,x}} \]
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