Integrand size = 198, antiderivative size = 30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=e^{\frac {x^2}{-16+x \left (5+e^x x \left (\frac {3}{5}+x+\frac {x}{\log (5)}\right )\right )}} \]
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Timed out. \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
\[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=\int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx \]
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Time = 60.96 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47
method | result | size |
risch | \(3125^{\frac {x^{2}}{5 \ln \left (5\right ) {\mathrm e}^{x} x^{3}+3 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{3}+25 x \ln \left (5\right )-80 \ln \left (5\right )}}\) | \(44\) |
parallelrisch | \({\mathrm e}^{\frac {5 x^{2} \ln \left (5\right )}{5 \ln \left (5\right ) {\mathrm e}^{x} x^{3}+3 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{3}+25 x \ln \left (5\right )-80 \ln \left (5\right )}}\) | \(46\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{{\left (5 \, x^{3} + {\left (5 \, x^{3} + 3 \, x^{2}\right )} \log \left (5\right )\right )} e^{x} + 5 \, {\left (5 \, x - 16\right )} \log \left (5\right )}} \]
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Time = 0.71 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=e^{\frac {5 x^{2} \log {\left (5 \right )}}{\left (25 x - 80\right ) \log {\left (5 \right )} + \left (5 x^{3} + \left (5 x^{3} + 3 x^{2}\right ) \log {\left (5 \right )}\right ) e^{x}}} \]
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Time = 0.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{{\left (5 \, x^{3} {\left (\log \left (5\right ) + 1\right )} + 3 \, x^{2} \log \left (5\right )\right )} e^{x} + 25 \, x \log \left (5\right ) - 80 \, \log \left (5\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx=5^{\frac {5 \, x^{2}}{5 \, x^{3} e^{x} \log \left (5\right ) + 5 \, x^{3} e^{x} + 3 \, x^{2} e^{x} \log \left (5\right ) + 25 \, x \log \left (5\right ) - 80 \, \log \left (5\right )}} \]
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Time = 15.61 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {5^{\frac {5 x^2}{(-80+25 x) \log (5)+e^x \left (5 x^3+\left (3 x^2+5 x^3\right ) \log (5)\right )}} \left (\left (-800 x+125 x^2\right ) \log ^2(5)+e^x \left (\left (-25 x^4-25 x^5\right ) \log (5)+\left (-40 x^4-25 x^5\right ) \log ^2(5)\right )\right )}{\left (6400-4000 x+625 x^2\right ) \log ^2(5)+e^x \left (\left (-800 x^3+250 x^4\right ) \log (5)+\left (-480 x^2-650 x^3+250 x^4\right ) \log ^2(5)\right )+e^{2 x} \left (25 x^6+\left (30 x^5+50 x^6\right ) \log (5)+\left (9 x^4+30 x^5+25 x^6\right ) \log ^2(5)\right )} \, dx={\mathrm {e}}^{\frac {5\,x^2\,\ln \left (5\right )}{5\,x^3\,{\mathrm {e}}^x-80\,\ln \left (5\right )+25\,x\,\ln \left (5\right )+3\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )+5\,x^3\,{\mathrm {e}}^x\,\ln \left (5\right )}} \]
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