Integrand size = 50, antiderivative size = 22 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x \left (-1+5^{-x} \left (x \left (-1+(-2+\log (5))^2\right )\right )^x\right ) \]
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\[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = -x+\int 5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int 5^{-x} \left (x (3-4 \log (5))+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int 5^{-x} \left (x \left (3-4 \log (5)+\log ^2(5)\right )\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int \left ((x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x+\frac {(x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x}{1-\log (5)}+\frac {(x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right )}{1-\log (5)}\right ) \, dx \\ & = -x+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right ) \, dx}{1-\log (5)}+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \, dx \\ & = -x+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}-\frac {\int \frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{x} \, dx}{1-\log (5)}+\frac {\log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right ) \int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \, dx \\ \end{align*}
Time = 0.60 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=-x+\frac {5^{-x} \left (x \left (3-4 \log (5)+\log ^2(5)\right )\right )^{1+x}}{3-4 \log (5)+\log ^2(5)} \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \({\mathrm e}^{x \ln \left (\frac {x \left (\ln \left (5\right )^{2}-4 \ln \left (5\right )+3\right )}{5}\right )} x -x\) | \(24\) |
risch | \(-x +\left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )^{x} x\) | \(25\) |
default | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
norman | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
parts | \(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\) | \(27\) |
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx={\left (\frac {1}{5} \, x \log \left (5\right )^{2} - \frac {4}{5} \, x \log \left (5\right ) + \frac {3}{5} \, x\right )}^{x} x - x \]
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Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{x \log {\left (- \frac {4 x \log {\left (5 \right )}}{5} + \frac {x \log {\left (5 \right )}^{2}}{5} + \frac {3 x}{5} \right )}} - x \]
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Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{\left (-x \log \left (5\right ) + x \log \left (x\right ) + x \log \left (\log \left (5\right ) - 1\right ) + x \log \left (\log \left (5\right ) - 3\right )\right )} - x \]
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Timed out. \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\text {Timed out} \]
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Time = 14.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x\,\left ({\left (\frac {3\,x}{5}-\frac {4\,x\,\ln \left (5\right )}{5}+\frac {x\,{\ln \left (5\right )}^2}{5}\right )}^x-1\right ) \]
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