Integrand size = 149, antiderivative size = 27 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^x}{\log \left (\frac {x^2}{16}+(-1+x)^2 \log \left (\frac {x}{5}\right )\right )} \]
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\[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {32 e^x}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {16 e^x}{x \left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {18 e^x x}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {32 e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {32 e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {e^x x^2}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {16 e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {32 e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {16 e^x x^2 \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )}\right ) \, dx \\ & = -\left (16 \int \frac {e^x}{x \left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx\right )+16 \int \frac {e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+16 \int \frac {e^x x^2 \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-18 \int \frac {e^x x}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+32 \int \frac {e^x}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+32 \int \frac {e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+\int \frac {e^x x^2}{\left (x^2+16 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 x^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx \\ & = -\left (16 \int \frac {e^x}{x \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx\right )+16 \int \frac {e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+16 \int \frac {e^x x^2 \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-18 \int \frac {e^x x}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+32 \int \frac {e^x}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+32 \int \frac {e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+\int \frac {e^x x^2}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx \\ & = -\left (16 \int \frac {e^x}{x \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx\right )+16 \int \frac {e^x x^2 \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-18 \int \frac {e^x x}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx+80 \text {Subst}\left (\int \frac {e^{5 x} \log (x)}{\left (25 x^2+16 (-1+5 x)^2 \log (x)\right ) \log \left (\frac {1}{16} \left (25 x^2+16 (-1+5 x)^2 \log (x)\right )\right )} \, dx,x,\frac {x}{5}\right )+160 \text {Subst}\left (\int \frac {e^{5 x}}{\left (25 x^2+16 (-1+5 x)^2 \log (x)\right ) \log ^2\left (\frac {1}{16} \left (25 x^2+16 (-1+5 x)^2 \log (x)\right )\right )} \, dx,x,\frac {x}{5}\right )+160 \text {Subst}\left (\int \frac {e^{5 x} \log (x)}{\left (25 x^2+16 (-1+5 x)^2 \log (x)\right ) \log ^2\left (\frac {1}{16} \left (25 x^2+16 (-1+5 x)^2 \log (x)\right )\right )} \, dx,x,\frac {x}{5}\right )+\int \frac {e^x x^2}{\left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \, dx \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^x}{\log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \]
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Time = 4.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{\ln \left (\frac {\left (16 x^{2}-32 x +16\right ) \ln \left (\frac {x}{5}\right )}{16}+\frac {x^{2}}{16}\right )}\) | \(29\) |
parallelrisch | \(\frac {{\mathrm e}^{x}}{\ln \left (\frac {\left (16 x^{2}-32 x +16\right ) \ln \left (\frac {x}{5}\right )}{16}+\frac {x^{2}}{16}\right )}\) | \(29\) |
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^{x}}{\log \left (\frac {1}{16} \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {1}{5} \, x\right )\right )} \]
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Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^{x}}{\log {\left (\frac {x^{2}}{16} + \left (x^{2} - 2 x + 1\right ) \log {\left (\frac {x}{5} \right )} \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=-\frac {e^{x}}{4 \, \log \left (2\right ) - \log \left (-x^{2} {\left (16 \, \log \left (5\right ) - 1\right )} + 32 \, x \log \left (5\right ) + 16 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (x\right ) - 16 \, \log \left (5\right )\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1234 vs. \(2 (22) = 44\).
Time = 0.61 (sec) , antiderivative size = 1234, normalized size of antiderivative = 45.70 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\text {Hanged} \]
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