Integrand size = 144, antiderivative size = 29 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {3-x}{20 e^x-x-4 \log (2) \left (e^x-\log (x)\right )} \]
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\[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{x^3+8 e^x x^2 \log (2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+e^{2 x} x \left (400+16 \log ^2(2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx \\ & = \int \frac {-4 e^x x^2 (-5+\log (2))-12 \log (2)+x \left (3+16 e^x (-5+\log (2))+\log (16)\right )-4 x \log (2) \log (x)}{x \left (x+4 e^x (-5+\log (2))-4 \log (2) \log (x)\right )^2} \, dx \\ & = \int \left (\frac {4-x}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)}+\frac {-4 x^2+x^3-12 \log (2)+3 x \left (1+\frac {\log (16)}{3}\right )+12 x \log (2) \log (x)-4 x^2 \log (2) \log (x)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}\right ) \, dx \\ & = \int \frac {4-x}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)} \, dx+\int \frac {-4 x^2+x^3-12 \log (2)+3 x \left (1+\frac {\log (16)}{3}\right )+12 x \log (2) \log (x)-4 x^2 \log (2) \log (x)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx \\ & = \int \left (-\frac {4 x}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {x^2}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}-\frac {12 \log (2)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {3+\log (16)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {12 \log (2) \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}-\frac {4 x \log (2) \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}\right ) \, dx+\int \left (\frac {4}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)}+\frac {x}{-x+20 e^x \left (1-\frac {\log (2)}{5}\right )+4 \log (2) \log (x)}\right ) \, dx \\ & = -\left (4 \int \frac {x}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx\right )+4 \int \frac {1}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)} \, dx-(4 \log (2)) \int \frac {x \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx-(12 \log (2)) \int \frac {1}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+(12 \log (2)) \int \frac {\log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+(3+\log (16)) \int \frac {1}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+\int \frac {x^2}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+\int \frac {x}{-x+20 e^x \left (1-\frac {\log (2)}{5}\right )+4 \log (2) \log (x)} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(29)=58\).
Time = 1.40 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.48 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x^2 \left (1+4 e^x (-5+\log (2))\right )-x \left (3+12 e^x (-5+\log (2))+\log (16)\right )+\log (4096)}{\left (x+4 e^x x (-5+\log (2))-4 \log (2)\right ) \left (x+4 e^x (-5+\log (2))-4 \log (2) \log (x)\right )} \]
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Time = 0.62 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {-3+x}{4 \,{\mathrm e}^{x} \ln \left (2\right )-4 \ln \left (2\right ) \ln \left (x \right )+x -20 \,{\mathrm e}^{x}}\) | \(25\) |
parallelrisch | \(\frac {15-5 x}{20 \ln \left (2\right ) \ln \left (x \right )-20 \,{\mathrm e}^{x} \ln \left (2\right )+25 \,{\mathrm e}^{x +2 \ln \left (2\right )}-5 x}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{{\left (\log \left (2\right ) - 5\right )} e^{\left (x + 2 \, \log \left (2\right )\right )} - 4 \, \log \left (2\right ) \log \left (x\right ) + x} \]
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Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{x + \left (-20 + 4 \log {\left (2 \right )}\right ) e^{x} - 4 \log {\left (2 \right )} \log {\left (x \right )}} \]
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Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{4 \, {\left (\log \left (2\right ) - 5\right )} e^{x} - 4 \, \log \left (2\right ) \log \left (x\right ) + x} \]
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Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{4 \, e^{x} \log \left (2\right ) - 4 \, \log \left (2\right ) \log \left (x\right ) + x - 20 \, e^{x}} \]
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Timed out. \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\int \frac {3\,x+\ln \left (2\right )\,\left (4\,x-12\right )-{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\left (20\,x-5\,x^2\right )+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (16\,x-4\,x^2\right )-4\,x\,\ln \left (2\right )\,\ln \left (x\right )}{25\,x\,{\mathrm {e}}^{2\,x+4\,\ln \left (2\right )}-{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\left (10\,x^2+40\,x\,{\mathrm {e}}^x\,\ln \left (2\right )\right )-\ln \left (x\right )\,\left (8\,x^2\,\ln \left (2\right )-40\,x\,{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\ln \left (2\right )+32\,x\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2\right )+x^3+8\,x^2\,{\mathrm {e}}^x\,\ln \left (2\right )+16\,x\,{\mathrm {e}}^{2\,x}\,{\ln \left (2\right )}^2+16\,x\,{\ln \left (2\right )}^2\,{\ln \left (x\right )}^2} \,d x \]
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