Integrand size = 116, antiderivative size = 28 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\frac {3}{\left (-13-x \left (4+e^5 \log (x)\right )\right ) \log ((-1+2 x) \log (2))} \]
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\[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-78-24 x-6 e^5 x \log (x)-\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx \\ & = \int \left (\frac {78}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}+\frac {24 x}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}+\frac {6 e^5 x \log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}+\frac {12 \left (1+\frac {e^5}{4}\right )}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}+\frac {24 \left (1+\frac {e^5}{4}\right ) x}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}+\frac {3 e^5 \log (x)}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}+\frac {6 e^5 x \log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}\right ) \, dx \\ & = 24 \int \frac {x}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+78 \int \frac {1}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (6 e^5\right ) \int \frac {x \log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (6 e^5\right ) \int \frac {x \log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 \left (4+e^5\right )\right ) \int \frac {1}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (6 \left (4+e^5\right )\right ) \int \frac {x}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx \\ & = 24 \int \left (\frac {1}{2 \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}+\frac {1}{2 (-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}\right ) \, dx+78 \int \frac {1}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (6 e^5\right ) \int \left (\frac {\log (x)}{2 \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}+\frac {\log (x)}{2 (-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))}\right ) \, dx+\left (6 e^5\right ) \int \left (\frac {\log (x)}{2 \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}+\frac {\log (x)}{2 (-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}\right ) \, dx+\left (3 \left (4+e^5\right )\right ) \int \frac {1}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (6 \left (4+e^5\right )\right ) \int \left (\frac {1}{2 \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}+\frac {1}{2 (-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))}\right ) \, dx \\ & = 12 \int \frac {1}{\left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+12 \int \frac {1}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+78 \int \frac {1}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{\left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log ^2(-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{\left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 e^5\right ) \int \frac {\log (x)}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 \left (4+e^5\right )\right ) \int \frac {1}{\left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 \left (4+e^5\right )\right ) \int \frac {1}{(1-2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx+\left (3 \left (4+e^5\right )\right ) \int \frac {1}{(-1+2 x) \left (13+4 x+e^5 x \log (x)\right )^2 \log (-\log (2)+2 x \log (2))} \, dx \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\frac {3}{\left (-13-4 x-e^5 x \log (x)\right ) \log ((-1+2 x) \log (2))} \]
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Time = 2.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {3}{\left (x \,{\mathrm e}^{5} \ln \left (x \right )+4 x +13\right ) \ln \left (\left (-1+2 x \right ) \ln \left (2\right )\right )}\) | \(27\) |
parallelrisch | \(-\frac {3}{\left (x \,{\mathrm e}^{5} \ln \left (x \right )+4 x +13\right ) \ln \left (\left (-1+2 x \right ) \ln \left (2\right )\right )}\) | \(27\) |
default | \(-\frac {3}{\left (\ln \left (x \right ) {\mathrm e}^{5+\ln \left (x \right )}+4 x +13\right ) \left (\ln \left (\ln \left (2\right )\right )+\ln \left (-1+2 x \right )\right )}\) | \(30\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{{\left (x e^{5} \log \left (x\right ) + 4 \, x + 13\right )} \log \left ({\left (2 \, x - 1\right )} \log \left (2\right )\right )} \]
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Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=- \frac {3}{\left (x e^{5} \log {\left (x \right )} + 4 x + 13\right ) \log {\left (\left (2 x - 1\right ) \log {\left (2 \right )} \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{x e^{5} \log \left (x\right ) \log \left (\log \left (2\right )\right ) + {\left (x e^{5} \log \left (x\right ) + 4 \, x + 13\right )} \log \left (2 \, x - 1\right ) + 4 \, x \log \left (\log \left (2\right )\right ) + 13 \, \log \left (\log \left (2\right )\right )} \]
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Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=-\frac {3}{x e^{5} \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right ) \log \left (x\right ) + 4 \, x \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right ) + 13 \, \log \left (2 \, x \log \left (2\right ) - \log \left (2\right )\right )} \]
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Timed out. \[ \int \frac {78+24 x+6 e^5 x \log (x)+\left (-12+24 x+e^5 (-3+6 x)+e^5 (-3+6 x) \log (x)\right ) \log ((-1+2 x) \log (2))}{\left (-169+234 x+192 x^2+32 x^3+e^5 \left (-26 x+44 x^2+16 x^3\right ) \log (x)+e^{10} \left (-x^2+2 x^3\right ) \log ^2(x)\right ) \log ^2((-1+2 x) \log (2))} \, dx=\int \frac {24\,x+\ln \left (\ln \left (2\right )\,\left (2\,x-1\right )\right )\,\left (24\,x+{\mathrm {e}}^5\,\left (6\,x-3\right )+{\mathrm {e}}^5\,\ln \left (x\right )\,\left (6\,x-3\right )-12\right )+6\,x\,{\mathrm {e}}^5\,\ln \left (x\right )+78}{{\ln \left (\ln \left (2\right )\,\left (2\,x-1\right )\right )}^2\,\left (234\,x+192\,x^2+32\,x^3+{\mathrm {e}}^5\,\ln \left (x\right )\,\left (16\,x^3+44\,x^2-26\,x\right )-{\mathrm {e}}^{10}\,{\ln \left (x\right )}^2\,\left (x^2-2\,x^3\right )-169\right )} \,d x \]
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