Integrand size = 237, antiderivative size = 32 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\frac {e^7+\log \left (\frac {\log (x)}{x}\right )}{\left (x+x^2\right ) \left (-\frac {4}{\log (5)}+\log (-1+x)\right )} \]
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\[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\log (5) \left (4 \left (-1+x^2\right )+\log (x) \left (4-4 x^2+e^7 \left (4+x^2 (-8+\log (5))+x (4+\log (5))\right )+\left (4+x^2 (-8+\log (5))+x (4+\log (5))\right ) \log \left (\frac {\log (x)}{x}\right )\right )+(-1+x) \log (5) \log (-1+x) \left (-1-x+\log (x) \left (1+e^7+x+2 e^7 x+(1+2 x) \log \left (\frac {\log (x)}{x}\right )\right )\right )\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2 \log (x)} \, dx \\ & = \log (5) \int \frac {4 \left (-1+x^2\right )+\log (x) \left (4-4 x^2+e^7 \left (4+x^2 (-8+\log (5))+x (4+\log (5))\right )+\left (4+x^2 (-8+\log (5))+x (4+\log (5))\right ) \log \left (\frac {\log (x)}{x}\right )\right )+(-1+x) \log (5) \log (-1+x) \left (-1-x+\log (x) \left (1+e^7+x+2 e^7 x+(1+2 x) \log \left (\frac {\log (x)}{x}\right )\right )\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2 \log (x)} \, dx \\ & = \log (5) \int \left (\frac {e^7 \left (4-x^2 (8-\log (5))+x (4+\log (5))\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2}+\frac {4}{(-1+x) (1+x)^2 (-4+\log (5) \log (-1+x))^2}-\frac {4}{(-1+x) x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2}-\frac {\left (1+e^7\right ) \log (5) \log (-1+x)}{x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2}-\frac {\left (1+2 e^7\right ) \log (5) \log (-1+x)}{x (1+x)^2 (-4+\log (5) \log (-1+x))^2}-\frac {4}{x^2 (1+x) (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {\log (5) \log (-1+x)}{x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {\log (5) \log (-1+x)}{x (1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {\left (4-8 x^2 \left (1-\frac {\log (5)}{8}\right )+4 x \left (1+\frac {\log (5)}{4}\right )-\log (5) \log (-1+x)-x \log (5) \log (-1+x)+2 x^2 \log (5) \log (-1+x)\right ) \log \left (\frac {\log (x)}{x}\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2}\right ) \, dx \\ & = \log (5) \int \frac {\left (4-8 x^2 \left (1-\frac {\log (5)}{8}\right )+4 x \left (1+\frac {\log (5)}{4}\right )-\log (5) \log (-1+x)-x \log (5) \log (-1+x)+2 x^2 \log (5) \log (-1+x)\right ) \log \left (\frac {\log (x)}{x}\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2} \, dx+(4 \log (5)) \int \frac {1}{(-1+x) (1+x)^2 (-4+\log (5) \log (-1+x))^2} \, dx-(4 \log (5)) \int \frac {1}{(-1+x) x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2} \, dx-(4 \log (5)) \int \frac {1}{x^2 (1+x) (-4+\log (5) \log (-1+x))^2 \log (x)} \, dx+\left (e^7 \log (5)\right ) \int \frac {4-x^2 (8-\log (5))+x (4+\log (5))}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2} \, dx+\log ^2(5) \int \frac {\log (-1+x)}{x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)} \, dx+\log ^2(5) \int \frac {\log (-1+x)}{x (1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)} \, dx-\left (\left (1+e^7\right ) \log ^2(5)\right ) \int \frac {\log (-1+x)}{x^2 (1+x)^2 (-4+\log (5) \log (-1+x))^2} \, dx-\left (\left (1+2 e^7\right ) \log ^2(5)\right ) \int \frac {\log (-1+x)}{x (1+x)^2 (-4+\log (5) \log (-1+x))^2} \, dx \\ & = \log (5) \int \frac {\left (4+x^2 (-8+\log (5))+x (4+\log (5))+\left (-1-x+2 x^2\right ) \log (5) \log (-1+x)\right ) \log \left (\frac {\log (x)}{x}\right )}{(1-x) x^2 (1+x)^2 (4-\log (5) \log (-1+x))^2} \, dx-(4 \log (5)) \int \left (\frac {1}{4 (-1+x) (-4+\log (5) \log (-1+x))^2}-\frac {1}{x^2 (-4+\log (5) \log (-1+x))^2}+\frac {1}{x (-4+\log (5) \log (-1+x))^2}-\frac {1}{2 (1+x)^2 (-4+\log (5) \log (-1+x))^2}-\frac {5}{4 (1+x) (-4+\log (5) \log (-1+x))^2}\right ) \, dx+(4 \log (5)) \int \left (-\frac {1}{2 (1+x)^2 (-4+\log (5) \log (-1+x))^2}+\frac {1}{2 \left (-1+x^2\right ) (-4+\log (5) \log (-1+x))^2}\right ) \, dx-(4 \log (5)) \int \left (\frac {1}{x^2 (-4+\log (5) \log (-1+x))^2 \log (x)}-\frac {1}{x (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {1}{(1+x) (-4+\log (5) \log (-1+x))^2 \log (x)}\right ) \, dx+\left (e^7 \log (5)\right ) \int \left (\frac {4}{x^2 (4-\log (5) \log (-1+x))^2}-\frac {4}{(1+x)^2 (4-\log (5) \log (-1+x))^2}+\frac {\log (5)}{x (4-\log (5) \log (-1+x))^2}-\frac {x \log (5)}{\left (-1+x^2\right ) (4-\log (5) \log (-1+x))^2}\right ) \, dx+\log ^2(5) \int \left (\frac {\log (-1+x)}{x (-4+\log (5) \log (-1+x))^2 \log (x)}-\frac {\log (-1+x)}{(1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)}-\frac {\log (-1+x)}{(1+x) (-4+\log (5) \log (-1+x))^2 \log (x)}\right ) \, dx+\log ^2(5) \int \left (\frac {\log (-1+x)}{x^2 (-4+\log (5) \log (-1+x))^2 \log (x)}-\frac {2 \log (-1+x)}{x (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {\log (-1+x)}{(1+x)^2 (-4+\log (5) \log (-1+x))^2 \log (x)}+\frac {2 \log (-1+x)}{(1+x) (-4+\log (5) \log (-1+x))^2 \log (x)}\right ) \, dx-\left (\left (1+e^7\right ) \log ^2(5)\right ) \int \left (\frac {4}{x^2 (1+x)^2 \log (5) (-4+\log (5) \log (-1+x))^2}+\frac {1}{x^2 (1+x)^2 \log (5) (-4+\log (5) \log (-1+x))}\right ) \, dx-\left (\left (1+2 e^7\right ) \log ^2(5)\right ) \int \left (\frac {4}{x (1+x)^2 \log (5) (-4+\log (5) \log (-1+x))^2}+\frac {1}{x (1+x)^2 \log (5) (-4+\log (5) \log (-1+x))}\right ) \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\frac {\log (5) \left (e^7+\log \left (\frac {\log (x)}{x}\right )\right )}{x (1+x) (-4+\log (5) \log (-1+x))} \]
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Time = 105.99 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25
method | result | size |
parallelrisch | \(\frac {4 \ln \left (5\right ) \ln \left (\frac {\ln \left (x \right )}{x}\right )+4 \ln \left (5\right ) {\mathrm e}^{7}}{4 x \left (1+x \right ) \left (\ln \left (5\right ) \ln \left (-1+x \right )-4\right )}\) | \(40\) |
risch | \(\frac {\ln \left (5\right ) \ln \left (\ln \left (x \right )\right )}{\left (1+x \right ) x \left (\ln \left (5\right ) \ln \left (-1+x \right )-4\right )}+\frac {\left (-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2}-i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{3}+2 \,{\mathrm e}^{7}-2 \ln \left (x \right )\right ) \ln \left (5\right )}{2 \left (1+x \right ) x \left (\ln \left (5\right ) \ln \left (-1+x \right )-4\right )}\) | \(143\) |
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Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\frac {e^{7} \log \left (5\right ) + \log \left (5\right ) \log \left (\frac {\log \left (x\right )}{x}\right )}{{\left (x^{2} + x\right )} \log \left (5\right ) \log \left (x - 1\right ) - 4 \, x^{2} - 4 \, x} \]
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Exception generated. \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=-\frac {e^{7} \log \left (5\right ) - \log \left (5\right ) \log \left (x\right ) + \log \left (5\right ) \log \left (\log \left (x\right )\right )}{4 \, x^{2} - {\left (x^{2} \log \left (5\right ) + x \log \left (5\right )\right )} \log \left (x - 1\right ) + 4 \, x} \]
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Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\frac {e^{7} \log \left (5\right ) - \log \left (5\right ) \log \left (x\right ) + \log \left (5\right ) \log \left (\log \left (x\right )\right )}{x^{2} \log \left (5\right ) \log \left (x - 1\right ) + x \log \left (5\right ) \log \left (x - 1\right ) - 4 \, x^{2} - 4 \, x} \]
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Time = 9.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4-4 x^2\right ) \log (5)+\left (-1+x^2\right ) \log ^2(5) \log (-1+x)+\left (\left (-4+4 x^2+e^7 \left (-4-4 x+8 x^2\right )\right ) \log (5)+e^7 \left (-x-x^2\right ) \log ^2(5)+\left (1-x^2+e^7 \left (1+x-2 x^2\right )\right ) \log ^2(5) \log (-1+x)\right ) \log (x)+\left (\left (-4-4 x+8 x^2\right ) \log (5)+\left (-x-x^2\right ) \log ^2(5)+\left (1+x-2 x^2\right ) \log ^2(5) \log (-1+x)\right ) \log (x) \log \left (\frac {\log (x)}{x}\right )}{\left (-16 x^2-16 x^3+16 x^4+16 x^5+\left (8 x^2+8 x^3-8 x^4-8 x^5\right ) \log (5) \log (-1+x)+\left (-x^2-x^3+x^4+x^5\right ) \log ^2(5) \log ^2(-1+x)\right ) \log (x)} \, dx=\frac {\ln \left (5\right )\,\left ({\mathrm {e}}^7+\ln \left (\frac {\ln \left (x\right )}{x}\right )\right )}{x\,\left (\ln \left (x-1\right )\,\ln \left (5\right )-4\right )\,\left (x+1\right )} \]
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