Integrand size = 69, antiderivative size = 26 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {1}{2 \left (x+\frac {x}{1+x^4 \log (5)}\right )}+\log \left (\log ^2(x)\right ) \]
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {1608, 28, 6820, 12, 14, 1498, 21, 30, 2339, 29} \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^3 \log (5)}{4 \left (x^4 \log (5)+2\right )}+\frac {1}{4 x}+2 \log (\log (x)) \]
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Rule 12
Rule 14
Rule 21
Rule 28
Rule 29
Rule 30
Rule 1498
Rule 1608
Rule 2339
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{x^2 \left (8+8 x^4 \log (5)+2 x^8 \log ^2(5)\right ) \log (x)} \, dx \\ & = \left (2 \log ^2(5)\right ) \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{x^2 \left (4 \log (5)+2 x^4 \log ^2(5)\right )^2 \log (x)} \, dx \\ & = \left (2 \log ^2(5)\right ) \int \frac {\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{\left (2+x^4 \log (5)\right )^2}+\frac {4 x}{\log (x)}}{4 x^2 \log ^2(5)} \, dx \\ & = \frac {1}{2} \int \frac {\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{\left (2+x^4 \log (5)\right )^2}+\frac {4 x}{\log (x)}}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{x^2 \left (2+x^4 \log (5)\right )^2}+\frac {4}{x \log (x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{x^2 \left (2+x^4 \log (5)\right )^2} \, dx+2 \int \frac {1}{x \log (x)} \, dx \\ & = \frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )-\frac {\int \frac {16 \log ^2(5)+8 x^4 \log ^3(5)}{x^2 \left (2+x^4 \log (5)\right )} \, dx}{32 \log ^2(5)} \\ & = \frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \log (\log (x))-\frac {1}{4} \int \frac {1}{x^2} \, dx \\ & = \frac {1}{4 x}+\frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \log (\log (x)) \\ \end{align*}
Time = 5.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {1}{4} \left (\frac {1}{x}+\frac {x^3 \log (5)}{2+x^4 \log (5)}+8 \log (\log (x))\right ) \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
method | result | size |
default | \(2 \ln \left (\ln \left (x \right )\right )+\frac {1}{4 x}+\frac {x^{3} \ln \left (5\right )}{4 x^{4} \ln \left (5\right )+8}\) | \(29\) |
parts | \(2 \ln \left (\ln \left (x \right )\right )+\frac {1}{4 x}+\frac {x^{3} \ln \left (5\right )}{4 x^{4} \ln \left (5\right )+8}\) | \(29\) |
norman | \(\frac {\frac {1}{2}+\frac {x^{4} \ln \left (5\right )}{2}}{x \left (x^{4} \ln \left (5\right )+2\right )}+2 \ln \left (\ln \left (x \right )\right )\) | \(30\) |
risch | \(\frac {x^{4} \ln \left (5\right )+1}{2 \left (x^{4} \ln \left (5\right )+2\right ) x}+2 \ln \left (\ln \left (x \right )\right )\) | \(30\) |
parallelrisch | \(\frac {8 \ln \left (5\right ) \ln \left (\ln \left (x \right )\right ) x^{5}+2+2 x^{4} \ln \left (5\right )+16 x \ln \left (\ln \left (x \right )\right )}{4 x \left (x^{4} \ln \left (5\right )+2\right )}\) | \(41\) |
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{4} \log \left (5\right ) + 4 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )} \log \left (\log \left (x\right )\right ) + 1}{2 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )}} \]
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Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=- \frac {- x^{4} \log {\left (5 \right )} - 1}{2 x^{5} \log {\left (5 \right )} + 4 x} + 2 \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{4} \log \left (5\right ) + 1}{2 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )}} + 2 \, \log \left (\log \left (x\right )\right ) \]
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Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{3} \log \left (5\right )}{4 \, {\left (x^{4} \log \left (5\right ) + 2\right )}} + \frac {1}{4 \, x} + 2 \, \log \left (\log \left (x\right )\right ) \]
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Time = 8.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=2\,\ln \left (\ln \left (x\right )\right )+\frac {\ln \left (5\right )\,x^4+1}{2\,\ln \left (5\right )\,x^5+4\,x} \]
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