Integrand size = 109, antiderivative size = 25 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {3+x}{x-x \left (4+x^2 \log ^2(2)\right )^2+\log (6)} \]
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\[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)}+\frac {-60 x+72 x^2 \log ^2(2)-16 x^3 \log ^2(2)+15 x^4 \log ^4(2)+5 (9+\log (6))}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2}\right ) \, dx \\ & = 4 \int \frac {1}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \, dx+\int \frac {-60 x+72 x^2 \log ^2(2)-16 x^3 \log ^2(2)+15 x^4 \log ^4(2)+5 (9+\log (6))}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2} \, dx \\ & = -\frac {3}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)}+4 \int \frac {1}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \, dx+\frac {\int \frac {-300 x \log ^4(2)-80 x^3 \log ^6(2)+25 \log ^4(2) \log (6)}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2} \, dx}{5 \log ^4(2)} \\ & = -\frac {3}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)}+4 \int \frac {1}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \, dx+\frac {\int \left (-\frac {300 x \log ^4(2)}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2}-\frac {80 x^3 \log ^6(2)}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2}+\frac {25 \log ^4(2) \log (6)}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2}\right ) \, dx}{5 \log ^4(2)} \\ & = -\frac {3}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)}+4 \int \frac {1}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \, dx-60 \int \frac {x}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2} \, dx-\left (16 \log ^2(2)\right ) \int \frac {x^3}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2} \, dx+(5 \log (6)) \int \frac {1}{\left (15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)\right )^2} \, dx \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {-3-x}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \]
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Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
gosper | \(-\frac {3+x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(33\) |
default | \(\frac {-3-x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(34\) |
risch | \(\frac {-3-x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (2\right )-\ln \left (3\right )+15 x}\) | \(38\) |
parallelrisch | \(\frac {-45+x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )}{15 x^{5} \ln \left (2\right )^{4}+120 x^{3} \ln \left (2\right )^{2}-15 \ln \left (6\right )+225 x}\) | \(53\) |
norman | \(\frac {-\frac {\left (\ln \left (6\right )+45\right ) x}{\ln \left (6\right )}-\frac {24 \ln \left (2\right )^{2} x^{3}}{\ln \left (6\right )}-\frac {3 \ln \left (2\right )^{4} x^{5}}{\ln \left (6\right )}}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(67\) |
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
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Time = 8.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {- x - 3}{x^{5} \log {\left (2 \right )}^{4} + 8 x^{3} \log {\left (2 \right )}^{2} + 15 x - \log {\left (6 \right )}} \]
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Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
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Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x+3}{{\ln \left (2\right )}^4\,x^5+8\,{\ln \left (2\right )}^2\,x^3+15\,x-\ln \left (6\right )} \]
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