Integrand size = 67, antiderivative size = 26 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {x (5+25 x)}{5 \left (\frac {3}{5}+x \log (3)-4 \log ^2(x)\right )} \]
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\[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (3+30 x+25 x^2 \log (3)+40 (1+5 x) \log (x)-20 (1+10 x) \log ^2(x)\right )}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx \\ & = 5 \int \frac {3+30 x+25 x^2 \log (3)+40 (1+5 x) \log (x)-20 (1+10 x) \log ^2(x)}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx \\ & = 5 \int \left (-\frac {5 (1+5 x) (x \log (3)-8 \log (x))}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2}+\frac {1+10 x}{3+5 x \log (3)-20 \log ^2(x)}\right ) \, dx \\ & = 5 \int \frac {1+10 x}{3+5 x \log (3)-20 \log ^2(x)} \, dx-25 \int \frac {(1+5 x) (x \log (3)-8 \log (x))}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx \\ & = 5 \int \left (\frac {1}{3+5 x \log (3)-20 \log ^2(x)}+\frac {10 x}{3+5 x \log (3)-20 \log ^2(x)}\right ) \, dx-25 \int \left (\frac {x \log (3)-8 \log (x)}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2}+\frac {5 x (x \log (3)-8 \log (x))}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2}\right ) \, dx \\ & = 5 \int \frac {1}{3+5 x \log (3)-20 \log ^2(x)} \, dx-25 \int \frac {x \log (3)-8 \log (x)}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx+50 \int \frac {x}{3+5 x \log (3)-20 \log ^2(x)} \, dx-125 \int \frac {x (x \log (3)-8 \log (x))}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx \\ & = 5 \int \frac {1}{3+5 x \log (3)-20 \log ^2(x)} \, dx-25 \int \left (\frac {x \log (3)}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2}-\frac {8 \log (x)}{\left (-3-5 x \log (3)+20 \log ^2(x)\right )^2}\right ) \, dx+50 \int \frac {x}{3+5 x \log (3)-20 \log ^2(x)} \, dx-125 \int \left (\frac {x^2 \log (3)}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2}-\frac {8 x \log (x)}{\left (-3-5 x \log (3)+20 \log ^2(x)\right )^2}\right ) \, dx \\ & = 5 \int \frac {1}{3+5 x \log (3)-20 \log ^2(x)} \, dx+50 \int \frac {x}{3+5 x \log (3)-20 \log ^2(x)} \, dx+200 \int \frac {\log (x)}{\left (-3-5 x \log (3)+20 \log ^2(x)\right )^2} \, dx+1000 \int \frac {x \log (x)}{\left (-3-5 x \log (3)+20 \log ^2(x)\right )^2} \, dx-(25 \log (3)) \int \frac {x}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx-(125 \log (3)) \int \frac {x^2}{\left (3+5 x \log (3)-20 \log ^2(x)\right )^2} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 x (1+5 x)}{3+5 x \log (3)-20 \log ^2(x)} \]
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Time = 0.94 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {5 \left (1+5 x \right ) x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(24\) |
default | \(\frac {25 x^{2}+5 x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(25\) |
norman | \(\frac {25 x^{2}+5 x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(26\) |
parallelrisch | \(\frac {75 x^{2}+15 x}{15 x \ln \left (3\right )-60 \ln \left (x \right )^{2}+9}\) | \(27\) |
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \]
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Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {- 25 x^{2} - 5 x}{- 5 x \log {\left (3 \right )} + 20 \log {\left (x \right )}^{2} - 3} \]
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Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \]
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Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \]
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Timed out. \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\int \frac {150\,x+\ln \left (x\right )\,\left (1000\,x+200\right )+125\,x^2\,\ln \left (3\right )-{\ln \left (x\right )}^2\,\left (1000\,x+100\right )+15}{25\,x^2\,{\ln \left (3\right )}^2+30\,x\,\ln \left (3\right )+400\,{\ln \left (x\right )}^4-{\ln \left (x\right )}^2\,\left (200\,x\,\ln \left (3\right )+120\right )+9} \,d x \]
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