Integrand size = 185, antiderivative size = 29 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=x \left (4+\frac {x (5+x)}{-3+x+3 x^2-\log (x)+\log (5 (3+x))}\right ) \]
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\[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=\int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+4 (3+x) \log ^2(x)+3 \left (-24+10 x+33 x^2+9 x^3\right ) \log (5 (3+x))+4 (3+x) \log ^2(5 (3+x))-(3+x) \log (x) \left (3 \left (-8+6 x+9 x^2\right )+8 \log (5 (3+x))\right )}{(3+x) \left (3-x-3 x^2+\log (x)-\log (5 (3+x))\right )^2} \, dx \\ & = \int \left (4-\frac {x \left (-15+12 x+98 x^2+49 x^3+6 x^4\right )}{(3+x) \left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}+\frac {x (10+3 x)}{-3+x+3 x^2-\log (x)+\log (5 (3+x))}\right ) \, dx \\ & = 4 x-\int \frac {x \left (-15+12 x+98 x^2+49 x^3+6 x^4\right )}{(3+x) \left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx+\int \frac {x (10+3 x)}{-3+x+3 x^2-\log (x)+\log (5 (3+x))} \, dx \\ & = 4 x-\int \left (-\frac {6}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}-\frac {3 x}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}+\frac {5 x^2}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}+\frac {31 x^3}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}+\frac {6 x^4}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}+\frac {18}{(3+x) \left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2}\right ) \, dx+\int \left (\frac {10 x}{-3+x+3 x^2-\log (x)+\log (5 (3+x))}+\frac {3 x^2}{-3+x+3 x^2-\log (x)+\log (5 (3+x))}\right ) \, dx \\ & = 4 x+3 \int \frac {x}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx+3 \int \frac {x^2}{-3+x+3 x^2-\log (x)+\log (5 (3+x))} \, dx-5 \int \frac {x^2}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx+6 \int \frac {1}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx-6 \int \frac {x^4}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx+10 \int \frac {x}{-3+x+3 x^2-\log (x)+\log (5 (3+x))} \, dx-18 \int \frac {1}{(3+x) \left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx-31 \int \frac {x^3}{\left (-3+x+3 x^2-\log (x)+\log (5 (3+x))\right )^2} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=4 x+\frac {x^2 (5+x)}{-3+x+3 x^2-\log (x)+\log (5 (3+x))} \]
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Time = 3.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
default | \(4 x +\frac {\left (5+x \right ) x^{2}}{3 x^{2}+\ln \left (5\right )+x +\ln \left (3+x \right )-\ln \left (x \right )-3}\) | \(32\) |
risch | \(4 x +\frac {x^{2} \left (5+x \right )}{\ln \left (5 x +15\right )-3+3 x^{2}-\ln \left (x \right )+x}\) | \(32\) |
parallelrisch | \(\frac {72-36 x -4 x \ln \left (x \right )-24 \ln \left (5 x +15\right )+13 x^{3}-63 x^{2}+24 \ln \left (x \right )+4 \ln \left (5 x +15\right ) x}{\ln \left (5 x +15\right )-3+3 x^{2}-\ln \left (x \right )+x}\) | \(63\) |
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Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=\frac {13 \, x^{3} + 9 \, x^{2} + 4 \, x \log \left (5 \, x + 15\right ) - 4 \, x \log \left (x\right ) - 12 \, x}{3 \, x^{2} + x + \log \left (5 \, x + 15\right ) - \log \left (x\right ) - 3} \]
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Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=4 x + \frac {x^{3} + 5 x^{2}}{3 x^{2} + x - \log {\left (x \right )} + \log {\left (5 x + 15 \right )} - 3} \]
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Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=\frac {13 \, x^{3} + 9 \, x^{2} + 4 \, x {\left (\log \left (5\right ) - 3\right )} + 4 \, x \log \left (x + 3\right ) - 4 \, x \log \left (x\right )}{3 \, x^{2} + x + \log \left (5\right ) + \log \left (x + 3\right ) - \log \left (x\right ) - 3} \]
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Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=4 \, x + \frac {x^{3} + 5 \, x^{2}}{3 \, x^{2} + x + \log \left (5 \, x + 15\right ) - \log \left (x\right ) - 3} \]
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Time = 10.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28 \[ \int \frac {108-111 x-267 x^2+6 x^3+143 x^4+39 x^5+\left (72-30 x-99 x^2-27 x^3\right ) \log (x)+(12+4 x) \log ^2(x)+\left (-72+30 x+99 x^2+27 x^3+(-24-8 x) \log (x)\right ) \log (15+5 x)+(12+4 x) \log ^2(15+5 x)}{27-9 x-57 x^2+x^3+33 x^4+9 x^5+\left (18-20 x^2-6 x^3\right ) \log (x)+(3+x) \log ^2(x)+\left (-18+20 x^2+6 x^3+(-6-2 x) \log (x)\right ) \log (15+5 x)+(3+x) \log ^2(15+5 x)} \, dx=\frac {19\,\ln \left (x\right )-19\,\ln \left (5\,x+15\right )-163\,x+48\,x\,\ln \left (5\,x+15\right )-48\,x\,\ln \left (x\right )+51\,x^2+156\,x^3+57}{12\,\left (x+\ln \left (5\,x+15\right )-\ln \left (x\right )+3\,x^2-3\right )} \]
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