Integrand size = 135, antiderivative size = 26 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=4-\frac {1}{x}+x-\frac {x}{3 (5-x-\log (8+x))} \]
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\[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+3 \left (8+x+8 x^2+x^3\right ) \log ^2(8+x)}{3 x^2 (8+x) (5-x-\log (8+x))^2} \, dx \\ & = \frac {1}{3} \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+3 \left (8+x+8 x^2+x^3\right ) \log ^2(8+x)}{x^2 (8+x) (5-x-\log (8+x))^2} \, dx \\ & = \frac {1}{3} \int \left (\frac {3 \left (1+x^2\right )}{x^2}-\frac {x (9+x)}{(8+x) (-5+x+\log (8+x))^2}+\frac {1}{-5+x+\log (8+x)}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {x (9+x)}{(8+x) (-5+x+\log (8+x))^2} \, dx\right )+\frac {1}{3} \int \frac {1}{-5+x+\log (8+x)} \, dx+\int \frac {1+x^2}{x^2} \, dx \\ & = \frac {1}{3} \int \frac {1}{-5+x+\log (8+x)} \, dx-\frac {1}{3} \int \left (\frac {1}{(-5+x+\log (8+x))^2}+\frac {x}{(-5+x+\log (8+x))^2}-\frac {8}{(8+x) (-5+x+\log (8+x))^2}\right ) \, dx+\int \left (1+\frac {1}{x^2}\right ) \, dx \\ & = -\frac {1}{x}+x-\frac {1}{3} \int \frac {1}{(-5+x+\log (8+x))^2} \, dx-\frac {1}{3} \int \frac {x}{(-5+x+\log (8+x))^2} \, dx+\frac {1}{3} \int \frac {1}{-5+x+\log (8+x)} \, dx+\frac {8}{3} \int \frac {1}{(8+x) (-5+x+\log (8+x))^2} \, dx \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {1}{3} \left (-\frac {3}{x}+3 x+\frac {x}{-5+x+\log (8+x)}\right ) \]
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Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x^{2}-1}{x}+\frac {x}{3 \ln \left (x +8\right )+3 x -15}\) | \(23\) |
norman | \(\frac {5+x^{3}-\frac {73 x}{3}+x^{2} \ln \left (x +8\right )+\frac {14 x \ln \left (x +8\right )}{3}-\ln \left (x +8\right )}{x \left (\ln \left (x +8\right )+x -5\right )}\) | \(43\) |
derivativedivides | \(\frac {-211 x +911-195 \ln \left (x +8\right )-38 \left (x +8\right )^{2}+3 \left (x +8\right )^{3}+3 \ln \left (x +8\right ) \left (x +8\right )^{2}}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(51\) |
default | \(\frac {-211 x +911-195 \ln \left (x +8\right )-38 \left (x +8\right )^{2}+3 \left (x +8\right )^{3}+3 \ln \left (x +8\right ) \left (x +8\right )^{2}}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(51\) |
parallelrisch | \(\frac {15+3 x^{3}+3 x^{2} \ln \left (x +8\right )-62 x^{2}-48 x \ln \left (x +8\right )+237 x -3 \ln \left (x +8\right )}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(52\) |
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {3 \, x^{3} - 14 \, x^{2} + 3 \, {\left (x^{2} - 1\right )} \log \left (x + 8\right ) - 3 \, x + 15}{3 \, {\left (x^{2} + x \log \left (x + 8\right ) - 5 \, x\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x + \frac {x}{3 x + 3 \log {\left (x + 8 \right )} - 15} - \frac {1}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {3 \, x^{3} - 14 \, x^{2} + 3 \, {\left (x^{2} - 1\right )} \log \left (x + 8\right ) - 3 \, x + 15}{3 \, {\left (x^{2} + x \log \left (x + 8\right ) - 5 \, x\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x + \frac {x}{3 \, {\left (x + \log \left (x + 8\right ) - 5\right )}} - \frac {1}{x} \]
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Time = 12.44 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x-\frac {\ln \left (x+8\right )+x\,\left (\frac {\ln \left (x+8\right )}{3}-\frac {2}{3}\right )-5}{x\,\left (x+\ln \left (x+8\right )-5\right )} \]
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