Integrand size = 26, antiderivative size = 60 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=-\frac {79}{273 (5+x)}+\frac {451 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2793 \sqrt {3}}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (5+x)}{24843}-\frac {481 \log \left (1+x+x^2\right )}{5586} \]
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Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6860, 648, 632, 210, 642} \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {451 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{2793 \sqrt {3}}-\frac {481 \log \left (x^2+x+1\right )}{5586}-\frac {79}{273 (x+5)}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (x+5)}{24843} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {79}{273 (5+x)^2}+\frac {2731}{24843 (5+x)}+\frac {400}{3211 (-3+2 x)}+\frac {-15-481 x}{2793 \left (1+x+x^2\right )}\right ) \, dx \\ & = -\frac {79}{273 (5+x)}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (5+x)}{24843}+\frac {\int \frac {-15-481 x}{1+x+x^2} \, dx}{2793} \\ & = -\frac {79}{273 (5+x)}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (5+x)}{24843}+\frac {451 \int \frac {1}{1+x+x^2} \, dx}{5586}-\frac {481 \int \frac {1+2 x}{1+x+x^2} \, dx}{5586} \\ & = -\frac {79}{273 (5+x)}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (5+x)}{24843}-\frac {481 \log \left (1+x+x^2\right )}{5586}-\frac {451 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )}{2793} \\ & = -\frac {79}{273 (5+x)}+\frac {451 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2793 \sqrt {3}}+\frac {200 \log (3-2 x)}{3211}+\frac {2731 \log (5+x)}{24843}-\frac {481 \log \left (1+x+x^2\right )}{5586} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {-\frac {819546}{5+x}+152438 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+176400 \log (3-2 x)+311334 \log (5+x)-243867 \log \left (1+x+x^2\right )}{2832102} \]
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Time = 0.38 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {79}{273 \left (5+x \right )}+\frac {2731 \ln \left (5+x \right )}{24843}-\frac {481 \ln \left (x^{2}+x +1\right )}{5586}+\frac {451 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{8379}+\frac {200 \ln \left (2 x -3\right )}{3211}\) | \(48\) |
risch | \(-\frac {79}{273 \left (5+x \right )}+\frac {2731 \ln \left (5+x \right )}{24843}-\frac {481 \ln \left (203401 x^{2}+203401 x +203401\right )}{5586}+\frac {451 \sqrt {3}\, \arctan \left (\frac {2 \left (451 x +\frac {451}{2}\right ) \sqrt {3}}{1353}\right )}{8379}+\frac {200 \ln \left (2 x -3\right )}{3211}\) | \(52\) |
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Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {152438 \, \sqrt {3} {\left (x + 5\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 243867 \, {\left (x + 5\right )} \log \left (x^{2} + x + 1\right ) + 176400 \, {\left (x + 5\right )} \log \left (2 \, x - 3\right ) + 311334 \, {\left (x + 5\right )} \log \left (x + 5\right ) - 819546}{2832102 \, {\left (x + 5\right )}} \]
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Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {200 \log {\left (x - \frac {3}{2} \right )}}{3211} + \frac {2731 \log {\left (x + 5 \right )}}{24843} - \frac {481 \log {\left (x^{2} + x + 1 \right )}}{5586} + \frac {451 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{8379} - \frac {79}{273 x + 1365} \]
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Time = 0.37 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {451}{8379} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {79}{273 \, {\left (x + 5\right )}} - \frac {481}{5586} \, \log \left (x^{2} + x + 1\right ) + \frac {200}{3211} \, \log \left (2 \, x - 3\right ) + \frac {2731}{24843} \, \log \left (x + 5\right ) \]
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Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {451}{8379} \, \sqrt {3} \arctan \left (-\sqrt {3} {\left (\frac {14}{x + 5} - 3\right )}\right ) - \frac {79}{273 \, {\left (x + 5\right )}} - \frac {481}{5586} \, \log \left (-\frac {9}{x + 5} + \frac {21}{{\left (x + 5\right )}^{2}} + 1\right ) + \frac {200}{3211} \, \log \left ({\left | -\frac {13}{x + 5} + 2 \right |}\right ) \]
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Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {1+16 x}{(5+x)^2 (-3+2 x) \left (1+x+x^2\right )} \, dx=\frac {200\,\ln \left (x-\frac {3}{2}\right )}{3211}+\frac {2731\,\ln \left (x+5\right )}{24843}-\frac {79}{273\,\left (x+5\right )}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {481}{5586}+\frac {\sqrt {3}\,451{}\mathrm {i}}{16758}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {481}{5586}+\frac {\sqrt {3}\,451{}\mathrm {i}}{16758}\right ) \]
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