Integrand size = 26, antiderivative size = 58 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=-\frac {83}{4 (4-x)^2}+\frac {41}{4 (4-x)}-\frac {3}{16} \arctan \left (1-\frac {x}{2}\right )-\frac {45}{16} \log (4-x)+\frac {45}{32} \log \left (8-4 x+x^2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 58, 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 = {1642, 648, 631, 210, 642} \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=-\frac {3}{16} \arctan \left (1-\frac {x}{2}\right )+\frac {45}{32} \log \left (x^2-4 x+8\right )+\frac {41}{4 (4-x)}-\frac {83}{4 (4-x)^2}-\frac {45}{16} \log (4-x) \]
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Rule 210
Rule 631
Rule 642
Rule 648
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {83}{2 (-4+x)^3}+\frac {41}{4 (-4+x)^2}-\frac {45}{16 (-4+x)}+\frac {3 (-28+15 x)}{16 \left (8-4 x+x^2\right )}\right ) \, dx \\ & = -\frac {83}{4 (4-x)^2}+\frac {41}{4 (4-x)}-\frac {45}{16} \log (4-x)+\frac {3}{16} \int \frac {-28+15 x}{8-4 x+x^2} \, dx \\ & = -\frac {83}{4 (4-x)^2}+\frac {41}{4 (4-x)}-\frac {45}{16} \log (4-x)+\frac {3}{8} \int \frac {1}{8-4 x+x^2} \, dx+\frac {45}{32} \int \frac {-4+2 x}{8-4 x+x^2} \, dx \\ & = -\frac {83}{4 (4-x)^2}+\frac {41}{4 (4-x)}-\frac {45}{16} \log (4-x)+\frac {45}{32} \log \left (8-4 x+x^2\right )+\frac {3}{16} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {x}{2}\right ) \\ & = -\frac {83}{4 (4-x)^2}+\frac {41}{4 (4-x)}-\frac {3}{16} \arctan \left (1-\frac {x}{2}\right )-\frac {45}{16} \log (4-x)+\frac {45}{32} \log \left (8-4 x+x^2\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=\frac {1}{32} \left (-\frac {664}{(-4+x)^2}-\frac {328}{-4+x}+6 \arctan \left (\frac {1}{2} (-2+x)\right )-90 \log (-4+x)+45 \log \left (8-4 x+x^2\right )\right ) \]
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Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.66
method | result | size |
risch | \(\frac {-\frac {41 x}{4}+\frac {81}{4}}{\left (x -4\right )^{2}}-\frac {45 \ln \left (x -4\right )}{16}+\frac {45 \ln \left (x^{2}-4 x +8\right )}{32}+\frac {3 \arctan \left (-1+\frac {x}{2}\right )}{16}\) | \(38\) |
default | \(\frac {45 \ln \left (x^{2}-4 x +8\right )}{32}+\frac {3 \arctan \left (-1+\frac {x}{2}\right )}{16}-\frac {83}{4 \left (x -4\right )^{2}}-\frac {41}{4 \left (x -4\right )}-\frac {45 \ln \left (x -4\right )}{16}\) | \(41\) |
parallelrisch | \(-\frac {-96 i \ln \left (x -2+2 i\right )-48 i \ln \left (x -2-2 i\right ) x +96 i \ln \left (x -2-2 i\right )+48 i \ln \left (x -2+2 i\right ) x +180 \ln \left (x -4\right ) x^{2}-90 \ln \left (x -2-2 i\right ) x^{2}-90 \ln \left (x -2+2 i\right ) x^{2}+6 i \ln \left (x -2-2 i\right ) x^{2}-6 i \ln \left (x -2+2 i\right ) x^{2}-1440 \ln \left (x -4\right ) x +720 \ln \left (x -2-2 i\right ) x +720 \ln \left (x -2+2 i\right ) x +81 x^{2}+2880 \ln \left (x -4\right )-1440 \ln \left (x -2-2 i\right )-1440 \ln \left (x -2+2 i\right )+8 x}{64 \left (x -4\right )^{2}}\) | \(157\) |
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Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=\frac {6 \, {\left (x^{2} - 8 \, x + 16\right )} \arctan \left (\frac {1}{2} \, x - 1\right ) + 45 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (x^{2} - 4 \, x + 8\right ) - 90 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (x - 4\right ) - 328 \, x + 648}{32 \, {\left (x^{2} - 8 \, x + 16\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=\frac {81 - 41 x}{4 x^{2} - 32 x + 64} - \frac {45 \log {\left (x - 4 \right )}}{16} + \frac {45 \log {\left (x^{2} - 4 x + 8 \right )}}{32} + \frac {3 \operatorname {atan}{\left (\frac {x}{2} - 1 \right )}}{16} \]
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Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.74 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=-\frac {41 \, x - 81}{4 \, {\left (x^{2} - 8 \, x + 16\right )}} + \frac {3}{16} \, \arctan \left (\frac {1}{2} \, x - 1\right ) + \frac {45}{32} \, \log \left (x^{2} - 4 \, x + 8\right ) - \frac {45}{16} \, \log \left (x - 4\right ) \]
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Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.67 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=-\frac {41 \, x - 81}{4 \, {\left (x - 4\right )}^{2}} + \frac {3}{16} \, \arctan \left (\frac {1}{2} \, x - 1\right ) + \frac {45}{32} \, \log \left (x^{2} - 4 \, x + 8\right ) - \frac {45}{16} \, \log \left ({\left | x - 4 \right |}\right ) \]
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Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int \frac {-20+8 x+5 x^3}{(-4+x)^3 \left (8-4 x+x^2\right )} \, dx=-\frac {45\,\ln \left (x-4\right )}{16}-\frac {\frac {41\,x}{4}-\frac {81}{4}}{x^2-8\,x+16}+\ln \left (x-2-2{}\mathrm {i}\right )\,\left (\frac {45}{32}-\frac {3}{32}{}\mathrm {i}\right )+\ln \left (x-2+2{}\mathrm {i}\right )\,\left (\frac {45}{32}+\frac {3}{32}{}\mathrm {i}\right ) \]
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