Integrand size = 47, antiderivative size = 24 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=\log (2)+x \left (4-80 \log \left (\frac {5 \left (x+x^2 (4+x)\right )}{x^2}\right )\right ) \]
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Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6860, 1671, 646, 31, 2603} \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=4 x-80 x \log \left (5 \left (x+\frac {1}{x}+4\right )\right ) \]
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Rule 31
Rule 646
Rule 1671
Rule 2603
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4 \left (-21-4 x+19 x^2\right )}{1+4 x+x^2}-80 \log \left (5 \left (4+\frac {1}{x}+x\right )\right )\right ) \, dx \\ & = -\left (4 \int \frac {-21-4 x+19 x^2}{1+4 x+x^2} \, dx\right )-80 \int \log \left (5 \left (4+\frac {1}{x}+x\right )\right ) \, dx \\ & = -80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right )-4 \int \left (19-\frac {40 (1+2 x)}{1+4 x+x^2}\right ) \, dx+80 \int \frac {-1+x^2}{1+4 x+x^2} \, dx \\ & = -76 x-80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right )+80 \int \left (1-\frac {2 (1+2 x)}{1+4 x+x^2}\right ) \, dx+160 \int \frac {1+2 x}{1+4 x+x^2} \, dx \\ & = 4 x-80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right )-160 \int \frac {1+2 x}{1+4 x+x^2} \, dx+\left (80 \left (2-\sqrt {3}\right )\right ) \int \frac {1}{2-\sqrt {3}+x} \, dx+\left (80 \left (2+\sqrt {3}\right )\right ) \int \frac {1}{2+\sqrt {3}+x} \, dx \\ & = 4 x+80 \left (2-\sqrt {3}\right ) \log \left (2-\sqrt {3}+x\right )+80 \left (2+\sqrt {3}\right ) \log \left (2+\sqrt {3}+x\right )-80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right )-\left (80 \left (2-\sqrt {3}\right )\right ) \int \frac {1}{2-\sqrt {3}+x} \, dx-\left (80 \left (2+\sqrt {3}\right )\right ) \int \frac {1}{2+\sqrt {3}+x} \, dx \\ & = 4 x-80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right ) \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=4 x-80 x \log \left (5 \left (4+\frac {1}{x}+x\right )\right ) \]
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Time = 0.48 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
norman | \(4 x -80 x \ln \left (\frac {5 x^{2}+20 x +5}{x}\right )\) | \(23\) |
risch | \(4 x -80 x \ln \left (\frac {5 x^{2}+20 x +5}{x}\right )\) | \(23\) |
parallelrisch | \(-32-80 x \ln \left (\frac {5 x^{2}+20 x +5}{x}\right )+4 x\) | \(23\) |
default | \(4 x -80 x \ln \left (5\right )-80 x \ln \left (\frac {x^{2}+4 x +1}{x}\right )\) | \(26\) |
parts | \(4 x -80 x \ln \left (5\right )-80 x \ln \left (\frac {x^{2}+4 x +1}{x}\right )\) | \(26\) |
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=-80 \, x \log \left (\frac {5 \, {\left (x^{2} + 4 \, x + 1\right )}}{x}\right ) + 4 \, x \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=- 80 x \log {\left (\frac {5 x^{2} + 20 x + 5}{x} \right )} + 4 x \]
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Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=-80 \, x {\left (\log \left (5\right ) - 1\right )} - 80 \, {\left (x + 2\right )} \log \left (x^{2} + 4 \, x + 1\right ) + 80 \, x \log \left (x\right ) - 76 \, x + 160 \, \log \left (x^{2} + 4 \, x + 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=-80 \, x \log \left (\frac {5 \, {\left (x^{2} + 4 \, x + 1\right )}}{x}\right ) + 4 \, x \]
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Time = 11.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {84+16 x-76 x^2+\left (-80-320 x-80 x^2\right ) \log \left (\frac {5+20 x+5 x^2}{x}\right )}{1+4 x+x^2} \, dx=4\,x-80\,x\,\ln \left (\frac {5\,x^2+20\,x+5}{x}\right ) \]
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