Integrand size = 18, antiderivative size = 43 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=-\frac {18-7 x}{20 \left (2+4 x-3 x^2\right )}-\frac {7 \text {arctanh}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}} \]
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Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {652, 632, 212} \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=-\frac {7 \text {arctanh}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}}-\frac {18-7 x}{20 \left (-3 x^2+4 x+2\right )} \]
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Rule 212
Rule 632
Rule 652
Rubi steps \begin{align*} \text {integral}& = -\frac {18-7 x}{20 \left (2+4 x-3 x^2\right )}-\frac {7}{20} \int \frac {1}{-2-4 x+3 x^2} \, dx \\ & = -\frac {18-7 x}{20 \left (2+4 x-3 x^2\right )}+\frac {7}{10} \text {Subst}\left (\int \frac {1}{40-x^2} \, dx,x,-4+6 x\right ) \\ & = -\frac {18-7 x}{20 \left (2+4 x-3 x^2\right )}-\frac {7 \text {arctanh}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.44 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=\frac {18-7 x}{20 \left (-2-4 x+3 x^2\right )}-\frac {7 \log \left (2+\sqrt {10}-3 x\right )}{40 \sqrt {10}}+\frac {7 \log \left (-2+\sqrt {10}+3 x\right )}{40 \sqrt {10}} \]
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Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {14 x -36}{40 \left (3 x^{2}-4 x -2\right )}+\frac {7 \sqrt {10}\, \operatorname {arctanh}\left (\frac {\left (6 x -4\right ) \sqrt {10}}{20}\right )}{200}\) | \(37\) |
risch | \(\frac {-\frac {7 x}{60}+\frac {3}{10}}{x^{2}-\frac {4}{3} x -\frac {2}{3}}+\frac {7 \sqrt {10}\, \ln \left (3 x -2+\sqrt {10}\right )}{400}-\frac {7 \sqrt {10}\, \ln \left (3 x -2-\sqrt {10}\right )}{400}\) | \(48\) |
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Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.58 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=\frac {7 \, \sqrt {10} {\left (3 \, x^{2} - 4 \, x - 2\right )} \log \left (\frac {9 \, x^{2} + 2 \, \sqrt {10} {\left (3 \, x - 2\right )} - 12 \, x + 14}{3 \, x^{2} - 4 \, x - 2}\right ) - 140 \, x + 360}{400 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.35 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=- \frac {7 x - 18}{60 x^{2} - 80 x - 40} + \frac {7 \sqrt {10} \log {\left (x - \frac {2}{3} + \frac {\sqrt {10}}{3} \right )}}{400} - \frac {7 \sqrt {10} \log {\left (x - \frac {\sqrt {10}}{3} - \frac {2}{3} \right )}}{400} \]
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Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=-\frac {7}{400} \, \sqrt {10} \log \left (\frac {3 \, x - \sqrt {10} - 2}{3 \, x + \sqrt {10} - 2}\right ) - \frac {7 \, x - 18}{20 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=-\frac {7}{400} \, \sqrt {10} \log \left (\frac {{\left | 6 \, x - 2 \, \sqrt {10} - 4 \right |}}{{\left | 6 \, x + 2 \, \sqrt {10} - 4 \right |}}\right ) - \frac {7 \, x - 18}{20 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {5-4 x}{\left (-2-4 x+3 x^2\right )^2} \, dx=\frac {7\,\sqrt {10}\,\mathrm {atanh}\left (\sqrt {10}\,\left (\frac {3\,x}{10}-\frac {1}{5}\right )\right )}{200}+\frac {\frac {7\,x}{60}-\frac {3}{10}}{-x^2+\frac {4\,x}{3}+\frac {2}{3}} \]
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