Optimal. Leaf size=43 \[ -\frac {1-2 x}{5 \left (x^2+1\right )}-\frac {14}{25} \log \left (x^2+1\right )-\frac {47}{25} \log (2-x)-\frac {46}{25} \tan ^{-1}(x) \]
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Rubi [A] time = 0.07, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1647, 1629, 635, 203, 260} \[ -\frac {1-2 x}{5 \left (x^2+1\right )}-\frac {14}{25} \log \left (x^2+1\right )-\frac {47}{25} \log (2-x)-\frac {46}{25} \tan ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 203
Rule 260
Rule 635
Rule 1629
Rule 1647
Rubi steps
\begin {align*} \int \frac {1-3 x^4}{(-2+x) \left (1+x^2\right )^2} \, dx &=-\frac {1-2 x}{5 \left (1+x^2\right )}-\frac {1}{2} \int \frac {-\frac {18}{5}-\frac {4 x}{5}+6 x^2}{(-2+x) \left (1+x^2\right )} \, dx\\ &=-\frac {1-2 x}{5 \left (1+x^2\right )}-\frac {1}{2} \int \left (\frac {94}{25 (-2+x)}+\frac {4 (23+14 x)}{25 \left (1+x^2\right )}\right ) \, dx\\ &=-\frac {1-2 x}{5 \left (1+x^2\right )}-\frac {47}{25} \log (2-x)-\frac {2}{25} \int \frac {23+14 x}{1+x^2} \, dx\\ &=-\frac {1-2 x}{5 \left (1+x^2\right )}-\frac {47}{25} \log (2-x)-\frac {28}{25} \int \frac {x}{1+x^2} \, dx-\frac {46}{25} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {1-2 x}{5 \left (1+x^2\right )}-\frac {46}{25} \tan ^{-1}(x)-\frac {47}{25} \log (2-x)-\frac {14}{25} \log \left (1+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.02, size = 57, normalized size = 1.33 \[ \frac {2 (x-2)+3}{5 \left ((x-2)^2+4 (x-2)+5\right )}-\frac {14}{25} \log \left ((x-2)^2+4 (x-2)+5\right )-\frac {47}{25} \log (x-2)-\frac {46}{25} \tan ^{-1}(x) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 47, normalized size = 1.09 \[ -\frac {46 \, {\left (x^{2} + 1\right )} \arctan \relax (x) + 14 \, {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 47 \, {\left (x^{2} + 1\right )} \log \left (x - 2\right ) - 10 \, x + 5}{25 \, {\left (x^{2} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 34, normalized size = 0.79 \[ \frac {2 \, x - 1}{5 \, {\left (x^{2} + 1\right )}} - \frac {46}{25} \, \arctan \relax (x) - \frac {14}{25} \, \log \left (x^{2} + 1\right ) - \frac {47}{25} \, \log \left ({\left | x - 2 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 34, normalized size = 0.79 \[ -\frac {46 \arctan \relax (x )}{25}-\frac {47 \ln \left (x -2\right )}{25}-\frac {14 \ln \left (x^{2}+1\right )}{25}-\frac {2 \left (-5 x +\frac {5}{2}\right )}{25 \left (x^{2}+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.10, size = 33, normalized size = 0.77 \[ \frac {2 \, x - 1}{5 \, {\left (x^{2} + 1\right )}} - \frac {46}{25} \, \arctan \relax (x) - \frac {14}{25} \, \log \left (x^{2} + 1\right ) - \frac {47}{25} \, \log \left (x - 2\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.05, size = 38, normalized size = 0.88 \[ \frac {\frac {2\,x}{5}-\frac {1}{5}}{x^2+1}-\frac {47\,\ln \left (x-2\right )}{25}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {14}{25}+\frac {23}{25}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {14}{25}-\frac {23}{25}{}\mathrm {i}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 37, normalized size = 0.86 \[ - \frac {1 - 2 x}{5 x^{2} + 5} - \frac {47 \log {\left (x - 2 \right )}}{25} - \frac {14 \log {\left (x^{2} + 1 \right )}}{25} - \frac {46 \operatorname {atan}{\relax (x )}}{25} \]
Verification of antiderivative is not currently implemented for this CAS.
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