Optimal. Leaf size=33 \[ -\frac {2 x+1}{2 \left (x^2+1\right )}-\frac {1}{2} \log \left (x^2+1\right )+\log (x)-2 \tan ^{-1}(x) \]
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Rubi [A] time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1805, 801, 635, 203, 260} \begin {gather*} -\frac {2 x+1}{2 \left (x^2+1\right )}-\frac {1}{2} \log \left (x^2+1\right )+\log (x)-2 \tan ^{-1}(x) \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 260
Rule 635
Rule 801
Rule 1805
Rubi steps
\begin {align*} \int \frac {1-3 x+2 x^2-x^3}{x \left (1+x^2\right )^2} \, dx &=-\frac {1+2 x}{2 \left (1+x^2\right )}-\frac {1}{2} \int \frac {-2+4 x}{x \left (1+x^2\right )} \, dx\\ &=-\frac {1+2 x}{2 \left (1+x^2\right )}-\frac {1}{2} \int \left (-\frac {2}{x}+\frac {2 (2+x)}{1+x^2}\right ) \, dx\\ &=-\frac {1+2 x}{2 \left (1+x^2\right )}+\log (x)-\int \frac {2+x}{1+x^2} \, dx\\ &=-\frac {1+2 x}{2 \left (1+x^2\right )}+\log (x)-2 \int \frac {1}{1+x^2} \, dx-\int \frac {x}{1+x^2} \, dx\\ &=-\frac {1+2 x}{2 \left (1+x^2\right )}-2 \tan ^{-1}(x)+\log (x)-\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.02, size = 33, normalized size = 1.00 \begin {gather*} \frac {-2 x-1}{2 \left (x^2+1\right )}-\frac {1}{2} \log \left (x^2+1\right )+\log (x)-2 \tan ^{-1}(x) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-3 x+2 x^2-x^3}{x \left (1+x^2\right )^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 1.54, size = 44, normalized size = 1.33 \begin {gather*} -\frac {4 \, {\left (x^{2} + 1\right )} \arctan \relax (x) + {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - 2 \, {\left (x^{2} + 1\right )} \log \relax (x) + 2 \, x + 1}{2 \, {\left (x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 30, normalized size = 0.91 \begin {gather*} -\frac {2 \, x + 1}{2 \, {\left (x^{2} + 1\right )}} - 2 \, \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 28, normalized size = 0.85 \begin {gather*} -2 \arctan \relax (x )+\ln \relax (x )-\frac {\ln \left (x^{2}+1\right )}{2}-\frac {x +\frac {1}{2}}{x^{2}+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.25, size = 29, normalized size = 0.88 \begin {gather*} -\frac {2 \, x + 1}{2 \, {\left (x^{2} + 1\right )}} - 2 \, \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.11, size = 33, normalized size = 1.00 \begin {gather*} \ln \relax (x)-\frac {x+\frac {1}{2}}{x^2+1}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{2}+1{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{2}-\mathrm {i}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 27, normalized size = 0.82 \begin {gather*} - \frac {2 x + 1}{2 x^{2} + 2} + \log {\relax (x )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - 2 \operatorname {atan}{\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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