Optimal. Leaf size=28 \[ -\frac {1}{2} \log \left (x^2+1\right )+\log (x)-\frac {1}{2} \tan ^{-1}(x)^2-\frac {\tan ^{-1}(x)}{x} \]
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Rubi [A] time = 0.06, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {4918, 4852, 266, 36, 29, 31, 4884} \[ -\frac {1}{2} \log \left (x^2+1\right )+\log (x)-\frac {1}{2} \tan ^{-1}(x)^2-\frac {\tan ^{-1}(x)}{x} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 4852
Rule 4884
Rule 4918
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx &=\int \frac {\tan ^{-1}(x)}{x^2} \, dx-\int \frac {\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\int \frac {1}{x \left (1+x^2\right )} \, dx\\ &=-\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )\\ &=-\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\log (x)-\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.01, size = 28, normalized size = 1.00 \[ -\frac {1}{2} \log \left (x^2+1\right )+\log (x)-\frac {1}{2} \tan ^{-1}(x)^2-\frac {\tan ^{-1}(x)}{x} \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 0.93, size = 29, normalized size = 1.04 \[ -\frac {x \arctan \relax (x)^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \relax (x) + 2 \, \arctan \relax (x)}{2 \, x} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \relax (x)}{{\left (x^{2} + 1\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.40, size = 25, normalized size = 0.89
method | result | size |
default | \(-\frac {\arctan \relax (x )}{x}-\frac {\arctan \relax (x )^{2}}{2}+\ln \relax (x )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(25\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}-\frac {\left (\ln \left (-i x +1\right ) x -2 i\right ) \ln \left (i x +1\right )}{4 x}-\frac {-x \ln \left (-i x +1\right )^{2}-8 x \ln \relax (x )+4 \ln \left (x^{2}+1\right ) x +4 i \ln \left (-i x +1\right )}{8 x}\) | \(79\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.99, size = 27, normalized size = 0.96 \[ -{\left (\frac {1}{x} + \arctan \relax (x)\right )} \arctan \relax (x) + \frac {1}{2} \, \arctan \relax (x)^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.09, size = 24, normalized size = 0.86 \[ \ln \relax (x)-\frac {\ln \left (x^2+1\right )}{2}-\frac {\mathrm {atan}\relax (x)}{x}-\frac {{\mathrm {atan}\relax (x)}^2}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.57, size = 22, normalized size = 0.79 \[ \log {\relax (x )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\relax (x )}}{2} - \frac {\operatorname {atan}{\relax (x )}}{x} \]
Verification of antiderivative is not currently implemented for this CAS.
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