Optimal. Leaf size=28 \[ -\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]
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Rubi [A]
time = 0.04, 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 = {5038, 4946,
272, 36, 29, 31, 5004} \begin {gather*} -\frac {1}{2} \text {ArcTan}(x)^2-\frac {\text {ArcTan}(x)}{x}-\frac {1}{2} \log \left (x^2+1\right )+\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 272
Rule 4946
Rule 5004
Rule 5038
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} \text {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} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \text {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 \begin {gather*} -\frac {\tan ^{-1}(x)}{x}-\frac {1}{2} \tan ^{-1}(x)^2+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 25, normalized size = 0.89
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(25\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}-\frac {\left (\ln \left (-i x +1\right ) x -2 i\right ) \ln \left (i x +1\right )}{4 x}-\frac {-x \ln \left (-i x +1\right )^{2}+4 i \ln \left (-i x +1\right )-8 x \ln \left (x \right )+4 x \ln \left (x^{2}+1\right )}{8 x}\) | \(79\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 3.09, size = 27, normalized size = 0.96 \begin {gather*} -{\left (\frac {1}{x} + \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.80, size = 29, normalized size = 1.04 \begin {gather*} -\frac {x \arctan \left (x\right )^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \left (x\right ) + 2 \, \arctan \left (x\right )}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.17, size = 22, normalized size = 0.79 \begin {gather*} \log {\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} - \frac {\operatorname {atan}{\left (x \right )}}{x} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \ln \left (x\right )-\frac {\ln \left (x^2+1\right )}{2}-\frac {\mathrm {atan}\left (x\right )}{x}-\frac {{\mathrm {atan}\left (x\right )}^2}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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