Optimal. Leaf size=89 \[ -\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3}{4} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}+i \tan ^{-1}(x)^2+2 \tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )+i \text {Li}_2\left (1-\frac {2}{1+i x}\right ) \]
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Rubi [A]
time = 0.17, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {5084, 5036,
4946, 327, 209, 5040, 4964, 2449, 2352, 5050, 205} \begin {gather*} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {1}{2} x^2 \text {ArcTan}(x)-\frac {\text {ArcTan}(x)}{2 \left (x^2+1\right )}+i \text {ArcTan}(x)^2+\frac {3 \text {ArcTan}(x)}{4}+2 \text {ArcTan}(x) \log \left (\frac {2}{1+i x}\right )+\frac {x}{4 \left (x^2+1\right )}-\frac {x}{2} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 209
Rule 327
Rule 2352
Rule 2449
Rule 4946
Rule 4964
Rule 5036
Rule 5040
Rule 5050
Rule 5084
Rubi steps
\begin {align*} \int \frac {x^5 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\int \frac {x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx+\int \frac {x^3 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\int x \tan ^{-1}(x) \, dx+\int \frac {x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx-2 \int \frac {x \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^2} \, dx-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\int \frac {\tan ^{-1}(x)}{i-x} \, dx\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\frac {1}{2} \int \frac {1}{1+x^2} \, dx-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )+\int \frac {\log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3}{4} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i x}\right )\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3}{4} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1+i x}\right )\right )\\ \end {align*}
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Mathematica [A]
time = 0.13, size = 70, normalized size = 0.79 \begin {gather*} \frac {1}{8} \left (-4 x+4 \left (1+x^2\right ) \tan ^{-1}(x)-8 i \tan ^{-1}(x)^2-2 \tan ^{-1}(x) \cos \left (2 \tan ^{-1}(x)\right )+16 \tan ^{-1}(x) \log \left (1+e^{2 i \tan ^{-1}(x)}\right )-8 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(x)}\right )+\sin \left (2 \tan ^{-1}(x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 149, normalized size = 1.67
method | result | size |
default | \(\frac {x^{2} \arctan \left (x \right )}{2}-\frac {\arctan \left (x \right )}{2 \left (x^{2}+1\right )}-\arctan \left (x \right ) \ln \left (x^{2}+1\right )-\frac {x}{2}+\frac {x}{4 x^{2}+4}+\frac {3 \arctan \left (x \right )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{2}+\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i \ln \left (x -i\right )^{2}}{4}+\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{2}-\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{2}-\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{2}-\frac {i \ln \left (x +i\right )^{2}}{4}\) | \(149\) |
risch | \(-\frac {x}{2}+\frac {5 \arctan \left (x \right )}{8}+\frac {i x^{2} \ln \left (-i x +1\right )}{4}-\frac {i \ln \left (i x +1\right )}{16 \left (i x -1\right )}+\frac {i \ln \left (-i x +1\right )}{-16 i x -16}+\frac {i}{8 i x +8}+\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{2}+\frac {i \ln \left (i x +1\right )}{8 i x +8}+\frac {i \ln \left (i x +1\right )^{2}}{4}-\frac {i \ln \left (-i x +1\right )}{8 \left (-i x +1\right )}-\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{2}-\frac {i}{8 \left (-i x +1\right )}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (-i x +1\right )^{2}}{4}-\frac {i x^{2} \ln \left (i x +1\right )}{4}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{2}+\frac {\ln \left (-i x +1\right ) x}{-16 i x -16}+\frac {\ln \left (i x +1\right ) x}{16 i x -16}\) | \(243\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RecursionError} \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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\mathrm {atan}\left (x\right )}{{\left (x^2+1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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