Optimal. Leaf size=79 \[ -\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right )-\frac {x}{4 \left (x^2+1\right )}+\frac {\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\frac {1}{4} \tan ^{-1}(x)-\log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]
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Rubi [A] time = 0.11, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {4964, 4920, 4854, 2402, 2315, 4930, 199, 203} \[ -\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )-\frac {x}{4 \left (x^2+1\right )}+\frac {\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\frac {1}{4} \tan ^{-1}(x)-\log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 2315
Rule 2402
Rule 4854
Rule 4920
Rule 4930
Rule 4964
Rubi steps
\begin {align*} \int \frac {x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\int \frac {x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx+\int \frac {x \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx-\int \frac {\tan ^{-1}(x)}{i-x} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\int \frac {\log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-i \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i x}\right )\\ &=-\frac {x}{4 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\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 )\\ \end {align*}
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Mathematica [A] time = 0.08, size = 64, normalized size = 0.81 \[ \frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \tan ^{-1}(x)}\right )+\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (1+e^{2 i \tan ^{-1}(x)}\right )-\frac {1}{8} \sin \left (2 \tan ^{-1}(x)\right )+\frac {1}{4} \tan ^{-1}(x) \cos \left (2 \tan ^{-1}(x)\right ) \]
Warning: Unable to verify antiderivative.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \arctan \relax (x)}{x^{4} + 2 \, x^{2} + 1}, x\right ) \]
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 \[ \int \frac {x^{3} \arctan \relax (x)}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.35, size = 139, normalized size = 1.76
method | result | size |
default | \(\frac {\arctan \relax (x ) \ln \left (x^{2}+1\right )}{2}+\frac {\arctan \relax (x )}{2 x^{2}+2}-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\arctan \relax (x )}{4}+\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right )^{2}}{8}-\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{4}+\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (x +i\right )^{2}}{8}\) | \(139\) |
risch | \(\frac {i \ln \left (-i x +1\right )}{-8 i x +8}+\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}-\frac {i}{8 \left (i x +1\right )}-\frac {i \ln \left (i x +1\right )^{2}}{8}-\frac {i \ln \left (i x +1\right )}{8 \left (i x +1\right )}-\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {\ln \left (-i x +1\right ) x}{16 \left (-i x -1\right )}+\frac {i \ln \left (i x +1\right )}{16 i x -16}-\frac {\arctan \relax (x )}{8}-\frac {i \ln \left (-i x +1\right )}{16 \left (-i x -1\right )}+\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{4}+\frac {i}{-8 i x +8}+\frac {i \ln \left (-i x +1\right )^{2}}{8}-\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {\ln \left (i x +1\right ) x}{16 \left (i x -1\right )}\) | \(214\) |
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 \[ \int \frac {x^{3} \arctan \relax (x)}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \]
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 \[ \int \frac {x^3\,\mathrm {atan}\relax (x)}{{\left (x^2+1\right )}^2} \,d x \]
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 \[ \text {Exception raised: RecursionError} \]
Verification of antiderivative is not currently implemented for this CAS.
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