Optimal. Leaf size=79 \[ -\frac {1}{32 \left (1+x^2\right )^2}+\frac {5}{32 \left (1+x^2\right )}+\frac {x^3 \tan ^{-1}(x)}{8 \left (1+x^2\right )^2}+\frac {3 x \tan ^{-1}(x)}{16 \left (1+x^2\right )}-\frac {3}{32} \tan ^{-1}(x)^2+\frac {x^4 \tan ^{-1}(x)^2}{4 \left (1+x^2\right )^2} \]
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Rubi [A]
time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.04, number of steps
used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5064, 5058,
5054, 5004} \begin {gather*} \frac {3 x \text {ArcTan}(x)}{16 \left (x^2+1\right )}+\frac {x^4 \text {ArcTan}(x)^2}{4 \left (x^2+1\right )^2}+\frac {x^3 \text {ArcTan}(x)}{8 \left (x^2+1\right )^2}-\frac {3 \text {ArcTan}(x)^2}{32}+\frac {3}{32 \left (x^2+1\right )}-\frac {x^4}{32 \left (x^2+1\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 5004
Rule 5054
Rule 5058
Rule 5064
Rubi steps
\begin {align*} \int \frac {x^3 \tan ^{-1}(x)^2}{\left (1+x^2\right )^3} \, dx &=\frac {x^4 \tan ^{-1}(x)^2}{4 \left (1+x^2\right )^2}-\frac {1}{2} \int \frac {x^4 \tan ^{-1}(x)}{\left (1+x^2\right )^3} \, dx\\ &=-\frac {x^4}{32 \left (1+x^2\right )^2}+\frac {x^3 \tan ^{-1}(x)}{8 \left (1+x^2\right )^2}+\frac {x^4 \tan ^{-1}(x)^2}{4 \left (1+x^2\right )^2}-\frac {3}{8} \int \frac {x^2 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {x^4}{32 \left (1+x^2\right )^2}+\frac {3}{32 \left (1+x^2\right )}+\frac {x^3 \tan ^{-1}(x)}{8 \left (1+x^2\right )^2}+\frac {3 x \tan ^{-1}(x)}{16 \left (1+x^2\right )}+\frac {x^4 \tan ^{-1}(x)^2}{4 \left (1+x^2\right )^2}-\frac {3}{16} \int \frac {\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {x^4}{32 \left (1+x^2\right )^2}+\frac {3}{32 \left (1+x^2\right )}+\frac {x^3 \tan ^{-1}(x)}{8 \left (1+x^2\right )^2}+\frac {3 x \tan ^{-1}(x)}{16 \left (1+x^2\right )}-\frac {3}{32} \tan ^{-1}(x)^2+\frac {x^4 \tan ^{-1}(x)^2}{4 \left (1+x^2\right )^2}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 47, normalized size = 0.59 \begin {gather*} \frac {4+5 x^2+2 x \left (3+5 x^2\right ) \tan ^{-1}(x)+\left (-3-6 x^2+5 x^4\right ) \tan ^{-1}(x)^2}{32 \left (1+x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.14, size = 78, normalized size = 0.99
method | result | size |
default | \(\frac {\arctan \left (x \right )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (x \right )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \left (x \right )^{2}}{32}-\frac {1}{32 \left (x^{2}+1\right )^{2}}+\frac {5}{32 \left (x^{2}+1\right )}\) | \(78\) |
risch | \(-\frac {\left (5 x^{4}-6 x^{2}-3\right ) \ln \left (i x +1\right )^{2}}{128 \left (x^{2}+1\right )^{2}}+\frac {\left (-6 x^{2} \ln \left (-i x +1\right )-3 \ln \left (-i x +1\right )+5 x^{4} \ln \left (-i x +1\right )-10 i x^{3}-6 i x \right ) \ln \left (i x +1\right )}{64 \left (x +i\right )^{2} \left (x -i\right )^{2}}-\frac {5 x^{4} \ln \left (-i x +1\right )^{2}-6 x^{2} \ln \left (-i x +1\right )^{2}-3 \ln \left (-i x +1\right )^{2}-20 i x^{3} \ln \left (-i x +1\right )-12 i x \ln \left (-i x +1\right )-20 x^{2}-16}{128 \left (x +i\right )^{2} \left (x -i\right )^{2}}\) | \(181\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 3.22, size = 94, normalized size = 1.19 \begin {gather*} \frac {1}{16} \, {\left (\frac {5 \, x^{3} + 3 \, x}{x^{4} + 2 \, x^{2} + 1} + 5 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {5 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} - 5 \, x^{2} - 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.01, size = 51, normalized size = 0.65 \begin {gather*} \frac {{\left (5 \, x^{4} - 6 \, x^{2} - 3\right )} \arctan \left (x\right )^{2} + 5 \, x^{2} + 2 \, {\left (5 \, x^{3} + 3 \, x\right )} \arctan \left (x\right ) + 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \operatorname {atan}^{2}{\left (x \right )}}{\left (x^{2} + 1\right )^{3}}\, dx \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 = 0.36, size = 56, normalized size = 0.71 \begin {gather*} -\frac {-5\,x^4\,{\mathrm {atan}\left (x\right )}^2+4\,x^4-10\,x^3\,\mathrm {atan}\left (x\right )+6\,x^2\,{\mathrm {atan}\left (x\right )}^2+3\,x^2-6\,x\,\mathrm {atan}\left (x\right )+3\,{\mathrm {atan}\left (x\right )}^2}{32\,{\left (x^2+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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