Optimal. Leaf size=37 \[ -\frac {x^2}{8 \left (1+x^4\right )^2}+\frac {x^2}{16 \left (1+x^4\right )}+\frac {1}{16} \tan ^{-1}\left (x^2\right ) \]
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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {281, 294, 205,
209} \begin {gather*} \frac {\text {ArcTan}\left (x^2\right )}{16}+\frac {x^2}{16 \left (x^4+1\right )}-\frac {x^2}{8 \left (x^4+1\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 209
Rule 281
Rule 294
Rubi steps
\begin {align*} \int \frac {x^5}{\left (1+x^4\right )^3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac {x^2}{8 \left (1+x^4\right )^2}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac {x^2}{8 \left (1+x^4\right )^2}+\frac {x^2}{16 \left (1+x^4\right )}+\frac {1}{16} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^2\right )\\ &=-\frac {x^2}{8 \left (1+x^4\right )^2}+\frac {x^2}{16 \left (1+x^4\right )}+\frac {1}{16} \tan ^{-1}\left (x^2\right )\\ \end {align*}
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Mathematica [A]
time = 0.01, size = 25, normalized size = 0.68 \begin {gather*} \frac {1}{16} \left (\frac {x^2 \left (-1+x^4\right )}{\left (1+x^4\right )^2}+\tan ^{-1}\left (x^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.05, size = 28, normalized size = 0.76
method | result | size |
meijerg | \(-\frac {x^{2} \left (-3 x^{4}+3\right )}{48 \left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\) | \(27\) |
risch | \(\frac {\frac {1}{16} x^{6}-\frac {1}{16} x^{2}}{\left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\) | \(27\) |
default | \(\frac {\frac {1}{8} x^{6}-\frac {1}{8} x^{2}}{2 \left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 2.78, size = 30, normalized size = 0.81 \begin {gather*} \frac {x^{6} - x^{2}}{16 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}} + \frac {1}{16} \, \arctan \left (x^{2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 38, normalized size = 1.03 \begin {gather*} \frac {x^{6} - x^{2} + {\left (x^{8} + 2 \, x^{4} + 1\right )} \arctan \left (x^{2}\right )}{16 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.05, size = 24, normalized size = 0.65 \begin {gather*} \frac {x^{6} - x^{2}}{16 x^{8} + 32 x^{4} + 16} + \frac {\operatorname {atan}{\left (x^{2} \right )}}{16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.61, size = 40, normalized size = 1.08 \begin {gather*} \frac {x^{2} - \frac {1}{x^{2}}}{16 \, {\left ({\left (x^{2} - \frac {1}{x^{2}}\right )}^{2} + 4\right )}} + \frac {1}{32} \, \arctan \left (\frac {x^{4} - 1}{2 \, x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.04, size = 32, normalized size = 0.86 \begin {gather*} \frac {\mathrm {atan}\left (x^2\right )}{16}-\frac {\frac {x^2}{16}-\frac {x^6}{16}}{x^8+2\,x^4+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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