Optimal. Leaf size=41 \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]
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Rubi [A] time = 0.31, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6694, 180} \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]
Antiderivative was successfully verified.
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Rule 180
Rule 6694
Rubi steps
\begin {align*} \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1+x) (2+x) (3+x) (4+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{6 (1+x)}-\frac {1}{2 (2+x)}+\frac {1}{2 (3+x)}-\frac {1}{6 (4+x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \log \left (2+x^2\right )+\frac {1}{4} \log \left (3+x^2\right )-\frac {1}{12} \log \left (4+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.01, size = 41, normalized size = 1.00 \[ \frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.01, size = 29, normalized size = 0.71 \[ \frac {1}{2} \tanh ^{-1}\left (2 x^2+5\right )-\frac {1}{6} \tanh ^{-1}\left (\frac {2 x^2}{3}+\frac {5}{3}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.97, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 34, normalized size = 0.83
method | result | size |
default | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
norman | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
risch | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 33, normalized size = 0.80 \[ -\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.06, size = 33, normalized size = 0.80 \[ \frac {\mathrm {atanh}\left (\frac {3072}{5\,\left (1280\,x^2+3072\right )}-\frac {1}{5}\right )}{2}-\frac {\mathrm {atanh}\left (\frac {1024}{405\,\left (\frac {640\,x^2}{243}+\frac {1024}{243}\right )}-\frac {3}{5}\right )}{6} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 32, normalized size = 0.78 \[ \frac {\log {\left (x^{2} + 1 \right )}}{12} - \frac {\log {\left (x^{2} + 2 \right )}}{4} + \frac {\log {\left (x^{2} + 3 \right )}}{4} - \frac {\log {\left (x^{2} + 4 \right )}}{12} \]
Verification of antiderivative is not currently implemented for this CAS.
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