Optimal. Leaf size=45 \[ \frac {1}{2} \tan ^{-1}(x)+\frac {\sqrt {x^6} \tan ^{-1}(x)}{2 x^3}+\frac {1}{2} \tanh ^{-1}(x)-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3} \]
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Rubi [A]
time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6857, 218, 212,
209, 15, 304} \begin {gather*} \frac {\sqrt {x^6} \text {ArcTan}(x)}{2 x^3}+\frac {\text {ArcTan}(x)}{2}-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3}+\frac {1}{2} \tanh ^{-1}(x) \end {gather*}
Antiderivative was successfully verified.
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Rule 15
Rule 209
Rule 212
Rule 218
Rule 304
Rule 6857
Rubi steps
\begin {align*} \int \frac {1-\frac {\sqrt {x^6}}{x}}{1-x^4} \, dx &=\int \left (\frac {1}{1-x^4}+\frac {\sqrt {x^6}}{x \left (-1+x^4\right )}\right ) \, dx\\ &=\int \frac {1}{1-x^4} \, dx+\int \frac {\sqrt {x^6}}{x \left (-1+x^4\right )} \, dx\\ &=\frac {1}{2} \int \frac {1}{1-x^2} \, dx+\frac {1}{2} \int \frac {1}{1+x^2} \, dx+\frac {\sqrt {x^6} \int \frac {x^2}{-1+x^4} \, dx}{x^3}\\ &=\frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \tanh ^{-1}(x)-\frac {\sqrt {x^6} \int \frac {1}{1-x^2} \, dx}{2 x^3}+\frac {\sqrt {x^6} \int \frac {1}{1+x^2} \, dx}{2 x^3}\\ &=\frac {1}{2} \tan ^{-1}(x)+\frac {\sqrt {x^6} \tan ^{-1}(x)}{2 x^3}+\frac {1}{2} \tanh ^{-1}(x)-\frac {\sqrt {x^6} \tanh ^{-1}(x)}{2 x^3}\\ \end {align*}
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Mathematica [A]
time = 0.00, size = 57, normalized size = 1.27 \begin {gather*} \frac {1}{2} \tan ^{-1}(x)-\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {x^6}}{x^4}\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {x^6}}{x^4}\right )-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (1+x) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.49, size = 35, normalized size = 0.78
method | result | size |
default | \(\frac {\sqrt {x^{6}}\, \left (\ln \left (-1+x \right )-\ln \left (1+x \right )+2 \arctan \left (x \right )\right )}{4 x^{3}}+\frac {\arctan \left (x \right )}{2}+\frac {\arctanh \left (x \right )}{2}\) | \(35\) |
meijerg | \(-\frac {x \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {\sqrt {x^{6}}\, \left (\ln \left (1-\left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{4}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}\) | \(80\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 2, normalized size = 0.04 \begin {gather*} \arctan \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 2, normalized size = 0.04 \begin {gather*} \arctan \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.03, size = 2, normalized size = 0.04 \begin {gather*} \operatorname {atan}{\left (x \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 2.40, size = 31, normalized size = 0.69 \begin {gather*} \frac {1}{2} \, {\left (\mathrm {sgn}\left (x\right ) + 1\right )} \arctan \left (x\right ) - \frac {1}{4} \, {\left (\mathrm {sgn}\left (x\right ) - 1\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, {\left (\mathrm {sgn}\left (x\right ) - 1\right )} \log \left ({\left | x - 1 \right |}\right ) \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 = -0.02 \begin {gather*} \int \frac {\frac {\sqrt {x^6}}{x}-1}{x^4-1} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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