Integrand size = 20, antiderivative size = 101 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {\left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {\arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]
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Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {455, 44, 57, 631, 210, 31} \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}-\frac {\log \left (x^2+3\right )}{48\ 2^{2/3}}+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]
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Rule 31
Rule 44
Rule 57
Rule 210
Rule 455
Rule 631
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {1}{24} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right ) \\ & = -\frac {\left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac {1}{16} \text {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {\text {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \\ & = -\frac {\left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{8\ 2^{2/3}} \\ & = -\frac {\left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {\tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.11 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{96} \left (-\frac {12 \left (1-x^2\right )^{2/3}}{3+x^2}+2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )+2 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )-\sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )\right ) \]
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Time = 7.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07
method | result | size |
pseudoelliptic | \(\frac {\left (x^{2}+3\right ) \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )-\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )\right ) 2^{\frac {1}{3}}-12 \left (-x^{2}+1\right )^{\frac {2}{3}}}{96 x^{2}+288}\) | \(108\) |
risch | \(\text {Expression too large to display}\) | \(655\) |
trager | \(\text {Expression too large to display}\) | \(753\) |
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Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {4 \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (x^{2} + 3\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {1}{3}} \sqrt {3} + 2 \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 2 \cdot 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - 24 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{192 \, {\left (x^{2} + 3\right )}} \]
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\[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\int \frac {x}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )^{2}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{96} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{192} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{96} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {{\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{96} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{192} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{96} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {{\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]
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Time = 5.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{64}-\frac {2^{2/3}}{64}\right )}{48}-\frac {{\left (1-x^2\right )}^{2/3}}{8\,\left (x^2+3\right )}+\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{64}-\frac {2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{96}-\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{64}-\frac {2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{96} \]
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