Integrand size = 22, antiderivative size = 183 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac {\arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {\log (x)}{54}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac {1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]
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Time = 0.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {457, 105, 156, 162, 57, 632, 210, 31, 631} \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\arctan \left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{18 \sqrt {3}}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (x^2+3\right )}-\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (x^2+3\right )}-\frac {\log \left (x^2+3\right )}{48\ 2^{2/3}}-\frac {1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac {\log (x)}{54} \]
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Rule 31
Rule 57
Rule 105
Rule 156
Rule 162
Rule 210
Rule 457
Rule 631
Rule 632
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x^2 (3+x)^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac {1}{6} \text {Subst}\left (\int \frac {1-\frac {4 x}{3}}{\sqrt [3]{1-x} x (3+x)^2} \, dx,x,x^2\right ) \\ & = -\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac {1}{72} \text {Subst}\left (\int \frac {4-\frac {5 x}{3}}{\sqrt [3]{1-x} x (3+x)} \, dx,x,x^2\right ) \\ & = -\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac {1}{54} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x} \, dx,x,x^2\right )+\frac {1}{24} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right ) \\ & = -\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac {\log (x)}{54}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac {1}{36} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {1}{36} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )+\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 {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac {\log (x)}{54}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac {1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac {1}{18} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right )-\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 {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \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 {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {\log (x)}{54}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac {1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{864} \left (-\frac {12 \left (1-x^2\right )^{2/3} \left (12+5 x^2\right )}{x^2 \left (3+x^2\right )}+18 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )-16 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )+18 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )-9 \sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )-16 \log \left (-1+\sqrt [3]{1-x^2}\right )+8 \log \left (1+\sqrt [3]{1-x^2}+\left (1-x^2\right )^{2/3}\right )\right ) \]
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Time = 4.79 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.48
method | result | size |
pseudoelliptic | \(\frac {9 x^{2} \left (x^{2}+3\right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )-2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )\right ) 2^{\frac {1}{3}}+16 x^{2} \sqrt {3}\, \left (x^{2}+3\right ) \arctan \left (\frac {\left (1+2 \left (-x^{2}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )+\left (16 \ln \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right )-8 \ln \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )\right ) x^{4}+\left (60 \left (-x^{2}+1\right )^{\frac {2}{3}}-24 \ln \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )+48 \ln \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right )\right ) x^{2}+144 \left (-x^{2}+1\right )^{\frac {2}{3}}}{864 \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right ) \left (x^{2}+3\right ) \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )}\) | \(271\) |
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Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {36 \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {1}{3}} \sqrt {3} + 2 \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 9 \cdot 4^{\frac {2}{3}} {\left (x^{4} + 3 \, x^{2}\right )} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 18 \cdot 4^{\frac {2}{3}} {\left (x^{4} + 3 \, x^{2}\right )} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - 32 \, \sqrt {3} {\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 16 \, {\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - 32 \, {\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) - 24 \, {\left (5 \, x^{2} + 12\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{1728 \, {\left (x^{4} + 3 \, x^{2}\right )}} \]
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\[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\int \frac {1}{x^{3} \sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )^{2}}\, dx \]
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\[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\int { \frac {1}{{\left (x^{2} + 3\right )}^{2} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} x^{3}} \,d x } \]
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Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{96} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{192} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{96} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {5 \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - 17 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{72 \, {\left ({\left (x^{2} - 1\right )}^{2} + 5 \, x^{2} - 1\right )}} + \frac {1}{108} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{54} \, \log \left (-{\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) \]
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Time = 5.40 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.23 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {2^{1/3}\,\ln \left (\frac {2^{2/3}\,\left (\frac {2^{1/3}\,\left (\frac {10935\,2^{2/3}}{64}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{48}-\frac {665}{128}\right )}{2304}+\frac {{\left (1-x^2\right )}^{1/3}}{576}\right )}{48}-\frac {\ln \left (\frac {985\,{\left (1-x^2\right )}^{1/3}}{373248}-\frac {985}{373248}\right )}{54}+\ln \left ({\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2\,\left (\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )\,\left (393660\,{\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )-\frac {665}{128}\right )+\frac {{\left (1-x^2\right )}^{1/3}}{576}\right )\,\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )-\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}-{\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2\,\left (\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )\,\left (393660\,{\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )+\frac {665}{128}\right )\right )\,\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )-\frac {\frac {17\,{\left (1-x^2\right )}^{2/3}}{72}-\frac {5\,{\left (1-x^2\right )}^{5/3}}{72}}{{\left (x^2-1\right )}^2+5\,x^2-1}+\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}+\frac {2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {10935\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{96}-\frac {665}{128}\right )}{9216}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{96}-\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}-\frac {2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {10935\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{96}+\frac {665}{128}\right )}{9216}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{96} \]
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