Integrand size = 24, antiderivative size = 163 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {9 \arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \]
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Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {451, 272, 44, 65, 304, 209, 212, 450} \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9 \arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2-3 x^2}+\sqrt {2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2} \]
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Rule 44
Rule 65
Rule 209
Rule 212
Rule 272
Rule 304
Rule 450
Rule 451
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4 x^3 \sqrt [4]{2-3 x^2}}+\frac {3}{16 x \sqrt [4]{2-3 x^2}}-\frac {9 x}{16 \sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {3}{16} \int \frac {1}{x \sqrt [4]{2-3 x^2}} \, dx+\frac {1}{4} \int \frac {1}{x^3 \sqrt [4]{2-3 x^2}} \, dx-\frac {9}{16} \int \frac {x}{\sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )} \, dx \\ & = \frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3}{64} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {x^2}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {1}{16} \text {Subst}\left (\int \frac {x^2}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right )-\frac {3}{16} \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {3}{16} \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {9 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {-4 \left (2-3 x^2\right )^{3/4}+9\ 2^{3/4} x^2 \arctan \left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+6 \sqrt [4]{2} x^2 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-9\ 2^{3/4} x^2 \text {arctanh}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+6 \sqrt [4]{2} x^2 \text {arctanh}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{64 x^2} \]
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Time = 16.48 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.19
method | result | size |
pseudoelliptic | \(\frac {18 \arctan \left (\frac {2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}}{2}\right ) 2^{\frac {3}{4}} x^{2}-9 \ln \left (\frac {\left (-3 x^{2}+2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} x^{2}-6 \ln \left (\frac {-2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}{2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}\right ) 2^{\frac {1}{4}} x^{2}-12 \arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+1\right ) 2^{\frac {1}{4}} x^{2}-12 \arctan \left (-1+2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}\right ) 2^{\frac {1}{4}} x^{2}-8 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{128 x^{2}}\) | \(194\) |
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Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {9 \cdot 2^{\frac {3}{4}} x^{2} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 9 i \cdot 2^{\frac {3}{4}} x^{2} \log \left (i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + 9 i \cdot 2^{\frac {3}{4}} x^{2} \log \left (-i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 9 \cdot 2^{\frac {3}{4}} x^{2} \log \left (-2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (6 i - 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (6 i + 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (-\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (6 i + 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (6 i - 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (-\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + 8 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{128 \, x^{2}} \]
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\[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{5} \sqrt [4]{2 - 3 x^{2}} - 4 x^{3} \sqrt [4]{2 - 3 x^{2}}}\, dx \]
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\[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} x^{3}} \,d x } \]
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none
Time = 0.33 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9}{64} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {9}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {9}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} - {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {3}{32} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {3}{32} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{64} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {3}{64} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{16 \, x^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9\,2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{64}-\frac {{\left (2-3\,x^2\right )}^{3/4}}{16\,x^2}+\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,9{}\mathrm {i}}{64}-\frac {{\left (-1\right )}^{1/4}\,2^{3/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,3{}\mathrm {i}}{32}-\frac {{\left (-1\right )}^{3/4}\,2^{3/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{3/4}\,2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,3{}\mathrm {i}}{32} \]
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