Integrand size = 24, antiderivative size = 191 \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {3}{4} \arctan \left (\sqrt [4]{-1+3 x^2}\right )+\frac {15 \arctan \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}}-\frac {15 \arctan \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}}-\frac {3}{4} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )+\frac {15 \log \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}}-\frac {15 \log \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}} \]
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Time = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {457, 105, 162, 65, 217, 1179, 642, 1176, 631, 210, 218, 212, 209} \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {3}{4} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {15 \arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt {2}}-\frac {15 \arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt {2}}-\frac {3}{4} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )-\frac {\sqrt [4]{3 x^2-1}}{4 x^2}+\frac {15 \log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt {2}}-\frac {15 \log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt {2}} \]
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Rule 65
Rule 105
Rule 162
Rule 209
Rule 210
Rule 212
Rule 217
Rule 218
Rule 457
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 (-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {1}{4} \text {Subst}\left (\int \frac {-\frac {15}{2}+\frac {27 x}{4}}{x (-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {15}{16} \text {Subst}\left (\int \frac {1}{x (-1+3 x)^{3/4}} \, dx,x,x^2\right )+\frac {9}{8} \text {Subst}\left (\int \frac {1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {5}{4} \text {Subst}\left (\int \frac {1}{\frac {1}{3}+\frac {x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {3}{2} \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {5}{8} \text {Subst}\left (\int \frac {1-x^2}{\frac {1}{3}+\frac {x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {5}{8} \text {Subst}\left (\int \frac {1+x^2}{\frac {1}{3}+\frac {x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {15}{16} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {15}{16} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {15 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{16 \sqrt {2}}+\frac {15 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{16 \sqrt {2}} \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )+\frac {15 \log \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}}-\frac {15 \log \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}}-\frac {15 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}}+\frac {15 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}} \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{4 x^2}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )+\frac {15 \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}}-\frac {15 \tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt {2}}-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )+\frac {15 \log \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}}-\frac {15 \log \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{16 \sqrt {2}} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {1}{16} \left (-\frac {4 \sqrt [4]{-1+3 x^2}}{x^2}-12 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-15 \sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+3 x^2}}{\sqrt {2} \sqrt [4]{-1+3 x^2}}\right )-12 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )-15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+3 x^2}}{1+\sqrt {-1+3 x^2}}\right )\right ) \]
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Time = 10.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(\frac {-15 \ln \left (\frac {-\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {3 x^{2}-1}-1}{\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {3 x^{2}-1}-1}\right ) \sqrt {2}\, x^{2}-30 \arctan \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}\, x^{2}-30 \arctan \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}\, x^{2}+12 \ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right ) x^{2}-12 \ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right ) x^{2}-24 \arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right ) x^{2}-8 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{32 x^{2}}\) | \(180\) |
trager | \(-\frac {\left (3 x^{2}-1\right )^{\frac {1}{4}}}{4 x^{2}}-\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{16}-\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )-2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{x^{2}}\right )}{16}+\frac {3 \ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{8}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {3 x^{2}-1}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}-2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{8}\) | \(314\) |
risch | \(\text {Expression too large to display}\) | \(917\) |
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Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {-\left (15 i + 15\right ) \, \sqrt {2} x^{2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (15 i - 15\right ) \, \sqrt {2} x^{2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \left (15 i - 15\right ) \, \sqrt {2} x^{2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (15 i + 15\right ) \, \sqrt {2} x^{2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - 24 \, x^{2} \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - 12 \, x^{2} \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + 12 \, x^{2} \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) - 8 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}}{32 \, x^{2}} \]
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\[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^{3} \cdot \left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{3}} \,d x } \]
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none
Time = 0.36 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {15}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {15}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {15}{32} \, \sqrt {2} \log \left (\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) + \frac {15}{32} \, \sqrt {2} \log \left (-\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) - \frac {{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}}{4 \, x^{2}} - \frac {3}{4} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{8} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{8} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
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Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {3\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{4}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4}-\frac {{\left (3\,x^2-1\right )}^{1/4}}{4\,x^2}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (3\,x^2-1\right )}^{1/4}\right )\,15{}\mathrm {i}}{8}-\frac {{\left (-1\right )}^{3/4}\,\mathrm {atan}\left ({\left (-1\right )}^{3/4}\,{\left (3\,x^2-1\right )}^{1/4}\right )\,15{}\mathrm {i}}{8} \]
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