Optimal. Leaf size=191 \[ -\frac {\sqrt [4]{3 x^2-1}}{4 x^2}+\frac {15 \log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt {2}}-\frac {15 \log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt {2}}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac {15 \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt {2}}-\frac {15 \tan ^{-1}\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt {2}}-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]
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Rubi [A] time = 0.15, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {446, 103, 156, 63, 211, 1165, 628, 1162, 617, 204, 212, 206, 203} \begin {gather*} -\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}}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac {15 \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt {2}}-\frac {15 \tan ^{-1}\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt {2}}-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 103
Rule 156
Rule 203
Rule 204
Rule 206
Rule 211
Rule 212
Rule 446
Rule 617
Rule 628
Rule 1162
Rule 1165
Rubi steps
\begin {align*} \int \frac {1}{x^3 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {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} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{x (-1+3 x)^{3/4}} \, dx,x,x^2\right )+\frac {9}{8} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{3}+\frac {x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {3}{2} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {3}{4} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {15}{16} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {15 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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*}
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Mathematica [A] time = 0.05, size = 181, normalized size = 0.95 \begin {gather*} \frac {1}{32} \left (-\frac {8 \sqrt [4]{3 x^2-1}}{x^2}+15 \sqrt {2} \log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )-15 \sqrt {2} \log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )-24 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+30 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )-30 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )-24 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.24, size = 140, normalized size = 0.73 \begin {gather*} -\frac {\sqrt [4]{3 x^2-1}}{4 x^2}-\frac {3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {15 \tan ^{-1}\left (\frac {\frac {\sqrt {3 x^2-1}}{\sqrt {2}}-\frac {1}{\sqrt {2}}}{\sqrt [4]{3 x^2-1}}\right )}{8 \sqrt {2}}-\frac {3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3 x^2-1}}{\sqrt {3 x^2-1}+1}\right )}{8 \sqrt {2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 252, normalized size = 1.32 \begin {gather*} \frac {60 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1} - \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) + 60 \, \sqrt {2} x^{2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {3 \, x^{2} - 1} + 4} - \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) - 15 \, \sqrt {2} x^{2} \log \left (4 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {3 \, x^{2} - 1} + 4\right ) + 15 \, \sqrt {2} x^{2} \log \left (-4 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {3 \, x^{2} - 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.50, size = 169, normalized size = 0.88 \begin {gather*} -\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 ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 5.96, size = 916, normalized size = 4.80
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.21, size = 81, normalized size = 0.42 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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