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 = 0.146251, antiderivative size = 191, normalized size of antiderivative = 1., 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} \[ -\frac{\sqrt [4]{3 x^2-1}}{4 x^2}+\frac{15 \log \left (\sqrt{3 x^2-1}-\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}-\frac{15 \log \left (\sqrt{3 x^2-1}+\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}-\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac{15 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt{2}}-\frac{15 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt{2}}-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]
Antiderivative was successfully verified.
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Rule 446
Rule 103
Rule 156
Rule 63
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rule 212
Rule 206
Rule 203
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*}
Mathematica [A] time = 0.0518076, size = 181, normalized size = 0.95 \[ \frac{1}{32} \left (-\frac{8 \sqrt [4]{3 x^2-1}}{x^2}+15 \sqrt{2} \log \left (\sqrt{3 x^2-1}-\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )-15 \sqrt{2} \log \left (\sqrt{3 x^2-1}+\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )-24 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+30 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{3 x^2-1}\right )-30 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )-24 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.078, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( 3\,{x}^{2}-2 \right ) } \left ( 3\,{x}^{2}-1 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (3 \, x^{2} - 1\right )}^{\frac{3}{4}}{\left (3 \, x^{2} - 2\right )} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.60238, size = 722, normalized size = 3.78 \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac{3}{4}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.22364, size = 228, normalized size = 1.19 \begin{align*} -\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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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