Optimal. Leaf size=158 \[ \frac{\left (1-x^2\right )^{2/3}}{24 \left (x^2+3\right )}+\frac{5 \log \left (x^2+3\right )}{144\ 2^{2/3}}+\frac{1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac{5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{24\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\log (x)}{18} \]
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Rubi [A] time = 0.110738, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {446, 103, 156, 55, 618, 204, 31, 617} \[ \frac{\left (1-x^2\right )^{2/3}}{24 \left (x^2+3\right )}+\frac{5 \log \left (x^2+3\right )}{144\ 2^{2/3}}+\frac{1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac{5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{24\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\log (x)}{18} \]
Antiderivative was successfully verified.
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Rule 446
Rule 103
Rule 156
Rule 55
Rule 618
Rule 204
Rule 31
Rule 617
Rubi steps
\begin{align*} \int \frac{1}{x \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x (3+x)^2} \, dx,x,x^2\right )\\ &=\frac{\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}+\frac{1}{24} \operatorname{Subst}\left (\int \frac{4-\frac{x}{3}}{\sqrt [3]{1-x} x (3+x)} \, dx,x,x^2\right )\\ &=\frac{\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}+\frac{1}{18} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x} \, dx,x,x^2\right )-\frac{5}{72} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac{\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac{\log (x)}{18}+\frac{5 \log \left (3+x^2\right )}{144\ 2^{2/3}}-\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac{5}{48} \operatorname{Subst}\left (\int \frac{1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )+\frac{5 \operatorname{Subst}\left (\int \frac{1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}\\ &=\frac{\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac{\log (x)}{18}+\frac{5 \log \left (3+x^2\right )}{144\ 2^{2/3}}+\frac{1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac{5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right )+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{24\ 2^{2/3}}\\ &=\frac{\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac{5 \tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{24\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1+2 \sqrt [3]{1-x^2}}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\log (x)}{18}+\frac{5 \log \left (3+x^2\right )}{144\ 2^{2/3}}+\frac{1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac{5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}\\ \end{align*}
Mathematica [A] time = 0.130308, size = 148, normalized size = 0.94 \[ \frac{1}{288} \left (\frac{12 \left (1-x^2\right )^{2/3}}{x^2+3}+5 \sqrt [3]{2} \log \left (x^2+3\right )+24 \log \left (1-\sqrt [3]{1-x^2}\right )-15 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )-10 \sqrt [3]{2} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )+16 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )-16 \log (x)\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ({x}^{2}+3 \right ) ^{2}}{\frac{1}{\sqrt [3]{-{x}^{2}+1}}}}\, 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 (x^{2} + 3\right )}^{2}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.59731, size = 711, normalized size = 4.5 \begin{align*} -\frac{20 \cdot 4^{\frac{1}{6}} \sqrt{3} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \arctan \left (\frac{1}{6} \cdot 4^{\frac{1}{6}}{\left (2 \, \sqrt{3} \left (-1\right )^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 4^{\frac{1}{3}} \sqrt{3}\right )}\right ) + 5 \cdot 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \log \left (4^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) - 10 \cdot 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \log \left (-4^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) - 32 \, \sqrt{3}{\left (x^{2} + 3\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^{2} + 3\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac{2}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}} + 1\right ) - 32 \,{\left (x^{2} + 3\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 1\right ) - 24 \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}}}{576 \,{\left (x^{2} + 3\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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