Optimal. Leaf size=272 \[ \frac{\log \left (2^{2/3}-\frac{1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac{\log \left (\frac{2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac{\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3\ 2^{2/3}}+\frac{1}{3} \sqrt [3]{2} \log \left (\frac{\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )-\frac{\log \left (\frac{(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac{2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}+2 \sqrt [3]{2}\right )}{6\ 2^{2/3}}+\frac{\sqrt [3]{2} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{\sqrt{3}}+\frac{\tan ^{-1}\left (\frac{\frac{\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt{3}}\right )}{2^{2/3} \sqrt{3}} \]
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Rubi [C] time = 0.0078488, antiderivative size = 21, normalized size of antiderivative = 0.08, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {429} \[ x F_1\left (\frac{1}{3};-\frac{1}{3},1;\frac{4}{3};x^3,-x^3\right ) \]
Warning: Unable to verify antiderivative.
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Rule 429
Rubi steps
\begin{align*} \int \frac{\sqrt [3]{1-x^3}}{1+x^3} \, dx &=x F_1\left (\frac{1}{3};-\frac{1}{3},1;\frac{4}{3};x^3,-x^3\right )\\ \end{align*}
Mathematica [C] time = 0.0882679, size = 109, normalized size = 0.4 \[ -\frac{4 x \sqrt [3]{1-x^3} F_1\left (\frac{1}{3};-\frac{1}{3},1;\frac{4}{3};x^3,-x^3\right )}{\left (x^3+1\right ) \left (x^3 \left (3 F_1\left (\frac{4}{3};-\frac{1}{3},2;\frac{7}{3};x^3,-x^3\right )+F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};x^3,-x^3\right )\right )-4 F_1\left (\frac{1}{3};-\frac{1}{3},1;\frac{4}{3};x^3,-x^3\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}+1}\sqrt [3]{-{x}^{3}+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{{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x^{3} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 21.8302, size = 902, normalized size = 3.32 \begin{align*} \frac{1}{18} \, \sqrt{3} 2^{\frac{1}{3}} \arctan \left (-\frac{6 \, \sqrt{3} 2^{\frac{2}{3}}{\left (x^{16} - 33 \, x^{13} + 110 \, x^{10} - 110 \, x^{7} + 33 \, x^{4} - x\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} - 24 \, \sqrt{3} 2^{\frac{1}{3}}{\left (x^{14} - 2 \, x^{11} - 6 \, x^{8} - 2 \, x^{5} + x^{2}\right )}{\left (-x^{3} + 1\right )}^{\frac{2}{3}} - \sqrt{3}{\left (x^{18} + 42 \, x^{15} - 417 \, x^{12} + 812 \, x^{9} - 417 \, x^{6} + 42 \, x^{3} + 1\right )}}{3 \,{\left (x^{18} - 102 \, x^{15} + 447 \, x^{12} - 628 \, x^{9} + 447 \, x^{6} - 102 \, x^{3} + 1\right )}}\right ) + \frac{1}{18} \cdot 2^{\frac{1}{3}} \log \left (-\frac{12 \,{\left (-x^{3} + 1\right )}^{\frac{2}{3}} x^{2} + 2^{\frac{2}{3}}{\left (x^{6} + 2 \, x^{3} + 1\right )} - 6 \cdot 2^{\frac{1}{3}}{\left (x^{4} - x\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) - \frac{1}{36} \cdot 2^{\frac{1}{3}} \log \left (\frac{12 \cdot 2^{\frac{2}{3}}{\left (x^{8} - 4 \, x^{5} + x^{2}\right )}{\left (-x^{3} + 1\right )}^{\frac{2}{3}} + 2^{\frac{1}{3}}{\left (x^{12} - 32 \, x^{9} + 78 \, x^{6} - 32 \, x^{3} + 1\right )} + 6 \,{\left (x^{10} - 11 \, x^{7} + 11 \, x^{4} - x\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x^{12} + 4 \, x^{9} + 6 \, x^{6} + 4 \, x^{3} + 1}\right ) \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{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )}}{\left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{x^{3} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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