Optimal. Leaf size=141 \[ \frac{1}{8} \left (1-x^3\right )^{8/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\left (1-x^3\right )^{2/3}+\frac{1}{2 \sqrt [3]{1-x^3}}-\frac{\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac{\tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}} \]
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Rubi [A] time = 0.113576, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {446, 87, 43, 627, 51, 55, 617, 204, 31} \[ \frac{1}{8} \left (1-x^3\right )^{8/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\left (1-x^3\right )^{2/3}+\frac{1}{2 \sqrt [3]{1-x^3}}-\frac{\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac{\tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}} \]
Antiderivative was successfully verified.
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Rule 446
Rule 87
Rule 43
Rule 627
Rule 51
Rule 55
Rule 617
Rule 204
Rule 31
Rubi steps
\begin{align*} \int \frac{x^{14}}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt [3]{1-x}}-\frac{x^2}{\sqrt [3]{1-x}}+\frac{1}{\sqrt [3]{1-x} \left (1-x^2\right )}\right ) \, dx,x,x^3\right )\\ &=\frac{1}{2} \left (1-x^3\right )^{2/3}-\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [3]{1-x}} \, dx,x,x^3\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} \left (1-x^2\right )} \, dx,x,x^3\right )\\ &=\frac{1}{2} \left (1-x^3\right )^{2/3}-\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{1}{\sqrt [3]{1-x}}-2 (1-x)^{2/3}+(1-x)^{5/3}\right ) \, dx,x,x^3\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\left (1-x^3\right )^{2/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\frac{1}{8} \left (1-x^3\right )^{8/3}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (1+x)} \, dx,x,x^3\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\left (1-x^3\right )^{2/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\frac{1}{8} \left (1-x^3\right )^{8/3}-\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\left (1-x^3\right )^{2/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\frac{1}{8} \left (1-x^3\right )^{8/3}-\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\left (1-x^3\right )^{2/3}-\frac{2}{5} \left (1-x^3\right )^{5/3}+\frac{1}{8} \left (1-x^3\right )^{8/3}+\frac{\tan ^{-1}\left (\frac{1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}}-\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ \end{align*}
Mathematica [C] time = 0.0234512, size = 53, normalized size = 0.38 \[ \frac{20 \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};\frac{1}{2} \left (1-x^3\right )\right )-5 x^9-x^6-23 x^3+29}{40 \sqrt [3]{1-x^3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{14}}{{x}^{3}+1} \left ( -{x}^{3}+1 \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.42169, size = 173, normalized size = 1.23 \begin{align*} \frac{1}{8} \,{\left (-x^{3} + 1\right )}^{\frac{8}{3}} + \frac{1}{12} \, \sqrt{3} 2^{\frac{2}{3}} \arctan \left (\frac{1}{6} \, \sqrt{3} 2^{\frac{2}{3}}{\left (2^{\frac{1}{3}} + 2 \,{\left (-x^{3} + 1\right )}^{\frac{1}{3}}\right )}\right ) - \frac{2}{5} \,{\left (-x^{3} + 1\right )}^{\frac{5}{3}} - \frac{1}{24} \cdot 2^{\frac{2}{3}} \log \left (2^{\frac{2}{3}} + 2^{\frac{1}{3}}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} +{\left (-x^{3} + 1\right )}^{\frac{2}{3}}\right ) + \frac{1}{12} \cdot 2^{\frac{2}{3}} \log \left (-2^{\frac{1}{3}} +{\left (-x^{3} + 1\right )}^{\frac{1}{3}}\right ) +{\left (-x^{3} + 1\right )}^{\frac{2}{3}} + \frac{1}{2 \,{\left (-x^{3} + 1\right )}^{\frac{1}{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.78337, size = 406, normalized size = 2.88 \begin{align*} \frac{10 \, \sqrt{6} 2^{\frac{1}{6}}{\left (x^{3} - 1\right )} \arctan \left (\frac{1}{6} \cdot 2^{\frac{1}{6}}{\left (\sqrt{6} 2^{\frac{1}{3}} + 2 \, \sqrt{6}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}\right )}\right ) - 5 \cdot 2^{\frac{2}{3}}{\left (x^{3} - 1\right )} \log \left (2^{\frac{2}{3}} + 2^{\frac{1}{3}}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} +{\left (-x^{3} + 1\right )}^{\frac{2}{3}}\right ) + 10 \cdot 2^{\frac{2}{3}}{\left (x^{3} - 1\right )} \log \left (-2^{\frac{1}{3}} +{\left (-x^{3} + 1\right )}^{\frac{1}{3}}\right ) + 3 \,{\left (5 \, x^{9} + x^{6} + 23 \, x^{3} - 49\right )}{\left (-x^{3} + 1\right )}^{\frac{2}{3}}}{120 \,{\left (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{x^{14}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac{4}{3}} \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(-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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