Integrand size = 22, antiderivative size = 94 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \]
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Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {427, 396, 252, 251} \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {12 a^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}}-\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3} \]
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Rule 251
Rule 252
Rule 396
Rule 427
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {\int \frac {6 a^2 b-12 a b^2 x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{5 b} \\ & = -\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {1}{5} \left (12 a^2\right ) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx \\ & = -\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {\left (12 a^2 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{5 \left (a+b x^3\right )^{2/3}} \\ & = -\frac {6}{5} a x \sqrt [3]{a+b x^3}-\frac {1}{5} x \left (a-b x^3\right ) \sqrt [3]{a+b x^3}+\frac {12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \\ \end{align*}
Time = 10.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {-7 a^2 x-6 a b x^4+b^2 x^7+12 a^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 \left (a+b x^3\right )^{2/3}} \]
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\[\int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
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Result contains complex when optimal does not.
Time = 1.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {a^{\frac {4}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {2 \sqrt [3]{a} b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {b^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {{\left (a-b\,x^3\right )}^2}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]
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