Integrand size = 22, antiderivative size = 93 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {57 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{77 a^3 \left (a+b x^3\right )^{2/3}} \]
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Time = 0.02 (sec) , antiderivative size = 93, 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 = {424, 393, 252, 251} \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\frac {57 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{77 a^3 \left (a+b x^3\right )^{2/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}} \]
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Rule 251
Rule 252
Rule 393
Rule 424
Rubi steps \begin{align*} \text {integral}& = \frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {\int \frac {12 a^2 b-6 a b^2 x^3}{\left (a+b x^3\right )^{14/3}} \, dx}{14 a b} \\ & = \frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {57}{77} \int \frac {1}{\left (a+b x^3\right )^{11/3}} \, dx \\ & = \frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {\left (57 \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{11/3}} \, dx}{77 a^3 \left (a+b x^3\right )^{2/3}} \\ & = \frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {57 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {11}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{77 a^3 \left (a+b x^3\right )^{2/3}} \\ \end{align*}
Time = 10.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\frac {x \left (2282 a^4+4879 a^3 b x^3+6270 a^2 b^2 x^6+3591 a b^3 x^9+798 b^4 x^{12}+798 \left (a+b x^3\right )^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{3080 a^3 \left (a+b x^3\right )^{14/3}} \]
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\[\int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {17}{3}}}d x\]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
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\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int \frac {{\left (a-b\,x^3\right )}^2}{{\left (b\,x^3+a\right )}^{17/3}} \,d x \]
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