Integrand size = 17, antiderivative size = 57 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]
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Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {348, 372, 371} \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (\frac {b x^{3/2}}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]
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Rule 348
Rule 371
Rule 372
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{x^9 \left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {\left (2 \left (1+\frac {b x^{3/2}}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{x^9 \left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx,x,\sqrt {x}\right )}{\left (a+b x^{3/2}\right )^{2/3}} \\ & = -\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \, _2F_1\left (-\frac {8}{3},\frac {2}{3};-\frac {5}{3};-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \\ \end{align*}
Time = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]
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\[\int \frac {1}{x^{5} \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]
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Result contains complex when optimal does not.
Time = 3.84 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {2 \Gamma \left (- \frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{3}, \frac {2}{3} \\ - \frac {5}{3} \end {matrix}\middle | {\frac {b x^{\frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} x^{4} \Gamma \left (- \frac {5}{3}\right )} \]
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\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]
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\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int \frac {1}{x^5\,{\left (a+b\,x^{3/2}\right )}^{2/3}} \,d x \]
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