Integrand size = 15, antiderivative size = 37 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,2 (1+m),3+2 m,-\frac {b \sqrt {x}}{a}\right )}{a^2 (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {348, 66} \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,2 (m+1),2 m+3,-\frac {b \sqrt {x}}{a}\right )}{a^2 (m+1)} \]
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Rule 66
Rule 348
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^{-1+2 (1+m)}}{(a+b x)^2} \, dx,x,\sqrt {x}\right ) \\ & = \frac {x^{1+m} \, _2F_1\left (2,2 (1+m);3+2 m;-\frac {b \sqrt {x}}{a}\right )}{a^2 (1+m)} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,2 (1+m),1+2 (1+m),-\frac {b \sqrt {x}}{a}\right )}{a^2 (1+m)} \]
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\[\int \frac {x^{m}}{\left (a +b \sqrt {x}\right )^{2}}d x\]
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\[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\int { \frac {x^{m}}{{\left (b \sqrt {x} + a\right )}^{2}} \,d x } \]
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Result contains complex when optimal does not.
Time = 1.04 (sec) , antiderivative size = 478, normalized size of antiderivative = 12.92 \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=- \frac {8 a m^{2} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {12 a m x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} + \frac {4 a m x^{m + 1} \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {4 a x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} + \frac {4 a x^{m + 1} \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {8 b m^{2} \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {12 b m \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} - \frac {4 b \sqrt {x} x^{m + 1} \Phi \left (\frac {b \sqrt {x} e^{i \pi }}{a}, 1, 2 m + 2\right ) \Gamma \left (2 m + 2\right )}{a^{3} \Gamma \left (2 m + 3\right ) + a^{2} b \sqrt {x} \Gamma \left (2 m + 3\right )} \]
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\[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\int { \frac {x^{m}}{{\left (b \sqrt {x} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\int { \frac {x^{m}}{{\left (b \sqrt {x} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^m}{\left (a+b \sqrt {x}\right )^2} \, dx=\int \frac {x^m}{{\left (a+b\,\sqrt {x}\right )}^2} \,d x \]
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