Integrand size = 23, antiderivative size = 57 \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=-\frac {2^{-3+p} \left (\frac {d-e x}{d}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^2 e (1+p)} \]
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Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {690, 71} \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=-\frac {2^{p-3} \left (\frac {d-e x}{d}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d^2 e (p+1)} \]
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Rule 71
Rule 690
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (\frac {d-e x}{d}\right )^{1+p} \left (\frac {1}{d}-\frac {e x}{d^2}\right )^{-1-p}\right ) \int \left (\frac {1}{d}-\frac {e x}{d^2}\right )^p \left (1+\frac {e x}{d}\right )^{-3+p} \, dx}{d^4} \\ & = -\frac {2^{-3+p} \left (\frac {d-e x}{d}\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^2 e (1+p)} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=-\frac {2^{-3+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (1-\frac {e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^3 e (1+p)} \]
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\[\int \frac {\left (1-\frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{\left (e x +d \right )^{3}}d x\]
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\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
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\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=\int \frac {\left (- \left (-1 + \frac {e x}{d}\right ) \left (1 + \frac {e x}{d}\right )\right )^{p}}{\left (d + e x\right )^{3}}\, dx \]
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\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
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\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx=\int \frac {{\left (1-\frac {e^2\,x^2}{d^2}\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]
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