Integrand size = 23, antiderivative size = 118 \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (3-p) (d+e x)^4}-\frac {2^{-2+p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^4 e^2 (3-p) (1+p)} \]
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Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {807, 692, 71} \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (3-p) (d+e x)^4}-\frac {2^{p-2} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d^4 e^2 (3-p) (p+1)} \]
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Rule 71
Rule 692
Rule 807
Rubi steps \begin{align*} \text {integral}& = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (3-p) (d+e x)^4}+\frac {2 \int \frac {\left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx}{e (3-p)} \\ & = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (3-p) (d+e x)^4}+\frac {\left (2 (d-e x)^{-1-p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p}\right ) \int (d-e x)^p \left (1+\frac {e x}{d}\right )^{-3+p} \, dx}{d^4 e (3-p)} \\ & = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (3-p) (d+e x)^4}-\frac {2^{-2+p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^4 e^2 (3-p) (1+p)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {2^{-4+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (-2 \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+\operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{d^3 e^2 (1+p)} \]
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\[\int \frac {x \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}d x\]
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\[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x}{{\left (e x + d\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {x \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \]
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