Integrand size = 25, antiderivative size = 163 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}-\frac {2^{-3+p} (7+p) \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (1+p)} \]
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Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1653, 807, 692, 71} \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (3-p) (d+e x)^4}-\frac {2^{p-3} (p+7) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (3-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (p+1)} \]
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Rule 71
Rule 692
Rule 807
Rule 1653
Rubi steps \begin{align*} \text {integral}& = \frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac {\int \frac {\left (3 d^2 e^2+2 d e^3 (1+p) x\right ) \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx}{e^4 (1-2 p)} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac {(d (7+p)) \int \frac {\left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx}{e^2 (1-2 p) (3-p)} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac {\left ((7+p) (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^3 e^2 (1-2 p) (3-p)} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}-\frac {2^{-3+p} (7+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^3 e^3 (1-2 p) (3-p) (1+p)} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {2^{-4+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )-4 \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+\operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{d^2 e^3 (1+p)} \]
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\[\int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}d x\]
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\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \]
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