Integrand size = 25, antiderivative size = 166 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}-2 e^3 (1+3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (1+p)} \]
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Time = 0.15 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1821, 778, 272, 67, 252, 251} \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {e^2 x^2}{d^2}\right )}{2 d (p+1)}-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 x^2}-2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right ) \]
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Rule 67
Rule 251
Rule 252
Rule 272
Rule 778
Rule 1821
Rubi steps \begin{align*} \text {integral}& = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^p \left (-6 d^4 e-2 d^3 e^2 (3-p) x-2 d^2 e^3 x^2\right )}{x^2} \, dx}{2 d^2} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}+\frac {\int \frac {\left (2 d^5 e^2 (3-p)-4 d^4 e^3 (1+3 p) x\right ) \left (d^2-e^2 x^2\right )^p}{x} \, dx}{2 d^4} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}+\left (d e^2 (3-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^p}{x} \, dx-\left (2 e^3 (1+3 p)\right ) \int \left (d^2-e^2 x^2\right )^p \, dx \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}+\frac {1}{2} \left (d e^2 (3-p)\right ) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^p}{x} \, dx,x,x^2\right )-\left (2 e^3 (1+3 p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{1+p}}{x}-2 e^3 (1+3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1-\frac {e^2 x^2}{d^2}\right )}{2 d (1+p)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (-6 d^3 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )+e x \left (2 d e (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\left (d^2-e^2 x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (3 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )+\operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1-\frac {e^2 x^2}{d^2}\right )\right )\right )\right )}{2 d (1+p) x} \]
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\[\int \frac {\left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3}}d x\]
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\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
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Result contains complex when optimal does not.
Time = 3.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=- \frac {d^{3} e^{2 p} x^{2 p - 2} e^{i \pi p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} - \frac {3 d^{2} d^{2 p} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} - \frac {3 d e^{2} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + d^{2 p} e^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} \]
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\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
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\[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^p}{x^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3}{x^3} \,d x \]
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