Optimal. Leaf size=166 \[ -\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )}{2 d (p+1)}-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 x^2}-2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right ) \]
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Rubi [A] time = 0.22, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1807, 764, 266, 65, 246, 245} \[ -2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )-\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )}{2 d (p+1)}-\frac {3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 65
Rule 245
Rule 246
Rule 266
Rule 764
Rule 1807
Rubi steps
\begin {align*} \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 {\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 ) \operatorname {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*}
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Mathematica [A] time = 0.10, size = 182, normalized size = 1.10 \[ \frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (e x \left (2 d e (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )-\left (d^2-e^2 x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (3 \, _2F_1\left (1,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )+\, _2F_1\left (2,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )\right )\right )-6 d^3 (p+1) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )\right )}{2 d (p+1) x} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3}{x^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 8.37, size = 177, normalized size = 1.07 \[ - \frac {d^{3} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 x^{2} \Gamma \left (2 - p\right )} - \frac {3 d^{2} d^{2 p} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} - \frac {3 d e^{2} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + d^{2 p} e^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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