Optimal. Leaf size=173 \[ \frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)}+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2}-\frac {2 e^3 (8-3 p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6} \]
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Rubi [A] time = 0.27, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {852, 1807, 764, 266, 65, 246, 245} \[ \frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)}-\frac {2 e^3 (8-3 p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6}+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 65
Rule 245
Rule 246
Rule 266
Rule 764
Rule 852
Rule 1807
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx &=\int \frac {(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^3} \, dx\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (6 d^4 e-2 d^3 e^2 (6-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 )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {\int \frac {\left (2 d^5 e^2 (6-p)-4 d^4 e^3 (8-3 p) x\right ) \left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx}{2 d^4}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\left (2 e^3 (8-3 p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx+\left (d e^2 (6-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {1}{2} \left (d e^2 (6-p)\right ) \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )-\frac {\left (2 e^3 (8-3 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 )^{-3+p} \, dx}{d^6}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {2 e^3 (8-3 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},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6}+\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (1,-2+p;-1+p;1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)}\\ \end {align*}
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Mathematica [A] time = 0.65, size = 341, normalized size = 1.97 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (\frac {24 d e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )}{p}+\frac {24 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}+\frac {4 d^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac {d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac {3 e^2 2^{p+3} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {3 e^2 2^{p+1} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {e^2 2^p (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}\right )}{8 d^6} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{3} x^{6} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{4} + d^{3} x^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3} x^{3}}\, dx \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{3}}\,{d x} \]
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 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^3\,{\left (d+e\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{3} \left (d + e x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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