Optimal. Leaf size=211 \[ \frac {e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}+\frac {e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2}-\frac {8 e^3 (4-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^7} \]
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Rubi [A] time = 0.37, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {852, 1807, 1652, 446, 79, 65, 12, 246, 245} \[ \frac {e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}-\frac {8 e^3 (4-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^7}+\frac {e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 65
Rule 79
Rule 245
Rule 246
Rule 446
Rule 852
Rule 1652
Rule 1807
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^3} \, dx\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (8 d^5 e-2 d^4 e^2 (10-p) x+8 d^3 e^3 x^2-2 d^2 e^4 x^3\right )}{x^2} \, dx}{2 d^2}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (2 d^6 e^2 (10-p)-16 d^5 e^3 (4-p) x+2 d^4 e^4 x^2\right )}{x} \, dx}{2 d^4}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {\int -16 d^5 e^3 (4-p) \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{2 d^4}+\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (2 d^6 e^2 (10-p)+2 d^4 e^4 x^2\right )}{x} \, dx}{2 d^4}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {\operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p} \left (2 d^6 e^2 (10-p)+2 d^4 e^4 x\right )}{x} \, dx,x,x^2\right )}{4 d^4}-\left (8 d e^3 (4-p)\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx\\ &=\frac {e^2 (11-p) \left (d^2-e^2 x^2\right )^{-3+p}}{2 (3-p)}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {1}{2} \left (e^2 (10-p)\right ) \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )-\frac {\left (8 e^3 (4-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 )^{-4+p} \, dx}{d^7}\\ &=\frac {e^2 (11-p) \left (d^2-e^2 x^2\right )^{-3+p}}{2 (3-p)}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac {8 e^3 (4-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},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^7}+\frac {e^2 (10-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 (2-p)}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 399, normalized size = 1.89 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (\frac {80 d e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )}{p}+\frac {64 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}+\frac {8 d^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac {d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac {5 e^2 2^{p+4} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {3 e^2 2^{p+3} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {3 e^2 2^{p+1} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {e^2 2^p (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}\right )}{16 d^7} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{4} x^{7} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{5} + 4 \, d^{3} e x^{4} + d^{4} x^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4} x^{3}}\, dx \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^3\,{\left (d+e\,x\right )}^4} \,d x \]
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 \[ \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{3} \left (d + e x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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