Optimal. Leaf size=207 \[ -\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac {4 e^2 \left (16-9 p+p^2\right ) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (1,-2+p;-1+p;1-\frac {e^2 x^2}{d^2}\right )}{d (2-p)} \]
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Rubi [A]
time = 0.18, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {866, 1821,
1666, 457, 80, 67, 396, 252, 251} \begin {gather*} -\frac {2 e \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 )}{d (2-p)}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x}+\frac {4 e^2 \left (p^2-9 p+16\right ) 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^6 (5-2 p)} \end {gather*}
Antiderivative was successfully verified.
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Rule 67
Rule 80
Rule 251
Rule 252
Rule 396
Rule 457
Rule 866
Rule 1666
Rule 1821
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^2} \, dx\\ &=-\frac {d^2 \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 (4 d^5 e-d^4 e^2 (13-2 p) x+4 d^3 e^3 x^2-d^2 e^4 x^3\right )}{x} \, dx}{d^2}\\ &=-\frac {d^2 \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 (4 d^5 e+4 d^3 e^3 x^2\right )}{x} \, dx}{d^2}-\frac {\int \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (13-2 p)-d^2 e^4 x^2\right ) \, dx}{d^2}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p} \left (4 d^5 e+4 d^3 e^3 x\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\frac {\left (4 d^2 e^2 \left (16-9 p+p^2\right )\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{5-2 p}\\ &=-\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-(2 d e) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )+\frac {\left (4 e^2 \left (16-9 p+p^2\right ) \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^6 (5-2 p)}\\ &=-\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac {4 e^2 \left (16-9 p+p^2\right ) 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^6 (5-2 p)}-\frac {2 e \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 )}{d (2-p)}\\ \end {align*}
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Mathematica [A]
time = 0.60, size = 337, normalized size = 1.63 \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^p \left (-16 d^2 p (1+p) \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 )+2^{5+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (1-p,1+p;2+p;\frac {d-e x}{2 d}\right )+3\ 2^{2+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )+2^{2+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )+2^p e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \, _2F_1\left (4-p,1+p;2+p;\frac {d-e x}{2 d}\right )-32 d e (1+p) \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} x \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )\right )}{16 d^6 p (1+p) x} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{2} \left (e x +d \right )^{4}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{2} \left (d + e x\right )^{4}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,{\left (d+e\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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