Optimal. Leaf size=207 \[ -\frac {2 e \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{d (2-p)}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x}+\frac {4 e^2 \left (p^2-9 p+16\right ) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)} \]
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Rubi [A] time = 0.28, 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 = {852, 1807, 1652, 446, 79, 65, 388, 246, 245} \[ \frac {4 e^2 \left (p^2-9 p+16\right ) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac {2 e \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac {e^2 x^2}{d^2}\right )}{d (2-p)}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x} \]
Antiderivative was successfully verified.
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Rule 65
Rule 79
Rule 245
Rule 246
Rule 388
Rule 446
Rule 852
Rule 1652
Rule 1807
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 {\operatorname {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) \operatorname {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 = 0.44, size = 337, normalized size = 1.63 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (-32 d e (p+1) x \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )-16 d^2 p (p+1) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )+e 2^{p+5} p x (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )+3 e 2^{p+2} p x (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )+e 2^{p+2} p x (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )+e 2^p p x (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac {d-e x}{2 d}\right )\right )}{16 d^6 p (p+1) x} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{4} x^{6} + 4 \, d e^{3} x^{5} + 6 \, d^{2} e^{2} x^{4} + 4 \, d^{3} e x^{3} + d^{4} x^{2}}, x\right ) \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{2}}\,{d x} \]
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 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4} x^{2}}\, dx \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{2}}\,{d x} \]
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 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]
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 \[ \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{2} \left (d + e x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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