Optimal. Leaf size=220 \[ -\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p-2}}{2 p+1}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^5 (p+1)}-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p-1}}{2 e^5 (1-p)}+\frac {2 (p+8) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {5}{2},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (2 p+1)} \]
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Rubi [A] time = 0.23, antiderivative size = 220, 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, 1652, 459, 365, 364, 446, 77} \[ \frac {2 (p+8) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {5}{2},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (2 p+1)}-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p-2}}{2 p+1}-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p-1}}{2 e^5 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e^5 (p+1)} \]
Antiderivative was successfully verified.
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Rule 77
Rule 364
Rule 365
Rule 446
Rule 459
Rule 852
Rule 1652
Rubi steps
\begin {align*} \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx &=\int x^4 (d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p} \, dx\\ &=\int x^4 \left (d^2-e^2 x^2\right )^{-3+p} \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^5 \left (d^2-e^2 x^2\right )^{-3+p} \left (-3 d^2 e-e^3 x^2\right ) \, dx\\ &=-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {1}{2} \operatorname {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-3+p} \left (-3 d^2 e-e^3 x\right ) \, dx,x,x^2\right )+\frac {\left (2 d^3 (8+p)\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{1+2 p}\\ &=-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {4 d^6 \left (d^2-e^2 x\right )^{-3+p}}{e^3}+\frac {9 d^4 \left (d^2-e^2 x\right )^{-2+p}}{e^3}-\frac {6 d^2 \left (d^2-e^2 x\right )^{-1+p}}{e^3}+\frac {\left (d^2-e^2 x\right )^p}{e^3}\right ) \, dx,x,x^2\right )+\frac {\left (2 (8+p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-3+p} \, dx}{d^3 (1+2 p)}\\ &=-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{-2+p}}{1+2 p}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{2 e^5 (1-p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^5 (1+p)}+\frac {2 (8+p) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},3-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^3 (1+2 p)}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 245, normalized size = 1.11 \[ -\frac {2^{p-3} \left (\frac {e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (24 d e (p+1) x \left (\frac {e x}{2 d}+\frac {1}{2}\right )^p \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (24 d \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )-8 d \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )+d \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )+4 d \left (\frac {e x}{2 d}+\frac {1}{2}\right )^p+4 e x \left (\frac {e x}{2 d}+\frac {1}{2}\right )^p\right )\right )}{e^5 (p+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3}}\, dx \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}}\,{d x} \]
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 \[ \int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]
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 \[ \int \frac {x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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