Optimal. Leaf size=265 \[ \frac {d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac {4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {5}{2},4-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)} \]
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Rubi [A] time = 0.31, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {852, 1652, 1267, 459, 365, 364, 446, 77} \[ -\frac {4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {5}{2},4-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}+\frac {d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p} \]
Antiderivative was successfully verified.
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Rule 77
Rule 364
Rule 365
Rule 446
Rule 459
Rule 852
Rule 1267
Rule 1652
Rubi steps
\begin {align*} \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx &=\int x^4 (d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx\\ &=\int x^5 \left (d^2-e^2 x^2\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x^2\right ) \, dx+\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (d^4+6 d^2 e^2 x^2+e^4 x^4\right ) \, dx\\ &=-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \operatorname {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x\right ) \, dx,x,x^2\right )-\frac {\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (1+2 p)-d^2 e^4 (13+12 p) x^2\right ) \, dx}{e^2 (1+2 p)}\\ &=\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {8 d^7 \left (d^2-e^2 x\right )^{-4+p}}{e^3}+\frac {20 d^5 \left (d^2-e^2 x\right )^{-3+p}}{e^3}-\frac {16 d^3 \left (d^2-e^2 x\right )^{-2+p}}{e^3}+\frac {4 d \left (d^2-e^2 x\right )^{-1+p}}{e^3}\right ) \, dx,x,x^2\right )-\frac {\left (4 d^4 \left (16+15 p+p^2\right )\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{1-4 p^2}\\ &=-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (4 \left (16+15 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 x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-4+p} \, dx}{d^4 \left (1-4 p^2\right )}\\ &=-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 \left (16+15 p+p^2\right ) 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},4-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 231, normalized size = 0.87 \[ \frac {2^{p-4} \left (\frac {e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (16 e (p+1) x \left (\frac {e x}{2 d}+\frac {1}{2}\right )^p \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (32 \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )-24 \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )+8 \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )-\, _2F_1\left (4-p,p+1;p+2;\frac {d-e x}{2 d}\right )\right )\right )}{e^5 (p+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{4}}\,{d x} \]
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 \[ \int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}\, dx \]
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 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{4}}\,{d x} \]
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 \[ \int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \]
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 \[ \int \frac {x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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