Optimal. Leaf size=173 \[ -\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3} \]
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Rubi [A] time = 0.23, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1809, 833, 780, 217, 203} \[ -\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5} \]
Antiderivative was successfully verified.
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Rule 203
Rule 217
Rule 780
Rule 833
Rule 1809
Rubi steps
\begin {align*} \int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^4 \left (-11 d^2 e^2-12 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{6 e^2}\\ &=-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x^3 \left (48 d^3 e^3+55 d^2 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{30 e^4}\\ &=-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^2 \left (-165 d^4 e^4-192 d^3 e^5 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{120 e^6}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x \left (384 d^5 e^5+495 d^4 e^6 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{360 e^8}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^4}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^4}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 103, normalized size = 0.60 \[ \frac {165 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (256 d^5+165 d^4 e x+128 d^3 e^2 x^2+110 d^2 e^3 x^3+96 d e^4 x^4+40 e^5 x^5\right )}{240 e^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.96, size = 105, normalized size = 0.61 \[ -\frac {330 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 96 \, d e^{4} x^{4} + 110 \, d^{2} e^{3} x^{3} + 128 \, d^{3} e^{2} x^{2} + 165 \, d^{4} e x + 256 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 84, normalized size = 0.49 \[ \frac {11}{16} \, d^{6} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) - \frac {1}{240} \, {\left (256 \, d^{5} e^{\left (-5\right )} + {\left (165 \, d^{4} e^{\left (-4\right )} + 2 \, {\left (64 \, d^{3} e^{\left (-3\right )} + {\left (55 \, d^{2} e^{\left (-2\right )} + 4 \, {\left (12 \, d e^{\left (-1\right )} + 5 \, x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 174, normalized size = 1.01 \[ -\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{5}}{6}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,x^{4}}{5 e}+\frac {11 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{4}}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x^{3}}{24 e^{2}}-\frac {8 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x^{2}}{15 e^{3}}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x}{16 e^{4}}-\frac {16 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}}{15 e^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.98, size = 153, normalized size = 0.88 \[ -\frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{5} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d x^{4}}{5 \, e} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x^{3}}{24 \, e^{2}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x^{2}}{15 \, e^{3}} + \frac {11 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{5}} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{4}} - \frac {16 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{15 \, e^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,{\left (d+e\,x\right )}^2}{\sqrt {d^2-e^2\,x^2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 13.48, size = 558, normalized size = 3.23 \[ d^{2} \left (\begin {cases} - \frac {3 i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{5}} + \frac {3 i d^{3} x}{8 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{3}}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{5}} - \frac {3 d^{3} x}{8 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{3}}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {8 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{6}} - \frac {4 d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {5 i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{7}} + \frac {5 i d^{5} x}{16 e^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d^{3} x^{3}}{48 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{5}}{24 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {5 d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{7}} - \frac {5 d^{5} x}{16 e^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d^{3} x^{3}}{48 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{5}}{24 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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