Optimal. Leaf size=142 \[ \frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2} \]
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Rubi [A] time = 0.18, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1809, 833, 780, 195, 217, 203} \[ \frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \]
Antiderivative was successfully verified.
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Rule 195
Rule 203
Rule 217
Rule 780
Rule 833
Rule 852
Rule 1809
Rubi steps
\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int x^2 (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx\\ &=-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int x^2 \left (-9 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^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {\int x \left (-24 d^3 e^3+45 d^2 e^4 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{30 e^4}\\ &=\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^4\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 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^2}\\ &=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 102, normalized size = 0.72 \[ \frac {45 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (64 d^5-45 d^4 e x+32 d^3 e^2 x^2+50 d^2 e^3 x^3-96 d e^4 x^4+40 e^5 x^5\right )}{240 e^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 106, normalized size = 0.75 \[ -\frac {90 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (40 \, e^{5} x^{5} - 96 \, d e^{4} x^{4} + 50 \, d^{2} e^{3} x^{3} + 32 \, d^{3} e^{2} x^{2} - 45 \, d^{4} e x + 64 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{3}} \]
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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 303, normalized size = 2.13 \[ -\frac {d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}+\frac {5 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{2}}+\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x}{16 e^{2}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4} x}{8 e^{2}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2} x}{24 e^{2}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{2} x}{12 e^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x}{6 e^{2}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d}{15 e^{3}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} d}{3 \left (x +\frac {d}{e}\right )^{2} e^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.03, size = 230, normalized size = 1.62 \[ \frac {i \, d^{6} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} + \frac {5 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{4 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x}{8 \, e^{2}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{2}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{4 \, e^{3}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e^{2}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{12 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{3}} \]
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 \[ \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 14.45, size = 541, normalized size = 3.81 \[ d^{2} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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