Optimal. Leaf size=140 \[ \frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2} \]
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Rubi [A] time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1639, 12, 785, 780, 195, 217, 203} \[ \frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 195
Rule 203
Rule 217
Rule 780
Rule 785
Rule 1639
Rubi steps
\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\int \frac {7 d e^3 x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{7 e^4}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {d \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{e}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac {\int x \left (d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}\\ &=\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^3 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e^2}\\ &=\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^5 \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \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 {d^5 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac {d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac {d^7 \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 = 0.10, size = 113, normalized size = 0.81 \[ \frac {105 d^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (96 d^6-105 d^5 e x+48 d^4 e^2 x^2+490 d^3 e^3 x^3-384 d^2 e^4 x^4-280 d e^5 x^5+240 e^6 x^6\right )}{1680 e^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 117, normalized size = 0.84 \[ -\frac {210 \, d^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (240 \, e^{6} x^{6} - 280 \, d e^{5} x^{5} - 384 \, d^{2} e^{4} x^{4} + 490 \, d^{3} e^{3} x^{3} + 48 \, d^{4} e^{2} x^{2} - 105 \, d^{5} e x + 96 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{1680 \, e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 282, normalized size = 2.01 \[ \frac {3 d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {5 d^{7} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{2}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} x}{16 e^{2}}+\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{5} x}{8 e^{2}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x}{24 e^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{3} x}{4 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}{6 e^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d^{2}}{5 e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{7 e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.04, size = 198, normalized size = 1.41 \[ -\frac {3 i \, d^{7} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} - \frac {5 \, d^{7} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5} x}{8 \, e^{2}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} x}{16 \, e^{2}} + \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{6}}{4 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} x}{24 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x}{6 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{5 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{7 \, e^{3}} \]
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 \[ \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{d+e\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 16.66, size = 653, normalized size = 4.66 \[ d^{3} \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 ) - d^{2} 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 ) - d 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 ) + e^{3} \left (\begin {cases} - \frac {8 d^{6} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{6}} - \frac {4 d^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{4}} - \frac {d^{2} x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{35 e^{2}} + \frac {x^{6} \sqrt {d^{2} - e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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