Optimal. Leaf size=108 \[ \frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {5 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]
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Rubi [A] time = 0.04, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {655, 671, 641, 195, 217, 203} \[ \frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {5 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]
Antiderivative was successfully verified.
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Rule 195
Rule 203
Rule 217
Rule 641
Rule 655
Rule 671
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{4} (5 d) \int (d-e x) \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{4} \left (5 d^2\right ) \int \sqrt {d^2-e^2 x^2} \, dx\\ &=\frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{8} \left (5 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{8} \left (5 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=\frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {5 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 80, normalized size = 0.74 \[ \frac {15 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (16 d^3+9 d^2 e x-16 d e^2 x^2+6 e^3 x^3\right )}{24 e} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 84, normalized size = 0.78 \[ -\frac {30 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, e^{3} x^{3} - 16 \, d e^{2} x^{2} + 9 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \]
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 = 194, normalized size = 1.80 \[ \frac {5 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}+\frac {5 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2} x}{8}+\frac {5 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{12}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{3 d e}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d \,e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.99, size = 119, normalized size = 1.10 \[ -\frac {5 i \, d^{4} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e} + \frac {5}{8} \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (e^{2} x + d e\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{12 \, e} \]
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 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 9.35, size = 350, normalized size = 3.24 \[ d^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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