Optimal. Leaf size=182 \[ -\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3} \]
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Rubi [A] time = 0.22, antiderivative size = 182, 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 = {852, 1635, 795, 671, 641, 217, 203} \[ -\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
Antiderivative was successfully verified.
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Rule 203
Rule 217
Rule 641
Rule 671
Rule 795
Rule 852
Rule 1635
Rubi steps
\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\int \frac {x^2 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\left (\frac {4 d^2}{e^2}-\frac {d x}{e}\right ) (d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {(19 d) \int \frac {(d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^2\right ) \int \frac {(d-e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx}{12 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^3\right ) \int \frac {d-e x}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 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 )}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 103, normalized size = 0.57 \[ \sqrt {d^2-e^2 x^2} \left (-\frac {8 d^4}{e^3 (d+e x)}-\frac {32 d^3}{3 e^3}+\frac {31 d^2 x}{8 e^2}-\frac {4 d x^2}{3 e}+\frac {x^3}{4}\right )-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.80, size = 124, normalized size = 0.68 \[ -\frac {448 \, d^{4} e x + 448 \, d^{5} - 570 \, {\left (d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, e^{4} x^{4} - 26 \, d e^{3} x^{3} + 61 \, d^{2} e^{2} x^{2} - 163 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, {\left (e^{4} x + d e^{3}\right )}} \]
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 [A] time = 0.01, size = 288, normalized size = 1.58 \[ -\frac {95 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {95 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2} x}{8 e^{2}}-\frac {95 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{12 e^{2}}-\frac {19 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{3 d \,e^{3}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} d}{\left (x +\frac {d}{e}\right )^{4} e^{7}}-\frac {19 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d \,e^{5}}-\frac {5 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{3} e^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.02, size = 363, normalized size = 1.99 \[ \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{2 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{4} x + d e^{3}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{3 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {5 i \, d^{4} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} - \frac {25 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (e^{4} x + d e^{3}\right )}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x}{8 \, e^{2}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e^{3}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{3}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{12 \, e^{3}} \]
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 \[ \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]
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 \[ \int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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