Optimal. Leaf size=147 \[ \frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5} \]
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Rubi [A] time = 0.14, antiderivative size = 147, 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 = {850, 833, 780, 217, 203} \begin {gather*} \frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5} \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 217
Rule 780
Rule 833
Rule 850
Rubi steps
\begin {align*} \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx &=\int \frac {x^4 (d-e x)}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x^3 \left (4 d^2 e-5 d e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {\int \frac {x^2 \left (15 d^3 e^2-16 d^2 e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{20 e^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x \left (32 d^4 e^3-45 d^3 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{60 e^6}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\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^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5}\\ \end {align*}
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Mathematica [A] time = 0.14, size = 91, normalized size = 0.62 \begin {gather*} \frac {45 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (64 d^4-45 d^3 e x+32 d^2 e^2 x^2-30 d e^3 x^3+24 e^4 x^4\right )}{120 e^5} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.26, size = 114, normalized size = 0.78 \begin {gather*} \frac {3 d^5 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^6}+\frac {\sqrt {d^2-e^2 x^2} \left (64 d^4-45 d^3 e x+32 d^2 e^2 x^2-30 d e^3 x^3+24 e^4 x^4\right )}{120 e^5} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.41, size = 95, normalized size = 0.65 \begin {gather*} -\frac {90 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 32 \, d^{2} e^{2} x^{2} - 45 \, d^{3} e x + 64 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 77, normalized size = 0.52 \begin {gather*} \frac {3}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) + \frac {1}{120} \, {\left (64 \, d^{4} e^{\left (-5\right )} - {\left (45 \, d^{3} e^{\left (-4\right )} - 2 \, {\left (16 \, d^{2} e^{\left (-3\right )} + 3 \, {\left (4 \, x e^{\left (-1\right )} - 5 \, d e^{\left (-2\right )}\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 208, normalized size = 1.41 \begin {gather*} \frac {d^{5} \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 )}{\sqrt {e^{2}}\, e^{4}}-\frac {5 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{4}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x}{8 e^{4}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4}}{e^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} x^{2}}{5 e^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d x}{4 e^{4}}-\frac {7 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2}}{15 e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.99, size = 125, normalized size = 0.85 \begin {gather*} \frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{5}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{5}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{4}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{15 \, e^{5}} \end {gather*}
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 \begin {gather*} \int \frac {x^4\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \end {gather*}
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 \begin {gather*} \int \frac {x^{4} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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