Optimal. Leaf size=123 \[ \frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]
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Rubi [A] time = 0.24, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {852, 1635, 1814, 12, 217, 203} \begin {gather*} -\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 203
Rule 217
Rule 852
Rule 1635
Rule 1814
Rubi steps
\begin {align*} \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {x^4 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x) \left (\frac {2 d^4}{e^4}-\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}-\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {11 d^4}{e^4}-\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 d^4}{e^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^4}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 106, normalized size = 0.86 \begin {gather*} \sqrt {d^2-e^2 x^2} \left (-\frac {d^2}{10 e^5 (d+e x)^3}+\frac {31 d}{60 e^5 (d+e x)^2}-\frac {1}{8 e^5 (e x-d)}-\frac {193}{120 e^5 (d+e x)}\right )-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.67, size = 114, normalized size = 0.93 \begin {gather*} -\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^6}-\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-17 d^2 e x+22 d e^2 x^2+26 e^3 x^3\right )}{15 e^5 (e x-d) (d+e x)^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.41, size = 171, normalized size = 1.39 \begin {gather*} -\frac {16 \, e^{4} x^{4} + 32 \, d e^{3} x^{3} - 32 \, d^{3} e x - 16 \, d^{4} - 30 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (26 \, e^{3} x^{3} + 22 \, d e^{2} x^{2} - 17 \, d^{2} e x - 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{9} x^{4} + 2 \, d e^{8} x^{3} - 2 \, d^{3} e^{6} x - d^{4} e^{5}\right )}} \end {gather*}
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 \begin {gather*} \mathit {sage}_{0} x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 198, normalized size = 1.61 \begin {gather*} \frac {4 x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {34 x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{4}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}}-\frac {d^{3}}{5 \left (x +\frac {d}{e}\right )^{2} \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{7}}+\frac {17 d^{2}}{15 \left (x +\frac {d}{e}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{6}}-\frac {2 d}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.02, size = 170, normalized size = 1.38 \begin {gather*} -\frac {d^{3}}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{7} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{5}\right )}} + \frac {17 \, d^{2}}{15 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{5}\right )}} + \frac {26 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{5}} - \frac {2 \, d}{\sqrt {-e^{2} x^{2} + d^{2}} 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}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,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}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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