Optimal. Leaf size=143 \[ -\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]
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Rubi [A] time = 0.31, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {852, 1805, 807, 266, 63, 208} \[ -\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 266
Rule 807
Rule 852
Rule 1805
Rubi steps
\begin {align*} \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^4+20 d^3 e x-27 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^4-60 d^3 e x+64 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^4+60 d^3 e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {(4 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^3}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {(2 e) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{d^3}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^3 e}\\ &=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 92, normalized size = 0.64 \[ -\frac {-60 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (15 d^3+149 d^2 e x+222 d e^2 x^2+94 e^3 x^3\right )}{x (d+e x)^3}+60 e \log (x)}{15 d^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.94, size = 181, normalized size = 1.27 \[ -\frac {104 \, e^{4} x^{4} + 312 \, d e^{3} x^{3} + 312 \, d^{2} e^{2} x^{2} + 104 \, d^{3} e x + 60 \, {\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (94 \, e^{3} x^{3} + 222 \, d e^{2} x^{2} + 149 \, d^{2} e x + 15 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{3} x^{4} + 3 \, d^{5} e^{2} x^{3} + 3 \, d^{6} e x^{2} + d^{7} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 1, normalized size = 0.01 \[ +\infty \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 361, normalized size = 2.52 \[ \frac {4 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{3}}+\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}-\frac {e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x}{d^{6}}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, e}{d^{5}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e}{d^{5}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{5 \left (x +\frac {d}{e}\right )^{4} d^{3} e^{3}}-\frac {11 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{15 \left (x +\frac {d}{e}\right )^{3} d^{4} e^{2}}-\frac {3 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{\left (x +\frac {d}{e}\right )^{2} d^{5} e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{6} x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )}^{4} x^{2}}\,{d x} \]
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 \[ \int \frac {\sqrt {d^2-e^2\,x^2}}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]
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 \[ \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{2} \left (d + e x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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