Optimal. Leaf size=110 \[ \frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]
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Rubi [A] time = 0.16, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {852, 1807, 813, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 203
Rule 208
Rule 217
Rule 266
Rule 813
Rule 844
Rule 852
Rule 1807
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^2} \, dx &=\int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^3} \, dx\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}-\frac {\int \frac {\left (4 d^3 e-d^2 e^2 x\right ) \sqrt {d^2-e^2 x^2}}{x^2} \, dx}{2 d^2}\\ &=\frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac {\int \frac {2 d^4 e^2+8 d^3 e^3 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{4 d^2}\\ &=\frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac {1}{2} \left (d^2 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx+\left (2 d e^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} \left (d^2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )+\left (2 d e^3\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 {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d^2 \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 )\\ &=\frac {e (4 d+e x) \sqrt {d^2-e^2 x^2}}{2 x}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^2}+2 d e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}
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Mathematica [A] time = 0.14, size = 102, normalized size = 0.93 \begin {gather*} \left (-\frac {d^2}{2 x^2}+\frac {2 d e}{x}+e^2\right ) \sqrt {d^2-e^2 x^2}-\frac {1}{2} d e^2 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+2 d e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{2} d e^2 \log (x) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.50, size = 128, normalized size = 1.16 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-d^2+4 d e x+2 e^2 x^2\right )}{2 x^2}+2 d \sqrt {-e^2} e \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )+d e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.41, size = 119, normalized size = 1.08 \begin {gather*} -\frac {8 \, d e^{2} x^{2} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - d e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 2 \, d e^{2} x^{2} - {\left (2 \, e^{2} x^{2} + 4 \, d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, x^{2}} \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 [B] time = 0.01, size = 456, normalized size = 4.15 \begin {gather*} -\frac {d^{2} e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}-\frac {7 d \,e^{3} \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 )}{4 \sqrt {e^{2}}}+\frac {15 d \,e^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{4 \sqrt {e^{2}}}+\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x}{4 d}-\frac {7 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{3} x}{4 d}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}{2}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3} x}{2 d^{3}}-\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{3} x}{6 d^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{6 d^{2}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3} x}{d^{5}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{10 d^{4}}-\frac {14 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{2}}{15 d^{4}}-\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^{4}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{d^{5} x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{2 d^{4} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.99, size = 111, normalized size = 1.01 \begin {gather*} 2 \, d e^{2} \arcsin \left (\frac {e x}{d}\right ) - \frac {1}{2} \, d e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e}{x} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{2 \, x^{2}} \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 {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^3\,{\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 [C] time = 10.17, size = 347, normalized size = 3.15 \begin {gather*} d^{2} \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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