Integrand size = 24, antiderivative size = 55 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=-\frac {\sqrt {3} \sqrt {2-e x}}{e (2+e x)}+\frac {\sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{2 e} \]
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Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {641, 43, 65, 212} \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\frac {\sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{2 e}-\frac {\sqrt {3} \sqrt {2-e x}}{e (e x+2)} \]
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Rule 43
Rule 65
Rule 212
Rule 641
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {6-3 e x}}{(2+e x)^2} \, dx \\ & = -\frac {\sqrt {3} \sqrt {2-e x}}{e (2+e x)}-\frac {3}{2} \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx \\ & = -\frac {\sqrt {3} \sqrt {2-e x}}{e (2+e x)}+\frac {\text {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{e} \\ & = -\frac {\sqrt {3} \sqrt {2-e x}}{e (2+e x)}+\frac {\sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{2 e} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\frac {\sqrt {3} \left (-\frac {2 \sqrt {4-e^2 x^2}}{(2+e x)^{3/2}}+\text {arctanh}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{2 e} \]
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Time = 2.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {\sqrt {-x^{2} e^{2}+4}\, \left (\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x +2 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )-2 \sqrt {-3 e x +6}\right ) \sqrt {3}}{2 \left (e x +2\right )^{\frac {3}{2}} \sqrt {-3 e x +6}\, e}\) | \(86\) |
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Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (43) = 86\).
Time = 0.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.11 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\frac {\sqrt {3} {\left (e^{2} x^{2} + 4 \, e x + 4\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x - 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{4 \, {\left (e^{3} x^{2} + 4 \, e^{2} x + 4 \, e\right )}} \]
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\[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\sqrt {3} \int \frac {\sqrt {- e^{2} x^{2} + 4}}{e^{2} x^{2} \sqrt {e x + 2} + 4 e x \sqrt {e x + 2} + 4 \sqrt {e x + 2}}\, dx \]
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\[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\int { \frac {\sqrt {-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac {5}{2}}} \,d x } \]
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none
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=-\frac {\sqrt {3} {\left (\frac {4 \, \sqrt {-e x + 2}}{e x + 2} - \log \left (\sqrt {-e x + 2} + 2\right ) + \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{4 \, e} \]
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Timed out. \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx=\int \frac {\sqrt {12-3\,e^2\,x^2}}{{\left (e\,x+2\right )}^{5/2}} \,d x \]
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