Integrand size = 24, antiderivative size = 141 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (2+e x)^{5/2}} \]
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Time = 0.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 665} \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (e x+2)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (e x+2)^{7/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (e x+2)^{9/2}}-\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (e x+2)^{11/2}} \]
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Rule 665
Rule 673
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}+\frac {3}{17} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{9/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}+\frac {6}{221} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}+\frac {2}{663} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (2+e x)^{5/2}} \\ \end{align*}
Time = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {(-2+e x) \sqrt [4]{4-e^2 x^2} \left (341+109 e x+22 e^2 x^2+2 e^3 x^3\right )}{1105\ 3^{3/4} e (2+e x)^{9/2}} \]
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Time = 2.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.37
method | result | size |
gosper | \(\frac {\left (e x -2\right ) \left (2 e^{3} x^{3}+22 x^{2} e^{2}+109 e x +341\right ) \left (-3 x^{2} e^{2}+12\right )^{\frac {1}{4}}}{3315 \left (e x +2\right )^{\frac {9}{2}} e}\) | \(52\) |
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Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {{\left (2 \, e^{4} x^{4} + 18 \, e^{3} x^{3} + 65 \, e^{2} x^{2} + 123 \, e x - 682\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2}}{3315 \, {\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \]
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Timed out. \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}}}{{\left (e x + 2\right )}^{\frac {11}{2}}} \,d x } \]
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Exception generated. \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\text {Exception raised: TypeError} \]
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Time = 10.51 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {{\left (12-3\,e^2\,x^2\right )}^{1/4}\,\left (\frac {41\,x}{1105\,e^4}-\frac {682}{3315\,e^5}+\frac {2\,x^4}{3315\,e}+\frac {6\,x^3}{1105\,e^2}+\frac {x^2}{51\,e^3}\right )}{\frac {16\,\sqrt {e\,x+2}}{e^4}+x^4\,\sqrt {e\,x+2}+\frac {32\,x\,\sqrt {e\,x+2}}{e^3}+\frac {8\,x^3\,\sqrt {e\,x+2}}{e}+\frac {24\,x^2\,\sqrt {e\,x+2}}{e^2}} \]
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