Optimal. Leaf size=309 \[ -\frac {\sqrt [4]{3} (e x+2)^{3/4} (2-e x)^{5/4}}{2 e}+\frac {3 \sqrt [4]{3} (e x+2)^{3/4} \sqrt [4]{2-e x}}{2 e}+\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}-\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}-\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}+\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}+\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e} \]
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Rubi [A] time = 0.34, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {675, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {\sqrt [4]{3} (e x+2)^{3/4} (2-e x)^{5/4}}{2 e}+\frac {3 \sqrt [4]{3} (e x+2)^{3/4} \sqrt [4]{2-e x}}{2 e}+\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}-\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}-\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}+\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}+\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e} \end {gather*}
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 204
Rule 211
Rule 240
Rule 617
Rule 628
Rule 675
Rule 1162
Rule 1165
Rubi steps
\begin {align*} \int \sqrt {2+e x} \sqrt [4]{12-3 e^2 x^2} \, dx &=\int \sqrt [4]{6-3 e x} (2+e x)^{3/4} \, dx\\ &=-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3}{2} \int \frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}} \, dx\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {9}{2} \int \frac {1}{(6-3 e x)^{3/4} \sqrt [4]{2+e x}} \, dx\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {6 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{4-\frac {x^4}{3}}} \, dx,x,\sqrt [4]{6-3 e x}\right )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {6 \operatorname {Subst}\left (\int \frac {1}{1+\frac {x^4}{3}} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {\sqrt {3} \operatorname {Subst}\left (\int \frac {\sqrt {3}-x^2}{1+\frac {x^4}{3}} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}-\frac {\sqrt {3} \operatorname {Subst}\left (\int \frac {\sqrt {3}+x^2}{1+\frac {x^4}{3}} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {\left (3 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}+2 x}{-\sqrt {3}-\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{2 \sqrt {2} e}+\frac {\left (3 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}-2 x}{-\sqrt {3}+\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{2 \sqrt {2} e}-\frac {\left (3 \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3}-\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{2 e}-\frac {\left (3 \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3}+\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{2 e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {2-e x}-\sqrt {2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e}-\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {2-e x}+\sqrt {2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e}-\frac {\left (3 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}+\frac {\left (3 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}+\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {2-e x}-\sqrt {2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e}-\frac {3 \sqrt [4]{3} \log \left (\frac {\sqrt {2-e x}+\sqrt {2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e}\\ \end {align*}
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Mathematica [C] time = 0.05, size = 60, normalized size = 0.19 \begin {gather*} \frac {8 \sqrt {2} (e x-2) \sqrt [4]{12-3 e^2 x^2} \, _2F_1\left (-\frac {3}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2}-\frac {e x}{4}\right )}{5 e \sqrt [4]{e x+2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.72, size = 228, normalized size = 0.74 \begin {gather*} \frac {\sqrt [4]{4 (e x+2)-(e x+2)^2} \left (\sqrt [4]{3} (e x+2)^{3/2}-\sqrt [4]{3} \sqrt {e x+2}\right )}{2 e}+\frac {3 \sqrt [4]{3} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x+2} \sqrt [4]{4 (e x+2)-(e x+2)^2}}{-e x+\sqrt {4 (e x+2)-(e x+2)^2}-2}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \tanh ^{-1}\left (\frac {\frac {e x+2}{\sqrt {2}}+\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {2}}}{\sqrt {e x+2} \sqrt [4]{4 (e x+2)-(e x+2)^2}}\right )}{\sqrt {2} e} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 542, normalized size = 1.75 \begin {gather*} \frac {12 \cdot 3^{\frac {1}{4}} \sqrt {2} e \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {3^{\frac {3}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} - 3^{\frac {3}{4}} \sqrt {2} {\left (e^{4} x + 2 \, e^{3}\right )} \sqrt {\frac {3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} + \sqrt {3} {\left (e^{3} x + 2 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + \sqrt {-3 \, e^{2} x^{2} + 12}}{e x + 2}} \frac {1}{e^{4}}^{\frac {3}{4}} + 3 \, e x + 6}{3 \, {\left (e x + 2\right )}}\right ) + 12 \cdot 3^{\frac {1}{4}} \sqrt {2} e \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {3^{\frac {3}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} - 3^{\frac {3}{4}} \sqrt {2} {\left (e^{4} x + 2 \, e^{3}\right )} \sqrt {-\frac {3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} - \sqrt {3} {\left (e^{3} x + 2 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - \sqrt {-3 \, e^{2} x^{2} + 12}}{e x + 2}} \frac {1}{e^{4}}^{\frac {3}{4}} - 3 \, e x - 6}{3 \, {\left (e x + 2\right )}}\right ) - 3 \cdot 3^{\frac {1}{4}} \sqrt {2} e \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (\frac {3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} + \sqrt {3} {\left (e^{3} x + 2 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + \sqrt {-3 \, e^{2} x^{2} + 12}}{e x + 2}\right ) + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} e \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (-\frac {3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} - \sqrt {3} {\left (e^{3} x + 2 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - \sqrt {-3 \, e^{2} x^{2} + 12}}{e x + 2}\right ) + 2 \, {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2} {\left (e x + 1\right )}}{4 \, e} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.15, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {e x +2}\, \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}\, dx \end {gather*}
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 \begin {gather*} \int {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2}\,{d x} \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 = -0.00 \begin {gather*} \int {\left (12-3\,e^2\,x^2\right )}^{1/4}\,\sqrt {e\,x+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*} \sqrt [4]{3} \int \sqrt {e x + 2} \sqrt [4]{- e^{2} x^{2} + 4}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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