Optimal. Leaf size=340 \[ -\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {5\ 3^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \log \left (\frac {\sqrt {6-3 e x}-\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {3} \sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e}+\frac {5\ 3^{3/4} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt {3} \sqrt {2+e x}}{\sqrt {2+e x}}\right )}{2 \sqrt {2} e} \]
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Rubi [A]
time = 0.22, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {689, 52, 65,
338, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {5\ 3^{3/4} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e}-\frac {(2-e x)^{3/4} (e x+2)^{9/4}}{3 \sqrt [4]{3} e}-\frac {3^{3/4} (2-e x)^{3/4} (e x+2)^{5/4}}{2 e}-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{e x+2}}{2 e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 52
Rule 65
Rule 210
Rule 303
Rule 338
Rule 631
Rule 642
Rule 689
Rule 1176
Rule 1179
Rubi steps
\begin {align*} \int \frac {(2+e x)^{5/2}}{\sqrt [4]{12-3 e^2 x^2}} \, dx &=\int \frac {(2+e x)^{9/4}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+3 \int \frac {(2+e x)^{5/4}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {15}{2} \int \frac {\sqrt [4]{2+e x}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {15}{2} \int \frac {1}{\sqrt [4]{6-3 e x} (2+e x)^{3/4}} \, dx\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {10 \text {Subst}\left (\int \frac {x^2}{\left (4-\frac {x^4}{3}\right )^{3/4}} \, dx,x,\sqrt [4]{6-3 e x}\right )}{e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {10 \text {Subst}\left (\int \frac {x^2}{1+\frac {x^4}{3}} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {5 \text {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 {5 \text {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 {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {15 \text {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 {15 \text {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 (5\ 3^{3/4}\right ) \text {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 (5\ 3^{3/4}\right ) \text {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 {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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 (5\ 3^{3/4}\right ) \text {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 (5\ 3^{3/4}\right ) \text {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 {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {5\ 3^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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 [A]
time = 0.56, size = 170, normalized size = 0.50 \begin {gather*} \frac {-\left (4-e^2 x^2\right )^{3/4} \left (71+17 e x+2 e^2 x^2\right )-45 \sqrt {4+2 e x} \tan ^{-1}\left (\frac {\sqrt {4+2 e x} \sqrt [4]{4-e^2 x^2}}{2+e x-\sqrt {4-e^2 x^2}}\right )+45 \sqrt {4+2 e x} \tanh ^{-1}\left (\frac {2+e x+\sqrt {4-e^2 x^2}}{\sqrt {4+2 e x} \sqrt [4]{4-e^2 x^2}}\right )}{6 \sqrt [4]{3} e \sqrt {2+e x}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +2\right )^{\frac {5}{2}}}{\left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}\, dx\]
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 571 vs.
\(2 (256) = 512\).
time = 3.58, size = 571, normalized size = 1.68 \begin {gather*} \frac {180 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (x e^{2} + 2 \, e\right )} \arctan \left (\frac {27^{\frac {3}{4}} \sqrt {2} {\left (x^{2} e^{5} - 4 \, e^{3}\right )} \sqrt {\frac {3 \, \sqrt {3} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} + 27^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 3 \, \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}} e^{\left (-3\right )} - 27 \, x^{2} e^{2} - 27^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} + 108}{27 \, {\left (x^{2} e^{2} - 4\right )}}\right ) e^{\left (-1\right )} + 180 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (x e^{2} + 2 \, e\right )} \arctan \left (\frac {27^{\frac {3}{4}} \sqrt {2} {\left (x^{2} e^{5} - 4 \, e^{3}\right )} \sqrt {\frac {3 \, \sqrt {3} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} - 27^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 3 \, \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}} e^{\left (-3\right )} + 27 \, x^{2} e^{2} - 27^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 108}{27 \, {\left (x^{2} e^{2} - 4\right )}}\right ) e^{\left (-1\right )} - 45 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (x e^{2} + 2 \, e\right )} e^{\left (-1\right )} \log \left (\frac {3 \, \sqrt {3} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} + 27^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 3 \, \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}\right ) + 45 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (x e^{2} + 2 \, e\right )} e^{\left (-1\right )} \log \left (\frac {3 \, \sqrt {3} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} - 27^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 3 \, \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}\right ) - 2 \, {\left (2 \, x^{2} e^{2} + 17 \, x e + 71\right )} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2}}{36 \, {\left (x e^{2} + 2 \, e\right )}} \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*} \frac {3^{\frac {3}{4}} \left (\int \frac {4 \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx + \int \frac {4 e x \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx + \int \frac {e^{2} x^{2} \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx\right )}{3} \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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x+2\right )}^{5/2}}{{\left (12-3\,e^2\,x^2\right )}^{1/4}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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