Optimal. Leaf size=326 \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt{3}\right )}{2 d e (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{3 a \tan (c+d x)}{2 d \sqrt{a \sec (c+d x)+a} (e \sec (c+d x))^{2/3}} \]
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Rubi [A] time = 0.243619, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3806, 51, 63, 218} \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{2 d e (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{3 a \tan (c+d x)}{2 d \sqrt{a \sec (c+d x)+a} (e \sec (c+d x))^{2/3}} \]
Antiderivative was successfully verified.
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Rule 3806
Rule 51
Rule 63
Rule 218
Rubi steps
\begin{align*} \int \frac{\sqrt{a+a \sec (c+d x)}}{(e \sec (c+d x))^{2/3}} \, dx &=-\frac{\left (a^2 e \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{(e x)^{5/3} \sqrt{a-a x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{a-a \sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{3 a \tan (c+d x)}{2 d (e \sec (c+d x))^{2/3} \sqrt{a+a \sec (c+d x)}}-\frac{\left (a^2 \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{(e x)^{2/3} \sqrt{a-a x}} \, dx,x,\sec (c+d x)\right )}{4 d \sqrt{a-a \sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{3 a \tan (c+d x)}{2 d (e \sec (c+d x))^{2/3} \sqrt{a+a \sec (c+d x)}}-\frac{\left (3 a^2 \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-\frac{a x^3}{e}}} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{4 d e \sqrt{a-a \sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{3 a \tan (c+d x)}{2 d (e \sec (c+d x))^{2/3} \sqrt{a+a \sec (c+d x)}}+\frac{3^{3/4} \sqrt{2+\sqrt{3}} a^2 F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{2 d e (a-a \sec (c+d x)) \sqrt{a+a \sec (c+d x)} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}\\ \end{align*}
Mathematica [C] time = 0.148028, size = 71, normalized size = 0.22 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{2}{3}}(c+d x) \sqrt{a (\sec (c+d x)+1)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{3},\frac{3}{2},1-\sec (c+d x)\right )}{d (e \sec (c+d x))^{2/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.24, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{a+a\sec \left ( dx+c \right ) } \left ( e\sec \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sec \left (d x + c\right ) + a}}{\left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{1}{3}}}{e \sec \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{2}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sec \left (d x + c\right ) + a}}{\left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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