Optimal. Leaf size=326 \[ \frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{2 d (\sec (c+d x)+1)}-\frac{3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
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Rubi [A] time = 0.269234, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3828, 3827, 50, 63, 225} \[ \frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{2 d (\sec (c+d x)+1)}-\frac{3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
Antiderivative was successfully verified.
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Rule 3828
Rule 3827
Rule 50
Rule 63
Rule 225
Rubi steps
\begin{align*} \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx &=\frac{(a+a \sec (c+d x))^{2/3} \int \sec (c+d x) (1+\sec (c+d x))^{2/3} \, dx}{(1+\sec (c+d x))^{2/3}}\\ &=-\frac{\left ((a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [6]{1+x}}{\sqrt{1-x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{2 d (1+\sec (c+d x))}-\frac{\left ((a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{2 d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{2 d (1+\sec (c+d x))}-\frac{\left (3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{2 d (1+\sec (c+d x))}-\frac{3^{3/4} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt{\frac{2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}\\ \end{align*}
Mathematica [C] time = 0.0492302, size = 66, normalized size = 0.2 \[ \frac{2 \sqrt [6]{2} \tan (c+d x) (a (\sec (c+d x)+1))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{3}{2},\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.092, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( a+a\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{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )\,{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 ({\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \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 \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{2}{3}} \sec{\left (c + d x \right )}\, 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{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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