Optimal. Leaf size=336 \[ -\frac{7\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 a d}-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.394185, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3800, 4001, 3828, 3827, 63, 225} \[ \frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 a d}-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a \sec (c+d x)+a}}-\frac{7\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 3800
Rule 4001
Rule 3828
Rule 3827
Rule 63
Rule 225
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx &=\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac{3 \int \frac{\sec (c+d x) \left (\frac{2 a}{3}-a \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac{7}{10} \int \frac{\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\\ &=-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac{\left (7 \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{10 \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac{(7 \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{10 d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac{(21 \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac{3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac{7\ 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 ) \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)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \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 = 0.161032, size = 95, normalized size = 0.28 \[ \frac{\tan (c+d x) \left (7 \sqrt [6]{2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{3}{2},\frac{1}{2} (1-\sec (c+d x))\right )+3 \sqrt [6]{\sec (c+d x)+1} (2 \sec (c+d x)-1)\right )}{10 d \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.119, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{3}{\frac{1}{\sqrt [3]{a+a\sec \left ( dx+c \right ) }}}}\, 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{\sec \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{1}{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{\sec \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}, 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{\sec ^{3}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sec{\left (c + d x \right )} + 1\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 \frac{\sec \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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