Optimal. Leaf size=413 \[ -\frac{343\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{880 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}+\frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{8/3}}{11 a d}-\frac{9 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{88 d}+\frac{147 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{440 d}+\frac{1029 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{880 d (\sec (c+d x)+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.507738, antiderivative size = 413, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3800, 4001, 3828, 3827, 50, 63, 225} \[ \frac{3 \tan (c+d x) (a \sec (c+d x)+a)^{8/3}}{11 a d}-\frac{9 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{88 d}+\frac{147 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{440 d}+\frac{1029 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{880 d (\sec (c+d x)+1)}-\frac{343\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt{\frac{(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{2}-\left (1-\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{880 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt{-\frac{\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt{3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3800
Rule 4001
Rule 3828
Rule 3827
Rule 50
Rule 63
Rule 225
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/3} \, dx &=\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac{3 \int \sec (c+d x) \left (\frac{8 a}{3}-a \sec (c+d x)\right ) (a+a \sec (c+d x))^{5/3} \, dx}{11 a}\\ &=-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac{49}{88} \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx\\ &=-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac{\left (49 a (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{5/3} \, dx}{88 (1+\sec (c+d x))^{2/3}}\\ &=-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac{\left (49 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{7/6}}{\sqrt{1-x}} \, dx,x,\sec (c+d x)\right )}{88 d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac{\left (343 a (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 )}{440 d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac{1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac{\left (343 a (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 )}{880 d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac{1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac{\left (1029 a (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 )}{440 d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac{1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac{9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac{3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac{343\ 3^{3/4} a 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)}{880 \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 = 0.302341, size = 96, normalized size = 0.23 \[ \frac{a \tan (c+d x) (a (\sec (c+d x)+1))^{2/3} \left (196 \sqrt [6]{2} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{2},\frac{3}{2},\frac{1}{2} (1-\sec (c+d x))\right )+3 (8 \sec (c+d x)+5) (\sec (c+d x)+1)^{13/6}\right )}{88 d (\sec (c+d x)+1)^{7/6}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.116, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{3} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right )^{4} + a \sec \left (d x + c\right )^{3}\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{3}} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]