3.3.73 \(\int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [273]

3.3.73.1 Optimal result

Integrand size = 25, antiderivative size = 739 \[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (\frac {5}{6},\frac {1}{2},1,\frac {11}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{5 d \sqrt {1-\sec (c+d x)}}-\frac {3 \left (1+\sqrt {3}\right ) B \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )}+\frac {3 \sqrt [3]{2} \sqrt [4]{3} B E\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{a+a \sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{d (1-\sec (c+d x)) (1+\sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}+\frac {3^{3/4} \left (1-\sqrt {3}\right ) B \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \sqrt [3]{a+a \sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2^{2/3} d (1-\sec (c+d x)) (1+\sec (c+d x))^{2/3} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

output

-3*B*(a+a*sec(d*x+c))^(1/3)*(1+3^(1/2))*tan(d*x+c)/d/(1+sec(d*x+c))^(2/3)/ 
(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))+3/5*A*AppellF1(5/6,1,1/2,11/6,1 
+sec(d*x+c),1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+c))^(1/3)*2^(1/2)*tan(d*x+c)/ 
d/(1-sec(d*x+c))^(1/2)+3*2^(1/3)*3^(1/4)*B*((2^(1/3)-(1+sec(d*x+c))^(1/3)* 
(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3 
)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/ 
2)))*EllipticE((1-(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1 
+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*sec 
(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x 
+c))^(1/3)+(1+sec(d*x+c))^(2/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)) 
)^2)^(1/2)*tan(d*x+c)/d/(1-sec(d*x+c))/(1+sec(d*x+c))^(2/3)/(-(1+sec(d*x+c 
))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3 
^(1/2)))^2)^(1/2)+1/2*3^(3/4)*B*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)) 
)^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+sec(d* 
x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*Ellipt 
icF((1-(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c) 
)^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*sec(d*x+c))^(1 
/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*(1-3^(1/2))*((2^(2/3)+2^(1/3)*(1+sec(d* 
x+c))^(1/3)+(1+sec(d*x+c))^(2/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2) 
))^2)^(1/2)*tan(d*x+c)*2^(1/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c))^(2/3)/(-...
 

3.3.73.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5094\) vs. \(2(739)=1478\).

Time = 18.28 (sec) , antiderivative size = 5094, normalized size of antiderivative = 6.89 \[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]

input

Integrate[(a + a*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x]),x]
 

output

Result too large to show
 

3.3.73.3 Rubi [A] (warning: unable to verify)

Time = 1.06 (sec) , antiderivative size = 752, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4412, 3042, 4266, 3042, 4265, 149, 25, 1012, 4315, 3042, 4314, 73, 837, 25, 27, 766, 2420}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt [3]{a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt [3]{a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4412

\(\displaystyle A \int \sqrt [3]{\sec (c+d x) a+a}dx+B \int \sec (c+d x) \sqrt [3]{\sec (c+d x) a+a}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\)

\(\Big \downarrow \) 4266

\(\displaystyle \frac {A \sqrt [3]{a \sec (c+d x)+a} \int \sqrt [3]{\sec (c+d x)+1}dx}{\sqrt [3]{\sec (c+d x)+1}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sqrt [3]{a \sec (c+d x)+a} \int \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}dx}{\sqrt [3]{\sec (c+d x)+1}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\)

\(\Big \downarrow \) 4265

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 149

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {6 A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {\cos (c+d x) (\sec (c+d x)+1)^{2/3}}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {6 A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int -\frac {\cos (c+d x) (\sec (c+d x)+1)^{2/3}}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}+B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\)

\(\Big \downarrow \) 1012

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {B \sqrt [3]{a \sec (c+d x)+a} \int \sec (c+d x) \sqrt [3]{\sec (c+d x)+1}dx}{\sqrt [3]{\sec (c+d x)+1}}+\frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sqrt [3]{a \sec (c+d x)+a} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}dx}{\sqrt [3]{\sec (c+d x)+1}}+\frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}\)

\(\Big \downarrow \) 4314

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {(\sec (c+d x)+1)^{2/3}}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 837

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \left (-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {1}{2} \int -\frac {2 (\sec (c+d x)+1)^{2/3}+2^{2/3} \left (1-\sqrt {3}\right )}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \left (\frac {1}{2} \int \frac {2^{2/3} \left (\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1\right )}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \left (\frac {\int \frac {\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \left (\frac {\int \frac {\sqrt [3]{2} (\sec (c+d x)+1)^{2/3}-\sqrt {3}+1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{\sqrt [3]{2}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2\ 2^{2/3} \sqrt [4]{3} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 2420

\(\displaystyle \frac {3 \sqrt {2} A \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{6},1,\frac {1}{2},\frac {11}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)}}-\frac {6 B \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \left (\frac {\frac {\left (1+\sqrt {3}\right ) \sqrt {1-\sec (c+d x)} \sqrt [6]{\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 )}-\frac {\sqrt [4]{3} \sqrt [6]{\sec (c+d x)+1} \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}} E\left (\arccos \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 )}{\sqrt [3]{2} \sqrt {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]{2}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [6]{\sec (c+d x)+1} \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}} \operatorname {EllipticF}\left (\arccos \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\ 2^{2/3} \sqrt [4]{3} \sqrt {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}}}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

input

Int[(a + a*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x]),x]
 

output

(3*Sqrt[2]*A*AppellF1[5/6, 1, 1/2, 11/6, 1 + Sec[c + d*x], (1 + Sec[c + d* 
x])/2]*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(5*d*Sqrt[1 - Sec[c + d*x] 
]) - (6*B*(a + a*Sec[c + d*x])^(1/3)*(-1/2*((1 - Sqrt[3])*EllipticF[ArcCos 
[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3 
])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/6)*( 
2^(1/3) - (1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d 
*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c 
 + d*x])^(1/3))^2])/(2^(2/3)*3^(1/4)*Sqrt[1 - Sec[c + d*x]]*Sqrt[-(((1 + S 
ec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + S 
qrt[3])*(1 + Sec[c + d*x])^(1/3))^2)]) + (((1 + Sqrt[3])*Sqrt[1 - Sec[c + 
d*x]]*(1 + Sec[c + d*x])^(1/6))/(2^(2/3)*(2^(1/3) - (1 + Sqrt[3])*(1 + Sec 
[c + d*x])^(1/3))) - (3^(1/4)*EllipticE[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 
 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3)) 
], (2 + Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/6)*(2^(1/3) - (1 + Sec[c + d*x]) 
^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d* 
x])^(2/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2])/(2^(1/3) 
*Sqrt[1 - Sec[c + d*x]]*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + S 
ec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2) 
]))/2^(1/3))*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(5 
/6))
 

3.3.73.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 149

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1 
) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && IntegerQ[p]
 

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 837

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 2420

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4265

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot 
[c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]]))   Subst[Int[(1 
 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0 
]
 

rule 4266

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 4314

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x 
]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 
)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, 
x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0]
 

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m 
]/(1 + (b/a)*Csc[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Csc[e + f*x])^m*(d 
*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

rule 4412

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[c   Int[(a + b*Csc[e + f*x])^m, x], x] + Sim 
p[d   Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[2*m]
 

3.3.73.4 Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]

input

int((a+a*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x)
 

output

int((a+a*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x)
 

3.3.73.5 Fricas [F(-1)]

Timed out. \[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

input

integrate((a+a*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="fricas")
 

output

Timed out
 

3.3.73.6 Sympy [F]

\[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right )\, dx \]

input

integrate((a+a*sec(d*x+c))**(1/3)*(A+B*sec(d*x+c)),x)
 

output

Integral((a*(sec(c + d*x) + 1))**(1/3)*(A + B*sec(c + d*x)), x)
 

3.3.73.7 Maxima [F]

\[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

input

integrate((a+a*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="maxima")
 

output

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(1/3), x)
 

3.3.73.8 Giac [F]

\[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

input

integrate((a+a*sec(d*x+c))^(1/3)*(A+B*sec(d*x+c)),x, algorithm="giac")
 

output

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(1/3), x)
 

3.3.73.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \]

input

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/3),x)
 

output

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/3), x)