3.7.28 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx\) [628]

3.7.28.1 Optimal result

Integrand size = 35, antiderivative size = 466 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a+a \sec (c+d x))^{7/3}}-\frac {3 (4 A-4 B-7 C) \tan (c+d x)}{55 a^2 d (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 \sqrt {2} A \operatorname {AppellF1}\left (-\frac {5}{6},\frac {1}{2},1,\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{5 a^2 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)}}+\frac {3^{3/4} (4 A-4 B-7 C) \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{55 \sqrt [3]{2} a^2 d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

output

-3/11*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(7/3)-3/55*(4*A-4*B-7*C)*tan(d 
*x+c)/a^2/d/(1+sec(d*x+c))/(a+a*sec(d*x+c))^(1/3)-3/5*A*AppellF1(-5/6,1,1/ 
2,1/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/a^2/d/(1+sec(d*x 
+c))/(a+a*sec(d*x+c))^(1/3)/(1-sec(d*x+c))^(1/2)+1/110*3^(3/4)*(4*A-4*B-7* 
C)*((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)))*EllipticF((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))*(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)*2^(2/3)/a^2/d/(1-sec(d*x+c))/(a+a*sec(d*x 
+c))^(1/3)/(-(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)
 

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

Leaf count is larger than twice the leaf count of optimal. \(3111\) vs. \(2(466)=932\).

Time = 21.07 (sec) , antiderivative size = 3111, normalized size of antiderivative = 6.68 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\text {Result too large to show} \]

input

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(7/ 
3),x]
 

output

(Cos[c + d*x]^2*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x]) 
^(7/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-6*Sec[(c + d*x)/2]*(20*A 
*Sin[(c + d*x)/2] - 9*B*Sin[(c + d*x)/2] - 2*C*Sin[(c + d*x)/2]))/55 - (3* 
Sec[(c + d*x)/2]^5*(A*Sin[(c + d*x)/2] - B*Sin[(c + d*x)/2] + C*Sin[(c + d 
*x)/2]))/22 + (3*Sec[(c + d*x)/2]^3*(25*A*Sin[(c + d*x)/2] - 14*B*Sin[(c + 
 d*x)/2] + 3*C*Sin[(c + d*x)/2]))/55))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A* 
Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(7/3)) + (2*2^(2/3)*Cos[c + d*x]^ 
2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(1 + Sec[c + d*x])^(7/3)*(A + B* 
Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Cos[c + d*x]*Sec[(c + d*x)/2]^2*(1 + S 
ec[c + d*x])^(2/3) + Sec[(c + d*x)/2]^2*((-3*A*(1 + Sec[c + d*x])^(2/3))/1 
1 + (4*B*(1 + Sec[c + d*x])^(2/3))/55 + (7*C*(1 + Sec[c + d*x])^(2/3))/55) 
)*Tan[(c + d*x)/2]*((-70*A + 4*B + 7*C)*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c 
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)* 
Tan[(c + d*x)/2]^2 + (27*(40*A + 4*B + 7*C)*AppellF1[1/2, 2/3, 1, 3/2, Tan 
[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(9*AppellF1[1/2, 
 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/ 
2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 
 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2 
)))/(165*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c 
 + d*x]))^(7/3)*((2^(2/3)*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c ...
 

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

Time = 1.15 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.07, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4540, 27, 3042, 4412, 3042, 4266, 3042, 4265, 149, 25, 1012, 4315, 3042, 4314, 61, 73, 766}

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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^{7/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/3}}dx\)

\(\Big \downarrow \) 4540

\(\displaystyle -\frac {3 \int -\frac {11 a A-a (4 A-4 B-7 C) \sec (c+d x)}{3 (\sec (c+d x) a+a)^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {11 a A-a (4 A-4 B-7 C) \sec (c+d x)}{(\sec (c+d x) a+a)^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {11 a A-a (4 A-4 B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 4412

\(\displaystyle \frac {11 a A \int \frac {1}{(\sec (c+d x) a+a)^{4/3}}dx-a (4 A-4 B-7 C) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 a A \int \frac {1}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 4266

\(\displaystyle \frac {\frac {11 A \sqrt [3]{\sec (c+d x)+1} \int \frac {1}{(\sec (c+d x)+1)^{4/3}}dx}{\sqrt [3]{a \sec (c+d x)+a}}-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {11 A \sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{4/3}}dx}{\sqrt [3]{a \sec (c+d x)+a}}-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 4265

\(\displaystyle \frac {-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx-\frac {11 A \tan (c+d x) \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{11/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 149

\(\displaystyle \frac {-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx-\frac {66 A \tan (c+d x) \int \frac {\cos ^7(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {66 A \tan (c+d x) \int -\frac {\cos ^7(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {-a (4 A-4 B-7 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{4/3}}dx-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {-\frac {(4 A-4 B-7 C) \sqrt [3]{\sec (c+d x)+1} \int \frac {\sec (c+d x)}{(\sec (c+d x)+1)^{4/3}}dx}{\sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {(4 A-4 B-7 C) \sqrt [3]{\sec (c+d x)+1} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{4/3}}dx}{\sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 4314

\(\displaystyle \frac {\frac {(4 A-4 B-7 C) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{11/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\frac {(4 A-4 B-7 C) \tan (c+d x) \left (\frac {1}{5} \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)-\frac {3 \sqrt {1-\sec (c+d x)}}{5 (\sec (c+d x)+1)^{5/6}}\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {(4 A-4 B-7 C) \tan (c+d x) \left (\frac {6}{5} \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {3 \sqrt {1-\sec (c+d x)}}{5 (\sec (c+d x)+1)^{5/6}}\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {\frac {(4 A-4 B-7 C) \tan (c+d x) \left (\frac {3^{3/4} \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt [3]{2} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}-\frac {3 \sqrt {1-\sec (c+d x)}}{5 (\sec (c+d x)+1)^{5/6}}\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}-\frac {33 \sqrt {2} A \sin (c+d x) \cos ^4(c+d x) \operatorname {AppellF1}\left (-\frac {5}{6},1,\frac {1}{2},\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}}{11 a^2}-\frac {3 (A-B+C) \tan (c+d x)}{11 d (a \sec (c+d x)+a)^{7/3}}\)

input

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(7/3),x]
 

output

(-3*(A - B + C)*Tan[c + d*x])/(11*d*(a + a*Sec[c + d*x])^(7/3)) + ((-33*Sq 
rt[2]*A*AppellF1[-5/6, 1, 1/2, 1/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x])/2 
]*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d* 
x])^(1/6)*(a + a*Sec[c + d*x])^(1/3)) + ((4*A - 4*B - 7*C)*((-3*Sqrt[1 - S 
ec[c + d*x]])/(5*(1 + Sec[c + d*x])^(5/6)) + (3^(3/4)*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])/(5*2^(1/3)*Sqrt[1 - Sec[c + d*x]]*Sqrt[-(((1 + Sec[c + d*x 
])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 
 + Sec[c + d*x])^(1/3))^2)]))*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(1 + 
 Sec[c + d*x])^(1/6)*(a + a*Sec[c + d*x])^(1/3)))/(11*a^2)
 

3.7.28.3.1 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 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 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 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]
 

rule 4540

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - 
b*B + a*C))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Sim 
p[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + 
 (b*B*(m + 1) - a*(A*(m + 1) - C*m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

3.7.28.4 Maple [F]

\[\int \frac {A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {7}{3}}}d x\]

input

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x)
 

output

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x)
 

3.7.28.5 Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\text {Timed out} \]

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algori 
thm="fricas")
 

output

Timed out
 

3.7.28.6 Sympy [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {7}{3}}}\, dx \]

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(7/3),x)
 

output

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))** 
(7/3), x)
 

3.7.28.7 Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\text {Timed out} \]

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algori 
thm="maxima")
 

output

Timed out
 

3.7.28.8 Giac [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {7}{3}}} \,d x } \]

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(7/3),x, algori 
thm="giac")
 

output

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(7/ 
3), x)
 

3.7.28.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{7/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{7/3}} \,d x \]

input

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(7/3),x)
 

output

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(7/3), x)