\(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [999]

Optimal result

Integrand size = 43, antiderivative size = 319 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b \left (45 a b B-a^2 (10 A-42 C)+3 b^2 (5 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 (5 a A-5 b B-9 a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}-\frac {2 b (5 A-3 C) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac {2 A (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:

2/5*(5*B*a^3-15*B*a*b^2+15*a^2*b*(A-C)-b^3*(5*A+3*C))*cos(d*x+c)^(1/2)*Ell 
ipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/d+2/3*(9*B*a^2*b+B*b^3 
+3*a*b^2*(3*A+C)+a^3*(A+3*C))*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2 
*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/15*b*(45*B*a*b-a^2*(10*A-42*C)+3*b^2*(5*A 
+3*C))*sec(d*x+c)^(1/2)*sin(d*x+c)/d-2/15*b^2*(5*A*a-5*B*b-9*C*a)*sec(d*x+ 
c)^(3/2)*sin(d*x+c)/d-2/15*b*(5*A-3*C)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2 
*sin(d*x+c)/d+2/3*A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.43 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (6 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+30 A b^3 \sin (c+d x)+90 a b^2 B \sin (c+d x)+90 a^2 b C \sin (c+d x)+18 b^3 C \sin (c+d x)+5 a^3 A \sin (2 (c+d x))+10 b^3 B \tan (c+d x)+30 a b^2 C \tan (c+d x)+6 b^3 C \sec (c+d x) \tan (c+d x)\right )}{15 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {9}{2}}(c+d x)} \] Input:

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

Output:

(2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(6*(5*a^ 
3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Sqrt[Cos[c + d*x]]* 
EllipticE[(c + d*x)/2, 2] + 10*(9*a^2*b*B + b^3*B + 3*a*b^2*(3*A + C) + a^ 
3*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 30*A*b^3*Sin[c 
 + d*x] + 90*a*b^2*B*Sin[c + d*x] + 90*a^2*b*C*Sin[c + d*x] + 18*b^3*C*Sin 
[c + d*x] + 5*a^3*A*Sin[2*(c + d*x)] + 10*b^3*B*Tan[c + d*x] + 30*a*b^2*C* 
Tan[c + d*x] + 6*b^3*C*Sec[c + d*x]*Tan[c + d*x]))/(15*d*(b + a*Cos[c + d* 
x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(9/2) 
)
 

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.01, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4582, 27, 3042, 4584, 27, 3042, 4564, 27, 3042, 4535, 3042, 4258, 3042, 3120, 4534, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {2}{3} \int \frac {(a+b \sec (c+d x))^2 \left (-b (5 A-3 C) \sec ^2(c+d x)+(3 b B+a (A+3 C)) \sec (c+d x)+3 (2 A b+a B)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {(a+b \sec (c+d x))^2 \left (-b (5 A-3 C) \sec ^2(c+d x)+(3 b B+a (A+3 C)) \sec (c+d x)+3 (2 A b+a B)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-b (5 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(3 b B+a (A+3 C)) \csc \left (c+d x+\frac {\pi }{2}\right )+3 (2 A b+a B)\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4584

\(\displaystyle \frac {1}{3} \left (\frac {2}{5} \int \frac {(a+b \sec (c+d x)) \left (-3 b (5 a A-5 b B-9 a C) \sec ^2(c+d x)+\left (5 (A+3 C) a^2+30 b B a+3 b^2 (5 A+3 C)\right ) \sec (c+d x)+a (35 A b-3 C b+15 a B)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {(a+b \sec (c+d x)) \left (-3 b (5 a A-5 b B-9 a C) \sec ^2(c+d x)+\left (5 (A+3 C) a^2+30 b B a+3 b^2 (5 A+3 C)\right ) \sec (c+d x)+a (35 A b-3 C b+15 a B)\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-3 b (5 a A-5 b B-9 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 (A+3 C) a^2+30 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (35 A b-3 C b+15 a B)\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4564

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left ((35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left ((A+3 C) a^3+9 b B a^2+3 b^2 (3 A+C) a+b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \sqrt {\sec (c+d x)}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\int \frac {(35 A b-3 C b+15 a B) a^2+b \left (-\left ((10 A-42 C) a^2\right )+45 b B a+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4534

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (3 \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left (\frac {1}{5} \left (\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)} \left (-\left (a^2 (10 A-42 C)\right )+45 a b B+3 b^2 (5 A+3 C)\right )}{d}+\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{d}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{d}-\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (5 a A-9 a C-5 b B)}{d}\right )-\frac {2 b (5 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d}\right )+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^3}{3 d \sqrt {\sec (c+d x)}}\)

Input:

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

Output:

(2*A*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((-2* 
b*(5*A - 3*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d 
) + ((6*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*Sqrt[C 
os[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*(9*a^2* 
b*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 3*C))*Sqrt[Cos[c + d*x]]*Ellipt 
icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*b*(45*a*b*B - a^2*(10*A - 4 
2*C) + 3*b^2*(5*A + 3*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d - (2*b^2*(5*a 
*A - 5*b*B - 9*a*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/d)/5)/3
 

Defintions of rubi rules used

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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4534

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4564

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ 
n/(f*(n + 2))), x] + Simp[1/(n + 2)   Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 
2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( 
n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] 
&&  !LtQ[n, -1]
 

rule 4582

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
 

rule 4584

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(m + n + 1)   Int[(a + b*Csc[e 
+ f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a 
*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C*m)*Csc 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 
 - b^2, 0] && GtQ[m, 0] &&  !LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1267\) vs. \(2(294)=588\).

Time = 15.12 (sec) , antiderivative size = 1268, normalized size of antiderivative = 3.97

method	result	size
parts	\(\text {Expression too large to display}\)	\(1268\)
default	\(\text {Expression too large to display}\)	\(1389\)

Input:

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, 
method=_RETURNVERBOSE)
 

Output:

2*(3*A*a^2*b+B*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2 
)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE 
(cos(1/2*d*x+1/2*c),2^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2 
)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d-2/3*(B*b^3+3 
*C*a*b^2)*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2 
)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x 
+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1 
/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*((2*cos(1/2*d*x+1/ 
2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x 
+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d-2*( 
A*b^3+3*B*a*b^2+3*C*a^2*b)*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c) 
^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x 
+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/ 
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c 
)^2-1)^(1/2)/d-2*(3*A*a*b^2+3*B*a^2*b+C*a^3)*((2*cos(1/2*d*x+1/2*c)^2-1)*s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/ 
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^ 
2-1)^(1/2)/d-2/5*C*b^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{3} + 9 i \, B a^{2} b + 3 i \, {\left (3 \, A + C\right )} a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{3} - 9 i \, B a^{2} b - 3 i \, {\left (3 \, A + C\right )} a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} - 15 i \, {\left (A - C\right )} a^{2} b + 15 i \, B a b^{2} + i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} + 15 i \, {\left (A - C\right )} a^{2} b - 15 i \, B a b^{2} - i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (5 \, A a^{3} \cos \left (d x + c\right )^{3} + 3 \, C b^{3} + 3 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

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

Output:

-1/15*(5*sqrt(2)*(I*(A + 3*C)*a^3 + 9*I*B*a^2*b + 3*I*(3*A + C)*a*b^2 + I* 
B*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c)) + 5*sqrt(2)*(-I*(A + 3*C)*a^3 - 9*I*B*a^2*b - 3*I*(3*A + C)*a*b^2 - 
I*B*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c)) + 3*sqrt(2)*(-5*I*B*a^3 - 15*I*(A - C)*a^2*b + 15*I*B*a*b^2 + I*(5 
*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(5*I*B*a^3 + 15*I*(A - C 
)*a^2*b - 15*I*B*a*b^2 - I*(5*A + 3*C)*b^3)*cos(d*x + c)^2*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(5* 
A*a^3*cos(d*x + c)^3 + 3*C*b^3 + 3*(15*C*a^2*b + 15*B*a*b^2 + (5*A + 3*C)* 
b^3)*cos(d*x + c)^2 + 5*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c)/sqr 
t(cos(d*x + c)))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co 
s(c + d*x))^(3/2),x)
 

Output:

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co 
s(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) a^{4}+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a^{3} b +\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{3} c +6 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{2} b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b^{3} c +3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \,b^{2} c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b^{4}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a^{2} b c +4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a \,b^{3} \] Input:

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

Output:

int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a**4 + 4*int(sqrt(sec(c + d*x))/ 
sec(c + d*x),x)*a**3*b + int(sqrt(sec(c + d*x)),x)*a**3*c + 6*int(sqrt(sec 
(c + d*x)),x)*a**2*b**2 + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*b**3*c 
 + 3*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a*b**2*c + int(sqrt(sec(c + 
 d*x))*sec(c + d*x)**2,x)*b**4 + 3*int(sqrt(sec(c + d*x))*sec(c + d*x),x)* 
a**2*b*c + 4*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a*b**3