\(\int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [953]

Optimal result

Integrand size = 41, antiderivative size = 505 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(a-b) \sqrt {a+b} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {\sqrt {a+b} \left (a^2 b (15 A+90 B-46 C)+30 a^3 C-2 b^3 (15 A-5 B+9 C)+2 a b^2 (45 A-35 B+17 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {a \sqrt {a+b} (5 A b+2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-1/15*(a-b)*(a+b)^(1/2)*(70*B*a*b-a^2*(15*A-46*C)+6*b^2*(5*A+3*C))*cot(d*x 
+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*( 
1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d+1/15*(a+b)^ 
(1/2)*(a^2*b*(15*A+90*B-46*C)+30*a^3*C-2*b^3*(15*A-5*B+9*C)+2*a*b^2*(45*A- 
35*B+17*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b) 
/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1 
/2)/b/d-a*(a+b)^(1/2)*(5*A*b+2*B*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c)) 
^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^( 
1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c) 
/d-1/15*b*(15*A*a-10*B*b-16*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d-1/5*b 
*(5*A-2*C)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(9288\) vs. \(2(505)=1010\).

Time = 28.41 (sec) , antiderivative size = 9288, normalized size of antiderivative = 18.39 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:

Integrate[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C* 
Sec[c + d*x]^2),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 2.31 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {3042, 4582, 27, 3042, 4544, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

\(\displaystyle \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4582

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (-b (15 a A-10 b B-16 a C) \sec ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (-b (15 a A-10 b B-16 a C) \sec ^2(c+d x)+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (-b (15 a A-10 b B-16 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (5 C a^2+10 b B a+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a (5 A b+2 a B)\right )dx-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {15 (5 A b+2 a B) a^2+b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 (5 A b+2 a B) a^2+\left (2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right )-b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 (5 A b+2 a B) a^2+\left (2 \left (15 C a^3+45 b B a^2+b^2 (45 A+17 C) a+5 b^3 B\right )-b \left (-\left ((15 A-46 C) a^2\right )+70 b B a+6 b^2 (5 A+3 C)\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+15 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+15 a^2 (2 a B+5 A b) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (b \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}+\frac {2 \sqrt {a+b} \cot (c+d x) \left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {30 a \sqrt {a+b} (2 a B+5 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}\)

Input:

Int[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c 
+ d*x]^2),x]
 

Output:

(A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/d + ((-2*b*(5*A - 2*C)*(a + b* 
Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b]*(70*a* 
b*B - a^2*(15*A - 46*C) + 6*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin 
[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[ 
c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (2*Sq 
rt[a + b]*(a^2*b*(15*A + 90*B - 46*C) + 30*a^3*C - 2*b^3*(15*A - 5*B + 9*C 
) + 2*a*b^2*(45*A - 35*B + 17*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b 
*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/ 
(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (30*a*Sqrt[a + b 
]*(5*A*b + 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec 
[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + 
 b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d)/3 - (2*b*(15*a*A - 10*b*B 
 - 16*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5)/2
 

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4544

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

rule 4546

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2626\) vs. \(2(464)=928\).

Time = 92.56 (sec) , antiderivative size = 2627, normalized size of antiderivative = 5.20

method	result	size
default	\(\text {Expression too large to display}\)	\(2627\)

Input:

int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

1/15/d*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+c)+b) 
*(30*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1 
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*EllipticF(-csc(d*x+c)+co 
t(d*x+c),((a-b)/(a+b))^(1/2))+15*A*a^2*b*cos(d*x+c)*sin(d*x+c)+60*(-cos(d* 
x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*( 
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1, 
((a-b)/(a+b))^(1/2))+15*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(cos(d*x+c)/(cos( 
d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*Ellip 
ticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+46*(cos(d*x+c)^2+2*cos(d* 
x+c)+1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2) 
)+90*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c 
)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticF(-csc(d*x+c)+ 
cot(d*x+c),((a-b)/(a+b))^(1/2))+2*sin(d*x+c)*(34+11*cos(d*x+c))*a^2*b*C+2* 
(9*cos(d*x+c)^2+14*cos(d*x+c)+14)*C*a*b^2*tan(d*x+c)+46*C*a^3*cos(d*x+c)*s 
in(d*x+c)+30*A*a*b^2*cos(d*x+c)*sin(d*x+c)+150*(-cos(d*x+c)^2-2*cos(d*x+c) 
-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x 
+c)+1))^(1/2)*a^2*b*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/ 
2))+30*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+ 
c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(-csc(d*x...
 

Fricas [F]

\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \] Input:

integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

integral((C*b^2*cos(d*x + c)*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + 
c)*sec(d*x + c)^3 + A*a^2*cos(d*x + c) + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x 
 + c)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)*sec(d*x + c))*sqrt(b 
*sec(d*x + c) + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)* 
*2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \] Input:

integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/ 
2)*cos(d*x + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:

int(cos(c + d*x)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c 
+ d*x)^2),x)
 

Output:

int(cos(c + d*x)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c 
+ d*x)^2), x)
 

Reduce [F]

\[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) b^{2} c +2 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a b c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) b^{3}+\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) a^{2} c +3 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )d x \right ) a^{3} \] Input:

int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

int(sqrt(sec(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**4,x)*b**2*c + 2*in 
t(sqrt(sec(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**3,x)*a*b*c + int(sqr 
t(sec(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**3,x)*b**3 + int(sqrt(sec( 
c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**2,x)*a**2*c + 3*int(sqrt(sec(c 
+ d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**2,x)*a*b**2 + 3*int(sqrt(sec(c + 
d*x)*b + a)*cos(c + d*x)*sec(c + d*x),x)*a**2*b + int(sqrt(sec(c + d*x)*b 
+ a)*cos(c + d*x),x)*a**3