\(\int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [947]

Optimal result

Integrand size = 43, antiderivative size = 442 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (5 A b+4 a B-8 b C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 b d}+\frac {\sqrt {a+b} (b (5 A+8 B-8 C)+2 a (A+2 B+8 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 d}-\frac {\sqrt {a+b} \left (3 A b^2+12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {(3 A b+4 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \] Output:

1/4*(a-b)*(a+b)^(1/2)*(5*A*b+4*B*a-8*C*b)*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+1/4*(a+b)^(1/2)*(b*(5*A+8*B-8*C)+2*a 
*(A+2*B+8*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)/d-1/4*(a+b)^(1/2)*(3*A*b^2+12*B*a*b+4*a^2*(A+2*C))*cot(d*x+c)*Ellipt 
icPi((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)/a/d+1/4*(3*A*b+4 
*B*a)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+b*sec(d*x+c) 
)^(3/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 19.01 (sec) , antiderivative size = 538, normalized size of antiderivative = 1.22 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} \left (\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (4 b C \sin (c+d x)+\frac {1}{2} a A \sin (2 (c+d x))\right )}{d (b+a \cos (c+d x))}-\frac {\sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (-a (a+b) (5 A b+4 a B-8 b C) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) (3 A b+2 a (A+2 B-4 C)) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (3 A b^2+12 a b B+4 a^2 (A+2 C)\right ) \left ((a-b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-2 a \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-a (5 A b+4 a B-8 b C) (b+a \cos (c+d x)) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2 a d (b+a \cos (c+d x))^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sec ^{\frac {3}{2}}(c+d x)}\right ) \] Input:

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

Output:

((Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(4*b*C*Sin[c + d*x] + (a*A*Sin[2 
*(c + d*x)])/2))/(d*(b + a*Cos[c + d*x])) - (Sqrt[Cos[c + d*x]*Sec[(c + d* 
x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)* 
(-(a*(a + b)*(5*A*b + 4*a*B - 8*b*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], ( 
a - b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x 
)/2]^2)/(a + b)]) + b*(a + b)*(3*A*b + 2*a*(A + 2*B - 4*C))*EllipticF[ArcS 
in[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos 
[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + (3*A*b^2 + 12*a*b*B + 4*a^2*(A + 
 2*C))*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*a 
*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[(c + d*x)/ 
2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - a*(5*A*b + 
4*a*B - 8*b*C)*(b + a*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2 
)*Sec[c + d*x]*Tan[(c + d*x)/2]))/(2*a*d*(b + a*Cos[c + d*x])^2*(Sec[(c + 
d*x)/2]^2)^(3/2)*Sec[c + d*x]^(3/2)))/2
 

Rubi [A] (verified)

Time = 1.96 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4582, 27, 3042, 4582, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{2} \int \frac {1}{2} \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (-b (A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+3 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (-b (A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+3 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{4} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \sec ^2(c+d x)+2 b (4 b B+a (A+8 C)) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \sec ^2(c+d x)+2 b (4 b B+a (A+8 C)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b (4 b B+a (A+8 C)) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2+(b (5 A b-8 C b+4 a B)+2 b (4 b B+a (A+8 C))) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b (4 a B+5 A b-8 b C) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2+(b (5 A b-8 C b+4 a B)+2 b (4 b B+a (A+8 C))) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (4 a B+5 A b-8 b C) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\)

Input:

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

Output:

(A*Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + (((2*(a - 
 b)*Sqrt[a + b]*(5*A*b + 4*a*B - 8*b*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*Sqrt[a 
+ b]*(b*(5*A + 8*B - 8*C) + 2*a*(A + 2*B + 8*C))*Cot[c + d*x]*EllipticF[Ar 
cSin[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 - (2*Sq 
rt[a + b]*(3*A*b^2 + 12*a*b*B + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[( 
a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S 
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) 
])/(a*d))/2 + ((3*A*b + 4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d)/4
 

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 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. \(1997\) vs. \(2(401)=802\).

Time = 15.92 (sec) , antiderivative size = 1998, normalized size of antiderivative = 4.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1998\)

Input:

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

Output:

-1/4/d*(4*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(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^2*EllipticE(-csc(d*x+c 
)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(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 
)*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(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*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b 
)/(a+b))^(1/2))+4*(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*b*EllipticE(-c 
sc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*sin(d*x+c)*cos(d*x+c)^2*(-cos( 
d*x+c)-1)*a^2*A-8*C*a*b*cos(d*x+c)*sin(d*x+c)+8*(cos(d*x+c)^2+2*cos(d*x+c) 
+1)*C*(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)*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+6* 
(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)*b^2*EllipticPi(-csc(d*x+c)+cot(d*x 
+c),-1,((a-b)/(a+b))^(1/2))+16*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(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 
*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+5*(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)*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(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)^(3/ 
2)*cos(d*x + c)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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