\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [742]

Optimal result

Integrand size = 35, antiderivative size = 460 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \] Output:

-2/5*(2*a^2*b^2*(5*A-4*C)+16*a^4*C-b^4*(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^5/(a+b)^(1/2)/d-2/5*(16*a^3*C+ 
12*a^2*b*C+2*a*b^2*(5*A+2*C)+b^3*(5*A+3*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^4/(a+b)^(1/2)/d-2*(A*b^2+C*a^2)*sec( 
d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2/5*a*(5*A*b^2+8* 
C*a^2-3*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A* 
b^2+6*C*a^2-C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2-b 
^2)/d
 

Mathematica [A] (warning: unable to verify)

Time = 19.44 (sec) , antiderivative size = 641, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {4 (b+a \cos (c+d x)) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (2 (a+b) \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (-16 a^3 C+12 a^2 b C-2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{5 b^4 \left (-a^2+b^2\right ) d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {(b+a \cos (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (-10 a^2 A b^2+5 A b^4-16 a^4 C+8 a^2 b^2 C+3 b^4 C\right ) \sin (c+d x)}{5 b^4 \left (-a^2+b^2\right )}+\frac {4 \left (a^2 A b^2 \sin (c+d x)+a^4 C \sin (c+d x)\right )}{b^3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}-\frac {12 a C \tan (c+d x)}{5 b^3}+\frac {4 C \sec (c+d x) \tan (c+d x)}{5 b^2}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{3/2}} \] Input:

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

Output:

(4*(b + a*Cos[c + d*x])*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + C*Sec[c 
 + d*x]^2)*(2*(a + b)*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C)) 
*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)* 
(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] 
+ 2*b*(a + b)*(-16*a^3*C + 12*a^2*b*C - 2*a*b^2*(5*A + 2*C) + b^3*(5*A + 3 
*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + 
 b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + 
b)] + (2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cos[c + d*x]*(b 
 + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(5*b^4*(-a^2 + b^ 
2)*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Sec[c + 
d*x]]*(a + b*Sec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^2*(A + C*Sec[c + 
 d*x]^2)*((4*(-10*a^2*A*b^2 + 5*A*b^4 - 16*a^4*C + 8*a^2*b^2*C + 3*b^4*C)* 
Sin[c + d*x])/(5*b^4*(-a^2 + b^2)) + (4*(a^2*A*b^2*Sin[c + d*x] + a^4*C*Si 
n[c + d*x]))/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d*x])) - (12*a*C*Tan[c + d*x 
])/(5*b^3) + (4*C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(A + 2*C + A*Cos 
[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 2.08 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4587, 27, 3042, 4580, 27, 3042, 4570, 27, 3042, 4493, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4587

\(\displaystyle -\frac {2 \int \frac {\sec ^2(c+d x) \left (-\left (\left (5 A b^2+\left (6 a^2-b^2\right ) C\right ) \sec ^2(c+d x)\right )-a b (A+C) \sec (c+d x)+4 \left (C a^2+A b^2\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (6 C a^2+5 A b^2-b^2 C\right ) \sec ^2(c+d x)\right )-a b (A+C) \sec (c+d x)+4 \left (C a^2+A b^2\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-6 C a^2-5 A b^2+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-a b (A+C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 \left (C a^2+A b^2\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4580

\(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \sec ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \sec ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4570

\(\displaystyle -\frac {-\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4493

\(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4319

\(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

\(\displaystyle -\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}}{b \left (a^2-b^2\right )}\)

Input:

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

Output:

(-2*(A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + 
 b*Sec[c + d*x]]) - ((-2*(5*A*b^2 + 6*a^2*C - b^2*C)*Sec[c + d*x]*Sqrt[a + 
 b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) - (((-2*(a - b)*Sqrt[a + b]*(2*a^2* 
b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSi 
n[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^2*d) - (2 
*(a - b)*Sqrt[a + b]*(16*a^3*C + 12*a^2*b*C + 2*a*b^2*(5*A + 2*C) + b^3*(5 
*A + 3*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))/b - (2*a*(5*A*b^2 + 8*a^2*C - 3*b^2*C)*S 
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(b*d))/(5*b))/(b*(a^2 - b^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 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 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 4493

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

rule 4570

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

rule 4580

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x 
_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 
1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[Csc[e + f*x]*(a + b*Csc[e 
+ f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* 
(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & 
& NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4587

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2165\) vs. \(2(428)=856\).

Time = 72.68 (sec) , antiderivative size = 2166, normalized size of antiderivative = 4.71

method	result	size
default	\(\text {Expression too large to display}\)	\(2166\)
parts	\(\text {Expression too large to display}\)	\(2169\)

Input:

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

Output:

-2/5/d/(a-b)/(a+b)/b^4*(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)*(8*sin(d*x+c)*(cos(d*x+c)-1)*a^4*b*C+(-3*cos(d*x+c)^3+5*c 
os(d*x+c)^2-cos(d*x+c)-1)*C*a^2*b^3*tan(d*x+c)*sec(d*x+c)+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^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a 
+b))^(1/2))+5*(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)*b^5*EllipticE(-csc(d 
*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+3*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(c 
os(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^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(-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*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c),((a- 
b)/(a+b))^(1/2))+(3*cos(d*x+c)^2-2*cos(d*x+c)-2)*C*a*b^4*tan(d*x+c)+5*A*a* 
b^4*cos(d*x+c)*sin(d*x+c)+2*(1+4*cos(d*x+c)^2+cos(d*x+c))*C*a^3*b^2*tan(d* 
x+c)+5*sin(d*x+c)*(cos(d*x+c)-1)*a^2*A*b^3+3*(-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)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))-10*A 
*a^3*b^2*cos(d*x+c)*sin(d*x+c)+5*(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 
*b^4*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+10*(cos(d*x+...
 

Fricas [F]

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

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

Output:

integral((C*sec(d*x + c)^5 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b 
^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)*c + int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)* 
*3)/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*a