\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [166]

Optimal result

Integrand size = 33, antiderivative size = 264 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(39*A-20*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/ 
d-1/32*(219*A-115*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c) 
)^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A-B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+ 
c))^(5/2)-1/16*(19*A-11*B)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2 
)-7/16*(9*A-5*B)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/16*(31*A-15*B)* 
cos(d*x+c)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.66 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {32 B \sin (c+d x)+152 A (1+\cos (c+d x)) \sin (c+d x)+120 B (1+\sec (c+d x)) \sin (c+d x)-280 B (1+\sec (c+d x))^2 \sin (c+d x)+16 A \sin (2 (c+d x))+760 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) (1+\sec (c+d x))^2 \tan (c+d x)+\frac {640 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}-\frac {460 \sqrt {2} B \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}-\frac {219 A \left (7 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-1+2 \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}}{128 d (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

-1/128*(32*B*Sin[c + d*x] + 152*A*(1 + Cos[c + d*x])*Sin[c + d*x] + 120*B* 
(1 + Sec[c + d*x])*Sin[c + d*x] - 280*B*(1 + Sec[c + d*x])^2*Sin[c + d*x] 
+ 16*A*Sin[2*(c + d*x)] + 760*A*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + 
 d*x]]*(1 + Sec[c + d*x])^2*Tan[c + d*x] + (640*B*ArcTanh[Sqrt[1 - Sec[c + 
 d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/Sqrt[1 - Sec[c + d*x]] - (460*S 
qrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x])^2*Tan[ 
c + d*x])/Sqrt[1 - Sec[c + d*x]] - (219*A*(7*ArcTanh[Sqrt[1 - Sec[c + d*x] 
]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-1 
+ 2*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*(1 + Sec[c + d*x])^2*Tan[c + d*x 
])/Sqrt[1 - Sec[c + d*x]])/(d*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4508, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4510, 25, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (4 a (3 A-B)-7 a (A-B) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (4 a (3 A-B)-7 a (A-B) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {4 a (3 A-B)-7 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {4 a^2 (31 A-15 B)-5 a^2 (19 A-11 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {\int -\frac {2 \cos (c+d x) \left (14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) \left (14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {14 a^3 (9 A-5 B)-3 a^3 (31 A-15 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B)-7 a^4 (9 A-5 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \sqrt {\sec (c+d x) a+a}dx-a^4 (219 A-115 B) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^4 (219 A-115 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^4 (219 A-115 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {8 a^4 (39 A-20 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^4 (219 A-115 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (219 A-115 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {8 a^{7/2} (39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^3 (9 A-5 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A-115 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + 
(-1/2*(a*(19*A - 11*B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^ 
(3/2)) + ((2*a^2*(31*A - 15*B)*Cos[c + d*x]*Sin[c + d*x])/(d*Sqrt[a + a*Se 
c[c + d*x]]) - (-(((8*a^(7/2)*(39*A - 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/ 
Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a^(7/2)*(219*A - 115*B)*ArcTan[(Sq 
rt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) + (14*a^3*( 
9*A - 5*B)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*a^2))/(8*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4261

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

rule 4282

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

rule 4408

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

rule 4508

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

rule 4510

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

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 776, normalized size of antiderivative = 2.94 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3 + 3*(219*A - 115*B)*cos(d*x 
 + c)^2 + 3*(219*A - 115*B)*cos(d*x + c) + 219*A - 115*B)*sqrt(-a)*log((2* 
sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin( 
d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2* 
cos(d*x + c) + 1)) + 8*((39*A - 20*B)*cos(d*x + c)^3 + 3*(39*A - 20*B)*cos 
(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*B)*sqrt(-a)*log((2* 
a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos( 
d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 4*(8*A*c 
os(d*x + c)^4 - 4*(5*A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B)*cos(d*x + c 
)^2 - 7*(9*A - 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos 
(d*x + c) + a^3*d), 1/32*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3 + 3*(219 
*A - 115*B)*cos(d*x + c)^2 + 3*(219*A - 115*B)*cos(d*x + c) + 219*A - 115* 
B)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x 
+ c)/(sqrt(a)*sin(d*x + c))) - 8*((39*A - 20*B)*cos(d*x + c)^3 + 3*(39*A - 
 20*B)*cos(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*B)*sqrt(a 
)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin 
(d*x + c))) + 2*(8*A*cos(d*x + c)^4 - 4*(5*A - 4*B)*cos(d*x + c)^3 - 5*(19 
*A - 11*B)*cos(d*x + c)^2 - 7*(9*A - 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos...
 

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((A + B*sec(c + d*x))*cos(c + d*x)**2/(a*(sec(c + d*x) + 1))**(5/2 
), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 685 vs. \(2 (229) = 458\).

Time = 0.94 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.59 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5)*tan 
(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(29*A*a^5 - 21*B*a^5 
)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(219*A - 115*B)* 
log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^ 
2)/(sqrt(-a)*a^2*sgn(cos(d*x + c))) + 8*(39*A - 20*B)*log(abs(309485009821 
345068724781056*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 
*c)^2 + a))^2 - 618970019642690137449562112*sqrt(2)*abs(a) - 9284550294640 
35206174343168*a)/abs(309485009821345068724781056*(sqrt(-a)*tan(1/2*d*x + 
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 618970019642690137449562 
112*sqrt(2)*abs(a) - 928455029464035206174343168*a))/(sqrt(-a)*a*abs(a)*sg 
n(cos(d*x + c))) - 32*sqrt(2)*(41*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a 
*tan(1/2*d*x + 1/2*c)^2 + a))^6*A - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq 
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B - 209*(sqrt(-a)*tan(1/2*d*x + 1/2*c 
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a + 76*(sqrt(-a)*tan(1/2*d*x 
+ 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a + 91*(sqrt(-a)*tan(1 
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*a^2 - 36*(sqrt( 
-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^2 - 
11*A*a^3 + 4*B*a^3)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x 
 + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a))^2*a + a^2)^2*sqrt(-a)*a*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x))/(sec(c 
 + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*b + int((sqrt(sec( 
c + d*x) + 1)*cos(c + d*x)**2)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*se 
c(c + d*x) + 1),x)*a))/a**3