\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [533]

Optimal result

Integrand size = 43, antiderivative size = 280 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(39*A-20*B+8*C)*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+43*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*s 
ec(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A-B+C)*cos(d*x+c)*sin(d*x+c)/d/(a 
+a*sec(d*x+c))^(5/2)-1/16*(19*A-11*B+3*C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*s 
ec(d*x+c))^(3/2)-1/16*(63*A-35*B+11*C)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^( 
1/2)+1/16*(31*A-15*B+7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/ 
2)
 

Mathematica [A] (warning: unable to verify)

Time = 6.94 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (\frac {\left ((-219 A+115 B-43 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+4 \sqrt {2} (39 A-20 B+8 C) \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {1}{4} (73 A-43 B+11 C+(89 A-55 B+15 C) \cos (c+d x)+2 (5 A-4 B) \cos (2 (c+d x))-2 A \cos (3 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{4 d (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*((((-219*A + 115*B - 43*C)*ArcSin[T 
an[(c + d*x)/2]] + 4*Sqrt[2]*(39*A - 20*B + 8*C)*ArcTan[Tan[(c + d*x)/2]/S 
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x] 
)]*Sqrt[1 + Sec[c + d*x]])/Sqrt[Sec[(c + d*x)/2]^2] + ((73*A - 43*B + 11*C 
 + (89*A - 55*B + 15*C)*Cos[c + d*x] + 2*(5*A - 4*B)*Cos[2*(c + d*x)] - 2* 
A*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Sqrt[Sec[c + d*x]]*(Sin[(c + d*x)/2 
] - Sin[(3*(c + d*x))/2]))/4))/(4*d*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 2.03 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.06, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {\int -\frac {2 \cos (c+d x) \left (2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) \left (2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B+8 C) \int \sqrt {\sec (c+d x) a+a}dx-a^4 (219 A-115 B+43 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {8 a^4 (39 A-20 B+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (219 A-115 B+43 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A-115 B+43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2) 
) + (-1/2*(a*(19*A - 11*B + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[ 
c + d*x])^(3/2)) + ((2*a^2*(31*A - 15*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/ 
(d*Sqrt[a + a*Sec[c + d*x]]) - (-(((8*a^(7/2)*(39*A - 20*B + 8*C)*ArcTan[( 
Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a^(7/2)*(219 
*A - 115*B + 43*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c 
 + d*x]])])/d)/a) + (2*a^3*(63*A - 35*B + 11*C)*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]
 

rule 4572

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*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e 
+ f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) 
   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - 
 A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - 
b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(744\) vs. \(2(245)=490\).

Time = 31.12 (sec) , antiderivative size = 745, normalized size of antiderivative = 2.66

method	result	size
default	\(-\frac {\left (\left (312 \cos \left (d x +c \right )^{3}+936 \cos \left (d x +c \right )^{2}+936 \cos \left (d x +c \right )+312\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-160 \cos \left (d x +c \right )^{3}-480 \cos \left (d x +c \right )^{2}-480 \cos \left (d x +c \right )-160\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (64 \cos \left (d x +c \right )^{3}+192 \cos \left (d x +c \right )^{2}+192 \cos \left (d x +c \right )+64\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (219 \cos \left (d x +c \right )^{3}+657 \cos \left (d x +c \right )^{2}+657 \cos \left (d x +c \right )+219\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-115 \cos \left (d x +c \right )^{3}-345 \cos \left (d x +c \right )^{2}-345 \cos \left (d x +c \right )-115\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (43 \cos \left (d x +c \right )^{3}+129 \cos \left (d x +c \right )^{2}+129 \cos \left (d x +c \right )+43\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-16 \cos \left (d x +c \right )^{3}+40 \cos \left (d x +c \right )^{2}+190 \cos \left (d x +c \right )+126\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-32 \cos \left (d x +c \right )^{2}-110 \cos \left (d x +c \right )-70\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (30 \cos \left (d x +c \right )+22\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(745\)

Input:

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

Output:

-1/32/d/a^3*((312*cos(d*x+c)^3+936*cos(d*x+c)^2+936*cos(d*x+c)+312)*A*(-co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(c 
sc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+(-160*cos(d*x+c 
)^3-480*cos(d*x+c)^2-480*cos(d*x+c)-160)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*co 
t(d*x+c)+cot(d*x+c)^2-1)^(1/2))+(64*cos(d*x+c)^3+192*cos(d*x+c)^2+192*cos( 
d*x+c)+64)*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+ 
c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2) 
)+(219*cos(d*x+c)^3+657*cos(d*x+c)^2+657*cos(d*x+c)+219)*2^(1/2)*A*(-cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d* 
x+c)+csc(d*x+c))+(-115*cos(d*x+c)^3-345*cos(d*x+c)^2-345*cos(d*x+c)-115)*2 
^(1/2)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(43*cos(d*x+c)^3+129*cos(d*x+c)^2+129*cos 
(d*x+c)+43)*2^(1/2)*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin(d*x+c)*cos(d*x+c)*(-16*c 
os(d*x+c)^3+40*cos(d*x+c)^2+190*cos(d*x+c)+126)*A+sin(d*x+c)*cos(d*x+c)*(- 
32*cos(d*x+c)^2-110*cos(d*x+c)-70)*B+sin(d*x+c)*cos(d*x+c)*(30*cos(d*x+c)+ 
22)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+ 
1)
 

Fricas [A] (verification not implemented)

Time = 43.87 (sec) , antiderivative size = 835, normalized size of antiderivative = 2.98 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/64*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^3 + 3*(219*A - 115*B 
+ 43*C)*cos(d*x + c)^2 + 3*(219*A - 115*B + 43*C)*cos(d*x + c) + 219*A - 1 
15*B + 43*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c 
os(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 + 8*C)* 
cos(d*x + c)^3 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B + 8 
*C)*cos(d*x + c) + 39*A - 20*B + 8*C)*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*cos(d*x + c)^4 - 4*(5 
*A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 - (63*A - 
35*B + 11*C)*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 + 43*C)*cos(d*x + c)^3 + 3*(219 
*A - 115*B + 43*C)*cos(d*x + c)^2 + 3*(219*A - 115*B + 43*C)*cos(d*x + c) 
+ 219*A - 115*B + 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 8*((39*A - 20*B + 8*C) 
*cos(d*x + c)^3 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B + 
8*C)*cos(d*x + c) + 39*A - 20*B + 8*C)*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 + 3*C)*cos(d*x +...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + 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. 702 vs. \(2 (245) = 490\).

Time = 1.30 (sec) , antiderivative size = 702, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 
/2),x, algorithm="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 + C* 
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 + 13*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt( 
2)*(219*A - 115*B + 43*C)*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 + 8*C)*log(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 - 61897001964269013744956211 
2*sqrt(2)*abs(a) - 928455029464035206174343168*a)/abs(30948500982134506872 
4781056*(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) - 928455029464035206174 
343168*a))/(sqrt(-a)*a*abs(a)*sgn(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*(sq 
rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B - 2 
09*(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(-...
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(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)**2)/(se 
c(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(s 
ec(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*sec(c + d*x) + 1),x)*a) 
)/a**3