\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [555]

Optimal result

Integrand size = 35, antiderivative size = 317 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(283 A-163 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(21 A-13 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(2671 A-1495 B) \sin (c+d x)}{240 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(787 A-475 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(157 A-85 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/32*(283*A-163*B)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2 
)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(5/2 
)/d-1/4*(A-B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(2 
1*A-13*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/240*(26 
71*A-1495*B)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-1/24 
0*(787*A-475*B)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1 
/80*(157*A-85*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 7.13 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.51 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-120 (283 A-163 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+4 (3491 A-1895 B+5 (887 A-479 B) \cos (c+d x)+16 (52 A-25 B) \cos (2 (c+d x))-40 A \cos (3 (c+d x))+40 B \cos (3 (c+d x))+12 A \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{960 a d \cos ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(Sec[(c + d*x)/2]*(-120*(283*A - 163*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + 
 d*x)/2]^4 + 4*(3491*A - 1895*B + 5*(887*A - 479*B)*Cos[c + d*x] + 16*(52* 
A - 25*B)*Cos[2*(c + d*x)] - 40*A*Cos[3*(c + d*x)] + 40*B*Cos[3*(c + d*x)] 
 + 12*A*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]))/(960*a*d*Cos[c + d*x]^(3/2)*( 
a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 2.11 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.09, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 3434, 3042, 4508, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (13 A-5 B)-8 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (13 A-5 B)-8 a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (13 A-5 B)-8 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B)-6 a^2 (21 A-13 B) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B)-6 a^2 (21 A-13 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B)-6 a^2 (21 A-13 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {a^3 (787 A-475 B)-4 a^3 (157 A-85 B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (787 A-475 B)-4 a^3 (157 A-85 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (787 A-475 B)-4 a^3 (157 A-85 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^4 (2671 A-1495 B)-2 a^4 (787 A-475 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (2671 A-1495 B)-2 a^4 (787 A-475 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (2671 A-1495 B)-2 a^4 (787 A-475 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^4 (283 A-163 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^4 (283 A-163 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {30 a^4 (283 A-163 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^4 (2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {15 \sqrt {2} a^{7/2} (283 A-163 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B)*Sin[c + d*x])/(d*Sec[ 
c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(21*A - 13*B)*Sin[c 
+ d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((2*a^2*(157*A 
 - 85*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - 
 ((2*a^3*(787*A - 475*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a* 
Sec[c + d*x]]) - ((-15*Sqrt[2]*a^(7/2)*(283*A - 163*B)*ArcTanh[(Sqrt[a]*Sq 
rt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2 
*a^4*(2671*A - 1495*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[ 
c + d*x]]))/(3*a))/(5*a))/(4*a^2))/(8*a^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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3434

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Csc[e + f*x])^m*((c + 
d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 4295

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

rule 4501

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

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 [A] (warning: unable to verify)

Time = 1.68 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\left (\left (4245 \cos \left (d x +c \right )^{3}+12735 \cos \left (d x +c \right )^{2}+12735 \cos \left (d x +c \right )+4245\right ) A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-2445 \cos \left (d x +c \right )^{3}-7335 \cos \left (d x +c \right )^{2}-7335 \cos \left (d x +c \right )-2445\right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (192 \cos \left (d x +c \right )^{4}-320 \cos \left (d x +c \right )^{3}+3136 \cos \left (d x +c \right )^{2}+9110 \cos \left (d x +c \right )+5342\right ) \sin \left (d x +c \right ) A +\left (320 \cos \left (d x +c \right )^{3}-1600 \cos \left (d x +c \right )^{2}-5030 \cos \left (d x +c \right )-2990\right ) \sin \left (d x +c \right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{480 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{3}}\)	\(308\)

Input:

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

Output:

1/480/d*((4245*cos(d*x+c)^3+12735*cos(d*x+c)^2+12735*cos(d*x+c)+4245)*A*(- 
2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1 
+cos(d*x+c)))^(1/2))+(-2445*cos(d*x+c)^3-7335*cos(d*x+c)^2-7335*cos(d*x+c) 
-2445)*B*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x 
+c))/(-1/(1+cos(d*x+c)))^(1/2))+(192*cos(d*x+c)^4-320*cos(d*x+c)^3+3136*co 
s(d*x+c)^2+9110*cos(d*x+c)+5342)*sin(d*x+c)*A+(320*cos(d*x+c)^3-1600*cos(d 
*x+c)^2-5030*cos(d*x+c)-2990)*sin(d*x+c)*B)*cos(d*x+c)^(1/2)*(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)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.83 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/960*(15*sqrt(2)*((283*A - 163*B)*cos(d*x + c)^3 + 3*(283*A - 163*B)*co 
s(d*x + c)^2 + 3*(283*A - 163*B)*cos(d*x + c) + 283*A - 163*B)*sqrt(a)*log 
(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + 
 c)^2 + 2*cos(d*x + c) + 1)) - 4*(96*A*cos(d*x + c)^4 - 160*(A - B)*cos(d* 
x + c)^3 + 32*(49*A - 25*B)*cos(d*x + c)^2 + 5*(911*A - 503*B)*cos(d*x + c 
) + 2671*A - 1495*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(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), 1/480*(15*sqrt(2)*((283*A - 163*B)*cos(d*x + c)^ 
3 + 3*(283*A - 163*B)*cos(d*x + c)^2 + 3*(283*A - 163*B)*cos(d*x + c) + 28 
3*A - 163*B)*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c) + a)) + 2* 
(96*A*cos(d*x + c)^4 - 160*(A - B)*cos(d*x + c)^3 + 32*(49*A - 25*B)*cos(d 
*x + c)^2 + 5*(911*A - 503*B)*cos(d*x + c) + 2671*A - 1495*B)*sqrt((a*cos( 
d*x + c) + a)/cos(d*x + c))*sqrt(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)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244922 vs. \(2 (270) = 540\).

Time = 5.20 (sec) , antiderivative size = 244922, normalized size of antiderivative = 772.62 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/480*((16*(3*cos(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) + 3*sin(5*d*x + 5*c) 
^2*sin(5/2*d*x + 5/2*c) - 25*(cos(5*d*x + 5*c)^2 + sin(5*d*x + 5*c)^2)*sin 
(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 300*(cos(5*d*x 
 + 5*c)^2 + sin(5*d*x + 5*c)^2)*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos( 
5/2*d*x + 5/2*c))))*cos(13/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5 
/2*c)))^4 + 4096*(3*cos(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) + 3*sin(5*d*x 
+ 5*c)^2*sin(5/2*d*x + 5/2*c) - 25*(cos(5*d*x + 5*c)^2 + sin(5*d*x + 5*c)^ 
2)*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 300*(cos 
(5*d*x + 5*c)^2 + sin(5*d*x + 5*c)^2)*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c) 
, cos(5/2*d*x + 5/2*c))))*cos(11/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d 
*x + 5/2*c)))^4 + 20736*(3*cos(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) + 3*sin 
(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) - 25*(cos(5*d*x + 5*c)^2 + sin(5*d*x 
+ 5*c)^2)*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3 
00*(cos(5*d*x + 5*c)^2 + sin(5*d*x + 5*c)^2)*sin(1/5*arctan2(sin(5/2*d*x + 
 5/2*c), cos(5/2*d*x + 5/2*c))))*cos(9/5*arctan2(sin(5/2*d*x + 5/2*c), cos 
(5/2*d*x + 5/2*c)))^4 + 4096*(3*cos(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) + 
3*sin(5*d*x + 5*c)^2*sin(5/2*d*x + 5/2*c) - 25*(cos(5*d*x + 5*c)^2 + sin(5 
*d*x + 5*c)^2)*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)) 
) + 300*(cos(5*d*x + 5*c)^2 + sin(5*d*x + 5*c)^2)*sin(1/5*arctan2(sin(5/2* 
d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*cos(7/5*arctan2(sin(5/2*d*x + 5/2...
 

Giac [A] (verification not implemented)

Time = 175.49 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} - \frac {29 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 21 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {6733 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 3685 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, {\left (1973 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1133 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (291 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 155 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}} - \frac {15 \, {\left (283 \, \sqrt {2} A - 163 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \] Input:

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

Output:

-1/480*((((15*(2*(sqrt(2)*A*a^2*sgn(cos(d*x + c)) - sqrt(2)*B*a^2*sgn(cos( 
d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^2 - (29*sqrt(2)*A*a^2*sgn(cos(d*x + c) 
) - 21*sqrt(2)*B*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (673 
3*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 3685*sqrt(2)*B*a^2*sgn(cos(d*x + c)))/ 
a^2)*tan(1/2*d*x + 1/2*c)^2 - 5*(1973*sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 11 
33*sqrt(2)*B*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(291* 
sqrt(2)*A*a^2*sgn(cos(d*x + c)) - 155*sqrt(2)*B*a^2*sgn(cos(d*x + c)))/a^2 
)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2) - 15*(283*sqrt 
(2)*A - 163*sqrt(2)*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan( 
1/2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^(5/2)*(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)*sqrt(cos(c + d*x))*cos(c + d*x)**2*s 
ec(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)*sqrt(cos(c + d*x))*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