\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [1286]

Optimal result

Integrand size = 45, antiderivative size = 333 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(283 A-163 B+75 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(21 A-13 B+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(2671 A-1495 B+735 C) \sin (c+d x)}{240 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(787 A-475 B+195 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(157 A-85 B+45 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/32*(283*A-163*B+75*C)*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+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)- 
1/16*(21*A-13*B+5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2 
)+1/240*(2671*A-1495*B+735*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d 
*x+c))^(1/2)-1/240*(787*A-475*B+195*C)*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+45*C)*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 = 8.93 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-120 (283 A-163 B+75 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+4 (3491 A-1895 B+975 C+5 (887 A-479 B+255 C) \cos (c+d x)+16 (52 A-25 B+15 C) \cos (2 (c+d x))-40 A \cos (3 (c+d x))+40 B \cos (3 (c+d x))+12 A \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{960 a d \cos ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(Sec[(c + d*x)/2]*(-120*(283*A - 163*B + 75*C)*ArcTanh[Sin[(c + d*x)/2]]*C 
os[(c + d*x)/2]^4 + 4*(3491*A - 1895*B + 975*C + 5*(887*A - 479*B + 255*C) 
*Cos[c + d*x] + 16*(52*A - 25*B + 15*C)*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 = 2.24 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.08, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4753, 3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B+45 C)-6 a^2 (21 A-13 B+5 C) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B+45 C)-6 a^2 (21 A-13 B+5 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (157 A-85 B+45 C)-6 a^2 (21 A-13 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {a^3 (787 A-475 B+195 C)-4 a^3 (157 A-85 B+45 C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (787 A-475 B+195 C)-4 a^3 (157 A-85 B+45 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (787 A-475 B+195 C)-4 a^3 (157 A-85 B+45 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^4 (2671 A-1495 B+735 C)-2 a^4 (787 A-475 B+195 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (2671 A-1495 B+735 C)-2 a^4 (787 A-475 B+195 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^4 (2671 A-1495 B+735 C)-2 a^4 (787 A-475 B+195 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^4 (283 A-163 B+75 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-15 a^4 (283 A-163 B+75 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {30 a^4 (283 A-163 B+75 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^4 (2671 A-1495 B+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (157 A-85 B+45 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (787 A-475 B+195 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (2671 A-1495 B+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {15 \sqrt {2} a^{7/2} (283 A-163 B+75 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{4 a^2}-\frac {a (21 A-13 B+5 C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*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 + 5 
*C)*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 + 45*C)*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 + 195*C)*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 - 1 
63*B + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sq 
rt[a + a*Sec[c + d*x]])])/d + (2*a^4*(2671*A - 1495*B + 735*C)*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 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]
 

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)]
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [A] (warning: unable to verify)

Time = 2.55 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\left (\left (-4245 \cos \left (d x +c \right )^{3}-12735 \cos \left (d x +c \right )^{2}-12735 \cos \left (d x +c \right )-4245\right ) A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (2445 \cos \left (d x +c \right )^{3}+7335 \cos \left (d x +c \right )^{2}+7335 \cos \left (d x +c \right )+2445\right ) B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-1125 \cos \left (d x +c \right )^{3}-3375 \cos \left (d x +c \right )^{2}-3375 \cos \left (d x +c \right )-1125\right ) C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (192 \cos \left (d x +c \right )^{4}-320 \cos \left (d x +c \right )^{3}+3136 \cos \left (d x +c \right )^{2}+9110 \cos \left (d x +c \right )+5342\right ) \sin \left (d x +c \right ) A +\left (320 \cos \left (d x +c \right )^{3}-1600 \cos \left (d x +c \right )^{2}-5030 \cos \left (d x +c \right )-2990\right ) \sin \left (d x +c \right ) B +\left (960 \cos \left (d x +c \right )^{2}+2550 \cos \left (d x +c \right )+1470\right ) \sin \left (d x +c \right ) C \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{480 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{3}}\)	\(417\)

Input:

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

Output:

1/480/d*((-4245*cos(d*x+c)^3-12735*cos(d*x+c)^2-12735*cos(d*x+c)-4245)*A*( 
-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(c 
os(d*x+c)+1))^(1/2))+(2445*cos(d*x+c)^3+7335*cos(d*x+c)^2+7335*cos(d*x+c)+ 
2445)*B*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(csc(d*x+c)-cot(d*x+c 
))/(-1/(cos(d*x+c)+1))^(1/2))+(-1125*cos(d*x+c)^3-3375*cos(d*x+c)^2-3375*c 
os(d*x+c)-1125)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(csc(d*x+c) 
-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))+(192*cos(d*x+c)^4-320*cos(d*x+c)^3 
+3136*cos(d*x+c)^2+9110*cos(d*x+c)+5342)*sin(d*x+c)*A+(320*cos(d*x+c)^3-16 
00*cos(d*x+c)^2-5030*cos(d*x+c)-2990)*sin(d*x+c)*B+(960*cos(d*x+c)^2+2550* 
cos(d*x+c)+1470)*sin(d*x+c)*C)*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 = 0.10 (sec) , antiderivative size = 622, normalized size of antiderivative = 1.87 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/960*(15*sqrt(2)*((283*A - 163*B + 75*C)*cos(d*x + c)^3 + 3*(283*A - 163 
*B + 75*C)*cos(d*x + c)^2 + 3*(283*A - 163*B + 75*C)*cos(d*x + c) + 283*A 
- 163*B + 75*C)*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 + 15*C)*cos(d*x + c) 
^2 + 5*(911*A - 503*B + 255*C)*cos(d*x + c) + 2671*A - 1495*B + 735*C)*sqr 
t((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 + 75*C)*cos(d*x + c)^3 + 3*(283*A - 16 
3*B + 75*C)*cos(d*x + c)^2 + 3*(283*A - 163*B + 75*C)*cos(d*x + c) + 283*A 
 - 163*B + 75*C)*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 + 
15*C)*cos(d*x + c)^2 + 5*(911*A - 503*B + 255*C)*cos(d*x + c) + 2671*A - 1 
495*B + 735*C)*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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(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. 250341 vs. \(2 (286) = 572\).

Time = 6.09 (sec) , antiderivative size = 250341, normalized size of antiderivative = 751.77 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

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 = 171.54 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/480*((((15*(2*(sqrt(2)*A*a^2*sgn(cos(d*x + c)) - sqrt(2)*B*a^2*sgn(cos( 
d*x + c)) + sqrt(2)*C*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)) + 
 13*sqrt(2)*C*a^2*sgn(cos(d*x + c)))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (6733*s 
qrt(2)*A*a^2*sgn(cos(d*x + c)) - 3685*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 17 
25*sqrt(2)*C*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)) - 1133*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 5 
49*sqrt(2)*C*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)) + 83 
*sqrt(2)*C*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 + 75*sqrt(2)*C) 
*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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+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)*sqrt(cos(c + d*x))*cos(c + d*x)**2*s 
ec(c + d*x)**2)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1) 
,x)*c + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*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)*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