\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [364]

Optimal result

Integrand size = 33, antiderivative size = 469 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 A-49 B)-6 a b^2 (60 A-19 B)+15 a^2 b (3 A-11 B)-10 a^3 B\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 A b-2 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 B (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \] Output:

-2/315*(a-b)*(a+b)^(1/2)*(45*A*a^3*b+435*A*a*b^3-10*B*a^4+279*B*a^2*b^2+14 
7*B*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a 
-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2) 
/b^3/d-2/315*(a-b)*(a+b)^(1/2)*(3*b^3*(25*A-49*B)-6*a*b^2*(60*A-19*B)+15*a 
^2*b*(3*A-11*B)-10*B*a^3)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b 
)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x 
+c))/(a-b))^(1/2)/b^2/d+2/315*(45*A*a^2*b+75*A*b^3-10*B*a^3+114*B*a*b^2)*( 
a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d+2/315*(45*A*a*b-10*B*a^2+49*B*b^2)*(a 
+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/63*(9*A*b-2*B*a)*(a+b*sec(d*x+c))^(5 
/2)*tan(d*x+c)/b/d+2/9*B*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 20.21 (sec) , antiderivative size = 660, normalized size of antiderivative = 1.41 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 (a+b) \left (-45 a^3 A b-435 a A b^3+10 a^4 B-279 a^2 b^2 B-147 b^4 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (-10 a^3 B+15 a^2 b (3 A+11 B)+6 a b^2 (60 A+19 B)+3 b^3 (25 A+49 B)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-45 a^3 A b-435 a A b^3+10 a^4 B-279 a^2 b^2 B-147 b^4 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^2 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \sin (c+d x)}{315 b^2}+\frac {2}{63} \sec ^3(c+d x) \left (9 A b^2 \sin (c+d x)+19 a b B \sin (c+d x)\right )+\frac {2}{315} \sec ^2(c+d x) \left (135 a A b \sin (c+d x)+75 a^2 B \sin (c+d x)+49 b^2 B \sin (c+d x)\right )+\frac {2 \sec (c+d x) \left (135 a^2 A b \sin (c+d x)+75 A b^3 \sin (c+d x)+5 a^3 B \sin (c+d x)+163 a b^2 B \sin (c+d x)\right )}{315 b}+\frac {2}{9} b^2 B \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \] Input:

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

Output:

(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*(a 
+ b)*(-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B - 147*b^4*B)*Sq 
rt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 
+ Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2 
*b*(a + b)*(-10*a^3*B + 15*a^2*b*(3*A + 11*B) + 6*a*b^2*(60*A + 19*B) + 3* 
b^3*(25*A + 49*B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c 
 + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], 
 (a - b)/(a + b)] + (-45*a^3*A*b - 435*a*A*b^3 + 10*a^4*B - 279*a^2*b^2*B 
- 147*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + 
 d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c + d*x)/2]^2]*Sec[ 
c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(45*a^3*A 
*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*B)*Sin[c + d*x])/(31 
5*b^2) + (2*Sec[c + d*x]^3*(9*A*b^2*Sin[c + d*x] + 19*a*b*B*Sin[c + d*x])) 
/63 + (2*Sec[c + d*x]^2*(135*a*A*b*Sin[c + d*x] + 75*a^2*B*Sin[c + d*x] + 
49*b^2*B*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(135*a^2*A*b*Sin[c + d*x] + 
75*A*b^3*Sin[c + d*x] + 5*a^3*B*Sin[c + d*x] + 163*a*b^2*B*Sin[c + d*x]))/ 
(315*b) + (2*b^2*B*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x] 
)^2)
 

Rubi [A] (verified)

Time = 2.03 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4498, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 4319, 4492}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} (7 b B+(9 A b-2 a B) \sec (c+d x))dx}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (7 b B+(9 A b-2 a B) \sec (c+d x))dx}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (7 b B+(9 A b-2 a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 A b+13 a B)+\left (-10 B a^2+45 A b a+49 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 A b+13 a B)+\left (-10 B a^2+45 A b a+49 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b (15 A b+13 a B)+\left (-10 B a^2+45 A b a+49 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (55 B a^2+120 A b a+49 b^2 B\right )+\left (-10 B a^3+45 A b a^2+114 b^2 B a+75 A b^3\right ) \sec (c+d x)\right )dx+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (55 B a^2+120 A b a+49 b^2 B\right )+\left (-10 B a^3+45 A b a^2+114 b^2 B a+75 A b^3\right ) \sec (c+d x)\right )dx+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (55 B a^2+120 A b a+49 b^2 B\right )+\left (-10 B a^3+45 A b a^2+114 b^2 B a+75 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (155 B a^3+405 A b a^2+261 b^2 B a+75 A b^3\right )+\left (-10 B a^4+45 A b a^3+279 b^2 B a^2+435 A b^3 a+147 b^4 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (155 B a^3+405 A b a^2+261 b^2 B a+75 A b^3\right )+\left (-10 B a^4+45 A b a^3+279 b^2 B a^2+435 A b^3 a+147 b^4 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (155 B a^3+405 A b a^2+261 b^2 B a+75 A b^3\right )+\left (-10 B a^4+45 A b a^3+279 b^2 B a^2+435 A b^3 a+147 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\left (-10 a^4 B+45 a^3 A b+279 a^2 b^2 B+435 a A b^3+147 b^4 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-10 a^3 B+a^2 (45 A b-165 b B)-6 a b^2 (60 A-19 B)+3 b^3 (25 A-49 B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\left (-10 a^4 B+45 a^3 A b+279 a^2 b^2 B+435 a A b^3+147 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-10 a^3 B+a^2 (45 A b-165 b B)-6 a b^2 (60 A-19 B)+3 b^3 (25 A-49 B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\left (-10 a^4 B+45 a^3 A b+279 a^2 b^2 B+435 a A b^3+147 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-10 a^3 B+a^2 (45 A b-165 b B)-6 a b^2 (60 A-19 B)+3 b^3 (25 A-49 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{7} \left (\frac {2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac {3}{5} \left (\frac {2 \left (-10 a^3 B+45 a^2 A b+114 a b^2 B+75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \left (-10 a^3 B+a^2 (45 A b-165 b B)-6 a b^2 (60 A-19 B)+3 b^3 (25 A-49 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (-10 a^4 B+45 a^3 A b+279 a^2 b^2 B+435 a A b^3+147 b^4 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )\right )+\frac {2 (9 A b-2 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}\)

Input:

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

Output:

(2*B*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((2*(9*A*b - 2*a*B 
)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*(45*a*A*b - 10*a^2* 
B + 49*b^2*B)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((-2*(a 
 - b)*Sqrt[a + b]*(45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 1 
47*b^4*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + 
b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + 
Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(3*b^3*(25*A - 
49*B) - 6*a*b^2*(60*A - 19*B) - 10*a^3*B + a^2*(45*A*b - 165*b*B))*Cot[c + 
 d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - 
 b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a 
 - b))])/(b*d))/3 + (2*(45*a^2*A*b + 75*A*b^3 - 10*a^3*B + 114*a*b^2*B)*Sq 
rt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/7)/(9*b)
 

Defintions of rubi rules used

rule 27

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

rule 3042

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

rule 4319

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

rule 4490

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

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B)   Int[Csc[e 
 + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B   Int[Csc[e + f*x]*((1 + 
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} 
, x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
 

rule 4498

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2423\) vs. \(2(431)=862\).

Time = 176.48 (sec) , antiderivative size = 2424, normalized size of antiderivative = 5.17

method	result	size
default	\(\text {Expression too large to display}\)	\(2424\)
parts	\(\text {Expression too large to display}\)	\(2437\)

Input:

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/315/d/b^2*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x 
+c)+b)*(7*(-21*cos(d*x+c)^4-7*cos(d*x+c)^3-7*cos(d*x+c)^2-5*cos(d*x+c)-5)* 
B*b^5*tan(d*x+c)*sec(d*x+c)^3+5*sin(d*x+c)*(1-cos(d*x+c))*B*a^4*b+15*(-5*c 
os(d*x+c)^3-5*cos(d*x+c)^2-3*cos(d*x+c)-3)*A*b^5*tan(d*x+c)*sec(d*x+c)^2-4 
5*A*a^4*b*cos(d*x+c)*sin(d*x+c)+155*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(1/(a+ 
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*a^3*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+279*(cos(d 
*x+c)^2+2*cos(d*x+c)+1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)* 
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c) 
,((a-b)/(a+b))^(1/2))+279*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*c 
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^3*b^2 
*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+279*(-cos(d*x+c)^2- 
2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x 
+c)/(1+cos(d*x+c)))^(1/2)*a^2*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/ 
(a+b))^(1/2))+(-147*cos(d*x+c)^4-212*cos(d*x+c)^3-212*cos(d*x+c)^2-130*cos 
(d*x+c)-130)*B*a*b^4*tan(d*x+c)*sec(d*x+c)^2+(-163*cos(d*x+c)^3-442*cos(d* 
x+c)^2-170*cos(d*x+c)-170)*B*a^2*b^3*tan(d*x+c)*sec(d*x+c)+15*(-5*cos(d*x+ 
c)^3-34*cos(d*x+c)^2-12*cos(d*x+c)-12)*A*a*b^4*tan(d*x+c)*sec(d*x+c)+45*si 
n(d*x+c)*(-4-3*cos(d*x+c))*A*a^3*b^2+15*(-29*cos(d*x+c)^2-18*cos(d*x+c)-18 
)*A*a^2*b^3*tan(d*x+c)+(-279*cos(d*x+c)^2-80*cos(d*x+c)-80)*B*a^3*b^2*t...
 

Fricas [F]

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

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

Output:

integral((B*b^2*sec(d*x + c)^5 + A*a^2*sec(d*x + c)^2 + (2*B*a*b + A*b^2)* 
sec(d*x + c)^4 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a 
), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{5}d x \right ) b^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}d x \right ) a^{3} \] Input:

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

Output:

int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5,x)*b**3 + 3*int(sqrt(sec(c + 
d*x)*b + a)*sec(c + d*x)**4,x)*a*b**2 + 3*int(sqrt(sec(c + d*x)*b + a)*sec 
(c + d*x)**3,x)*a**2*b + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**2,x)*a 
**3