\(\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [830]

Optimal result

Integrand size = 40, antiderivative size = 469 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 b B+435 a b^3 B-10 a^4 C+279 a^2 b^2 C+147 b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 B-49 C)-6 a b^2 (60 B-19 C)+15 a^2 b (3 B-11 C)-10 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}+\frac {2 \left (45 a^2 b B+75 b^3 B-10 a^3 C+114 a b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 20.42 (sec) , antiderivative size = 660, normalized size of antiderivative = 1.41 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 (a+b) \left (-45 a^3 b B-435 a b^3 B+10 a^4 C-279 a^2 b^2 C-147 b^4 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (-10 a^3 C+15 a^2 b (3 B+11 C)+6 a b^2 (60 B+19 C)+3 b^3 (25 B+49 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-45 a^3 b B-435 a b^3 B+10 a^4 C-279 a^2 b^2 C-147 b^4 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^2 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 \left (45 a^3 b B+435 a b^3 B-10 a^4 C+279 a^2 b^2 C+147 b^4 C\right ) \sin (c+d x)}{315 b^2}+\frac {2}{63} \sec ^3(c+d x) \left (9 b^2 B \sin (c+d x)+19 a b C \sin (c+d x)\right )+\frac {2}{315} \sec ^2(c+d x) \left (135 a b B \sin (c+d x)+75 a^2 C \sin (c+d x)+49 b^2 C \sin (c+d x)\right )+\frac {2 \sec (c+d x) \left (135 a^2 b B \sin (c+d x)+75 b^3 B \sin (c+d x)+5 a^3 C \sin (c+d x)+163 a b^2 C \sin (c+d x)\right )}{315 b}+\frac {2}{9} b^2 C \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \] Input:

Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),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*b*B - 435*a*b^3*B + 10*a^4*C - 279*a^2*b^2*C - 147*b^4*C)*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*C + 15*a^2*b*(3*B + 11*C) + 6*a*b^2*(60*B + 19*C) + 3* 
b^3*(25*B + 49*C))*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*b*B - 435*a*b^3*B + 10*a^4*C - 279*a^2*b^2*C 
- 147*b^4*C)*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*b 
*B + 435*a*b^3*B - 10*a^4*C + 279*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(31 
5*b^2) + (2*Sec[c + d*x]^3*(9*b^2*B*Sin[c + d*x] + 19*a*b*C*Sin[c + d*x])) 
/63 + (2*Sec[c + d*x]^2*(135*a*b*B*Sin[c + d*x] + 75*a^2*C*Sin[c + d*x] + 
49*b^2*C*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(135*a^2*b*B*Sin[c + d*x] + 
75*b^3*B*Sin[c + d*x] + 5*a^3*C*Sin[c + d*x] + 163*a*b^2*C*Sin[c + d*x]))/ 
(315*b) + (2*b^2*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x] 
)^2)
 

Rubi [A] (verified)

Time = 2.13 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.475, Rules used = {3042, 4560, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

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

rule 4560

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

Maple [B] (verified)

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

Time = 122.00 (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)*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth 
od=_RETURNVERBOSE)
 

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)*((279*cos(d*x+c)^2+80*cos(d*x+c)+80)*C*a^3*b^2*tan(d*x+c)+5*sin(d*x+ 
c)*(cos(d*x+c)-1)*a^4*b*C+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)*C*b^5*tan(d*x+c)*sec(d*x+c)^3+(163*cos(d*x+c)^3+442*cos(d 
*x+c)^2+170*cos(d*x+c)+170)*C*a^2*b^3*tan(d*x+c)*sec(d*x+c)+(147*cos(d*x+c 
)^4+212*cos(d*x+c)^3+212*cos(d*x+c)^2+130*cos(d*x+c)+130)*C*a*b^4*tan(d*x+ 
c)*sec(d*x+c)^2+10*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+ 
c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^5*EllipticE( 
-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+147*(cos(d*x+c)^2+2*cos(d*x+c) 
+1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x 
+c)+1))^(1/2)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+45 
*B*a^4*b*cos(d*x+c)*sin(d*x+c)+147*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2 
)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+155*(-cos(d*x+ 
c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),(( 
a-b)/(a+b))^(1/2))+261*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d 
*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^4*Elli 
pticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+10*(-cos(d*x+c)^2-2*cos( 
d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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