\(\int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [815]

Optimal result

Integrand size = 42, antiderivative size = 397 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (14 a^2 b B-63 b^3 B-8 a^3 C-19 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}+\frac {2 (a-b) \sqrt {a+b} \left (b^2 (63 B-25 C)+2 a b (7 B-3 C)-8 a^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 \left (7 a b B-4 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}+\frac {2 (7 b B+a C) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{35 b d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{7 d} \] Output:

2/105*(a-b)*(a+b)^(1/2)*(14*B*a^2*b-63*B*b^3-8*C*a^3-19*C*a*b^2)*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^4/d+2/105*(a-b) 
*(a+b)^(1/2)*(b^2*(63*B-25*C)+2*a*b*(7*B-3*C)-8*C*a^2)*cot(d*x+c)*Elliptic 
F((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+2/105*(7*B*a*b-4*C*a^ 
2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/35*(7*B*b+C*a)*sec(d 
*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d+2/7*C*sec(d*x+c)^2*(a+b*sec(d* 
x+c))^(1/2)*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 16.49 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.34 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (2 (a+b) \left (-14 a^2 b B+63 b^3 B+8 a^3 C+19 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (8 a^2 C-2 a b (7 B+3 C)+b^2 (63 B+25 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-14 a^2 b B+63 b^3 B+8 a^3 C+19 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^3 d (b+a \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 \left (-14 a^2 b B+63 b^3 B+8 a^3 C+19 a b^2 C\right ) \sin (c+d x)}{105 b^3}+\frac {2 \sec ^2(c+d x) (7 b B \sin (c+d x)+a C \sin (c+d x))}{35 b}+\frac {2 \sec (c+d x) \left (7 a b B \sin (c+d x)-4 a^2 C \sin (c+d x)+25 b^2 C \sin (c+d x)\right )}{105 b^2}+\frac {2}{7} C \sec ^2(c+d x) \tan (c+d x)\right )}{d} \] Input:

Integrate[Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(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]]*Sqrt[a + b*Sec[c + d*x]]*(2*(a + 
 b)*(-14*a^2*b*B + 63*b^3*B + 8*a^3*C + 19*a*b^2*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]))]*El 
lipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(a + b)*(8*a^2*C 
- 2*a*b*(7*B + 3*C) + b^2*(63*B + 25*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[Arc 
Sin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-14*a^2*b*B + 63*b^3*B + 8*a^3* 
C + 19*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[( 
c + d*x)/2]))/(105*b^3*d*(b + a*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sqr 
t[Sec[c + d*x]]) + (Sqrt[a + b*Sec[c + d*x]]*((2*(-14*a^2*b*B + 63*b^3*B + 
 8*a^3*C + 19*a*b^2*C)*Sin[c + d*x])/(105*b^3) + (2*Sec[c + d*x]^2*(7*b*B* 
Sin[c + d*x] + a*C*Sin[c + d*x]))/(35*b) + (2*Sec[c + d*x]*(7*a*b*B*Sin[c 
+ d*x] - 4*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^2) + (2*C*S 
ec[c + d*x]^2*Tan[c + d*x])/7))/d
 

Rubi [A] (verified)

Time = 1.91 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4519, 27, 3042, 4580, 27, 3042, 4570, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} (B+C \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4519

\(\displaystyle \frac {2}{7} \int \frac {\sec ^2(c+d x) \left ((7 b B+a C) \sec ^2(c+d x)+(7 a B+5 b C) \sec (c+d x)+4 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {\sec ^2(c+d x) \left ((7 b B+a C) \sec ^2(c+d x)+(7 a B+5 b C) \sec (c+d x)+4 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left ((7 b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(7 a B+5 b C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 4580

\(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\sec (c+d x) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+23 a C) \sec (c+d x)+2 a (7 b B+a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {\int \frac {\sec (c+d x) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+23 a C) \sec (c+d x)+2 a (7 b B+a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-4 C a^2+7 b B a+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (21 b B+23 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (7 b B+a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {1}{7} \left (\frac {\frac {2 \int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-19 b^2 C a-63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-19 b^2 C a-63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^2+49 b B a+25 b^2 C\right )+\left (8 C a^3-14 b B a^2+19 b^2 C a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {1}{7} \left (\frac {\frac {(a-b) \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {\frac {(a-b) \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{7} \left (\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{7} \left (\frac {\frac {2 \left (-4 a^2 C+7 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}+\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 C+a b (14 B-6 C)+b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 C+14 a^2 b B-19 a b^2 C-63 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}}{5 b}+\frac {2 (a C+7 b B) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\right )+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*d) + ((2*(7* 
b*B + a*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) + ( 
((2*(a - b)*Sqrt[a + b]*(14*a^2*b*B - 63*b^3*B - 8*a^3*C - 19*a*b^2*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]*(b^2*(63*B - 25*C) + a*b*(14 
*B - 6*C) - 8*a^2*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*b) + (2*(7*a*b*B - 4*a^2*C 
+ 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b))/7
 

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 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 4519

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

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]
 

rule 4570

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

rule 4580

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x 
_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 
1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[Csc[e + f*x]*(a + b*Csc[e 
+ f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* 
(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & 
& NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1868\) vs. \(2(363)=726\).

Time = 70.60 (sec) , antiderivative size = 1869, normalized size of antiderivative = 4.71

method	result	size
default	\(\text {Expression too large to display}\)	\(1869\)
parts	\(\text {Expression too large to display}\)	\(1889\)

Input:

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

2/105/d/b^3*(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)*(14*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(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^3*b*EllipticE(-csc(d* 
x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+14*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*( 
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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+63*(c 
os(d*x+c)^2+2*cos(d*x+c)+1)*B*(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^3*EllipticE(-csc(d*x+c)+cot(d*x+ 
c),((a-b)/(a+b))^(1/2))+63*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^4*Ell 
ipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(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))/(co 
s(d*x+c)+1))^(1/2)*a^4*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2 
))+8*(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^3*b*EllipticE(-csc(d*x+c)+c 
ot(d*x+c),((a-b)/(a+b))^(1/2))+19*(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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+19*(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^3*EllipticE(-csc(d*x+c)+cot(d*x+c),(...
 

Fricas [F]

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

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**(1/2)*(B*sec(d*x+c)+C*sec(d*x+c) 
**2),x)
 

Output:

Integral((B + C*sec(c + d*x))*sqrt(a + b*sec(c + d*x))*sec(c + d*x)**3, x)
 

Maxima [F]

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

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)*sec 
(d*x + c)^2, x)
 

Giac [F]

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

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(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))*sqrt(b*sec(d*x + c) + a)*sec 
(d*x + c)^2, x)
 

Mupad [F(-1)]

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

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2))/cos(c 
 + d*x)^2,x)
 

Output:

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2))/cos(c 
 + d*x)^2, x)
 

Reduce [F]

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

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(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)**4,x)*c + int(sqrt(sec(c + d*x)* 
b + a)*sec(c + d*x)**3,x)*b