\(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [837]

Optimal result

Integrand size = 42, antiderivative size = 411 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (56 a^2 b B+63 b^3 B-48 a^3 C-44 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^5 d}-\frac {2 \sqrt {a+b} \left (b^3 (63 B-25 C)-48 a^3 C+4 a^2 b (14 B+3 C)-2 a b^2 (7 B+22 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}-\frac {2 \left (28 a b B-24 a^2 C-25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{35 b^2 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{7 b d} \] Output:

-2/105*(a-b)*(a+b)^(1/2)*(56*B*a^2*b+63*B*b^3-48*C*a^3-44*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^5/d-2/105*(a+ 
b)^(1/2)*(b^3*(63*B-25*C)-48*a^3*C+4*a^2*b*(14*B+3*C)-2*a*b^2*(7*B+22*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^4/d-2/ 
105*(28*B*a*b-24*C*a^2-25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/d+2 
/35*(7*B*b-6*C*a)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/7*C 
*sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 17.74 (sec) , antiderivative size = 550, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-56 a^2 b B-63 b^3 B+48 a^3 C+44 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b \left (2 a b^2 (7 B-22 C)+4 a^2 b (14 B-3 C)-48 a^3 C+b^3 (63 B+25 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-56 a^2 b B-63 b^3 B+48 a^3 C+44 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^4 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \sec (c+d x)}}+\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (\frac {2 \left (56 a^2 b B+63 b^3 B-48 a^3 C-44 a b^2 C\right ) \sin (c+d x)}{105 b^4}+\frac {2 \sec ^2(c+d x) (7 b B \sin (c+d x)-6 a C \sin (c+d x))}{35 b^2}+\frac {2 \sec (c+d x) \left (-28 a b B \sin (c+d x)+24 a^2 C \sin (c+d x)+25 b^2 C \sin (c+d x)\right )}{105 b^3}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{7 b}\right )}{d \sqrt {a+b \sec (c+d x)}} \] Input:

Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b* 
Sec[c + d*x]],x]
 

Output:

(2*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-5 
6*a^2*b*B - 63*b^3*B + 48*a^3*C + 44*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]))]*Elliptic 
E[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(2*a*b^2*(7*B - 22*C) + 
 4*a^2*b*(14*B - 3*C) - 48*a^3*C + b^3*(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[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-56*a^2*b*B - 63*b 
^3*B + 48*a^3*C + 44*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^4*d*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[a + b 
*Sec[c + d*x]]) + ((b + a*Cos[c + d*x])*Sec[c + d*x]*((2*(56*a^2*b*B + 63* 
b^3*B - 48*a^3*C - 44*a*b^2*C)*Sin[c + d*x])/(105*b^4) + (2*Sec[c + d*x]^2 
*(7*b*B*Sin[c + d*x] - 6*a*C*Sin[c + d*x]))/(35*b^2) + (2*Sec[c + d*x]*(-2 
8*a*b*B*Sin[c + d*x] + 24*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(10 
5*b^3) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(7*b)))/(d*Sqrt[a + b*Sec[c + d 
*x]])
 

Rubi [A] (verified)

Time = 1.94 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.05, 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, 4521, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (-\left (\left (-24 C a^2+28 b B a-25 b^2 C\right ) \sec ^2(c+d x)\right )+b (21 b B+2 a C) \sec (c+d x)+2 a (7 b B-6 a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (-\left (\left (-24 C a^2+28 b B a-25 b^2 C\right ) \sec ^2(c+d x)\right )+b (21 b B+2 a C) \sec (c+d x)+2 a (7 b B-6 a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (24 C a^2-28 b B a+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (21 b B+2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (7 b B-6 a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {\frac {2 \int \frac {\sec (c+d x) \left (b \left (-12 C a^2+14 b B a+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-44 b^2 C a+63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^2+14 b B a+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-44 b^2 C a+63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^2+14 b B a+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-44 b^2 C a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\frac {\left (-48 a^3 C+56 a^2 b B-44 a b^2 C+63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-48 a^3 C+4 a^2 b (14 B+3 C)-2 a b^2 (7 B+22 C)+b^3 (63 B-25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-48 a^3 C+56 a^2 b B-44 a b^2 C+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-48 a^3 C+4 a^2 b (14 B+3 C)-2 a b^2 (7 B+22 C)+b^3 (63 B-25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {\left (-48 a^3 C+56 a^2 b B-44 a b^2 C+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (-48 a^3 C+4 a^2 b (14 B+3 C)-2 a b^2 (7 B+22 C)+b^3 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {-\frac {2 \sqrt {a+b} \left (-48 a^3 C+4 a^2 b (14 B+3 C)-2 a b^2 (7 B+22 C)+b^3 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^3 C+56 a^2 b B-44 a b^2 C+63 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}-\frac {2 \left (-24 a^2 C+28 a b B-25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

Input:

Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c 
+ d*x]],x]
 

Output:

(2*C*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*b*d) + ((2*( 
7*b*B - 6*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]*(56*a^2*b*B + 63*b^3*B - 48*a^3*C - 44*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*Sqrt[a + b]*(b^3*(63*B - 25*C) - 48*a^3*C 
+ 4*a^2*b*(14*B + 3*C) - 2*a*b^2*(7*B + 22*C))*Cot[c + d*x]*EllipticF[ArcS 
in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se 
c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/(3*b 
) - (2*(28*a*b*B - 24*a^2*C - 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d 
*x])/(3*b*d))/(5*b))/(7*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 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 4521

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

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. \(1869\) vs. \(2(377)=754\).

Time = 85.29 (sec) , antiderivative size = 1870, normalized size of antiderivative = 4.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1870\)
parts	\(\text {Expression too large to display}\)	\(1889\)

Input:

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

2/105/d/b^4*(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)*(56*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(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^3*b*EllipticE(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+56*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(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^2*b^2*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)/(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)/(cos( 
d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^4*Ellip 
ticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+48*(-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)/(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 
))+48*(-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^3*b*EllipticE(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+44*(-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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+44*(-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*b^3*EllipticE(-csc(d*x+c)+cot(d*x+...
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 
),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**( 
1/2),x)
 

Output:

Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/sqrt(a + b*sec(c + d*x)), x)
 

Maxima [F]

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

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 
),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sec(d*x + c)^3/sqrt(b*sec(d* 
x + c) + a), x)
 

Giac [F]

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

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sec(d*x + c)^3/sqrt(b*sec(d* 
x + c) + a), x)
 

Mupad [F(-1)]

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

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x 
))^(1/2)),x)
 

Output:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x 
))^(1/2)), x)
 

Reduce [F]

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

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

Output:

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