\(\int \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1032]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F(-1)]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 346 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(8 a A+4 b B+3 a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{4 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (8 A b^2+4 a b B-a^2 C+4 b^2 C\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{4 b d \sqrt {a+b \sec (c+d x)}}-\frac {(4 b B+a C) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{4 b d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {(4 b B+a C) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*(8*A*a+4*B*b+3*C*a)*((b+a*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2 
*d*x+1/2*c,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/d/(a+b*sec(d*x+c))^(1 
/2)+1/4*(8*A*b^2+4*B*a*b-C*a^2+4*C*b^2)*((b+a*cos(d*x+c))/(a+b))^(1/2)*Ell 
ipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/b/d 
/(a+b*sec(d*x+c))^(1/2)-1/4*(4*B*b+C*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/ 
2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/b/d/((b+a*cos(d*x+c))/(a+b))^(1 
/2)/sec(d*x+c)^(1/2)+1/4*(4*B*b+C*a)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/ 
2)*sin(d*x+c)/b/d+1/2*C*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c) 
/d
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.02 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.38 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {8 a (4 A+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{(a+b) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (16 A b^2+4 a b B-3 a^2 C+8 b^2 C\right ) \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b (a+b) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 i (4 b B+a C) \sqrt {-\frac {a (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {a (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 b (a+b) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )+a \operatorname {EllipticPi}\left (1-\frac {a}{b},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )\right )\right )}{a \sqrt {\frac {1}{a-b}} b^2 \sqrt {b+a \cos (c+d x)}}+\frac {4 (4 b B+a C) \tan (c+d x)}{b}+8 C \sec (c+d x) \tan (c+d x)\right )}{8 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sec ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*a*(4 
*A + C)*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/((a + b)*Sqrt[(b + a*Cos[c 
+ d*x])/(a + b)]) + (2*(16*A*b^2 + 4*a*b*B - 3*a^2*C + 8*b^2*C)*EllipticPi 
[2, (c + d*x)/2, (2*a)/(a + b)])/(b*(a + b)*Sqrt[(b + a*Cos[c + d*x])/(a + 
 b)]) - ((2*I)*(4*b*B + a*C)*Sqrt[-((a*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt 
[(a*(1 + Cos[c + d*x]))/(a - b)]*Csc[c + d*x]*(-2*b*(a + b)*EllipticE[I*Ar 
cSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a* 
(2*b*EllipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a 
 + b)/(a + b)] + a*EllipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b 
 + a*Cos[c + d*x]]], (-a + b)/(a + b)])))/(a*Sqrt[(a - b)^(-1)]*b^2*Sqrt[b 
 + a*Cos[c + d*x]]) + (4*(4*b*B + a*C)*Tan[c + d*x])/b + 8*C*Sec[c + d*x]* 
Tan[c + d*x]))/(8*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[ 
c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 3.37 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.578, Rules used = {3042, 4584, 27, 3042, 4590, 27, 3042, 4596, 3042, 4346, 3042, 3286, 3042, 3284, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4584

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {\sec (c+d x)} \left ((4 b B+a C) \sec ^2(c+d x)+2 (2 A b+C b+2 a B) \sec (c+d x)+a (4 A+C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\sec (c+d x)} \left ((4 b B+a C) \sec ^2(c+d x)+2 (2 A b+C b+2 a B) \sec (c+d x)+a (4 A+C)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((4 b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 (2 A b+C b+2 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+a (4 A+C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4590

\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {-\left (\left (-C a^2+4 b B a+8 A b^2+4 b^2 C\right ) \sec ^2(c+d x)\right )-2 a b (4 A+C) \sec (c+d x)+a (4 b B+a C)}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{b}+\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {-\left (\left (-C a^2+4 b B a+8 A b^2+4 b^2 C\right ) \sec ^2(c+d x)\right )-2 a b (4 A+C) \sec (c+d x)+a (4 b B+a C)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {\left (C a^2-4 b B a-8 A b^2-4 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )+a (4 b B+a C)}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4596

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx-\left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4346

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {a \cos (c+d x)+b} \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}}dx}{\sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {\sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {\int \frac {a (4 b B+a C)-2 a b (4 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4523

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx+(a C+4 b B) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a C+4 b B) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4343

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(a C+4 b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(a C+4 b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(a C+4 b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(a C+4 b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-b (8 a A+3 a C+4 b B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4345

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-\frac {b \sqrt {\sec (c+d x)} (8 a A+3 a C+4 b B) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-\frac {b \sqrt {\sec (c+d x)} (8 a A+3 a C+4 b B) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-\frac {b \sqrt {\sec (c+d x)} (8 a A+3 a C+4 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-\frac {b \sqrt {\sec (c+d x)} (8 a A+3 a C+4 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{b d}-\frac {-\frac {2 \sqrt {\sec (c+d x)} \left (a^2 (-C)+4 a b B+8 A b^2+4 b^2 C\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}-\frac {2 b \sqrt {\sec (c+d x)} (8 a A+3 a C+4 b B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a C+4 b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{2 b}\right )+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{2 d}\)

Input:

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

Output:

(C*Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2 
*((-2*b*(8*a*A + 4*b*B + 3*a*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti 
cF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a + b*Sec[c + d 
*x]]) - (2*(8*A*b^2 + 4*a*b*B - a^2*C + 4*b^2*C)*Sqrt[(b + a*Cos[c + d*x]) 
/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d 
*Sqrt[a + b*Sec[c + d*x]]) + (2*(4*b*B + a*C)*EllipticE[(c + d*x)/2, (2*a) 
/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]* 
Sqrt[Sec[c + d*x]]))/b + ((4*b*B + a*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[ 
c + d*x]]*Sin[c + d*x])/(b*d))/4
 

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 3132
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3284
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 4343
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S 
qrt[b + a*Sin[e + f*x]])   Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4345
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S 
qrt[a + b*Csc[e + f*x]])   Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[ 
{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4346
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)], x_Symbol] :> Simp[d*Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x 
]]/Sqrt[a + b*Csc[e + f*x]])   Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f*x]] 
), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4523
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d 
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Simp[A/a   I 
nt[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Simp[(A*b - a*B) 
/(a*d)   Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ 
[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
 

rule 4584
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(m + n + 1)   Int[(a + b*Csc[e 
+ f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a 
*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C*m)*Csc 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 
 - b^2, 0] && GtQ[m, 0] &&  !LeQ[n, -1]
 

rule 4590
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 
)*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) 
   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( 
A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 
 - b^2, 0] && GtQ[n, 0]
 

rule 4596
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)]), x_Symbol] :> Simp[C/d^2   Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*C 
sc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[ 
a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - 
 b^2, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 30.14 (sec) , antiderivative size = 1775, normalized size of antiderivative = 5.13

method result size
parts \(\text {Expression too large to display}\) \(1775\)
default \(\text {Expression too large to display}\) \(1804\)

Input:

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

Output:

2*A/d/((a-b)/(a+b))^(1/2)*sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)*(Ellipti 
cF(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))*a-Ell 
ipticF(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))*b 
+2*EllipticPi(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(a+b)/(a-b),I/(( 
a-b)/(a+b))^(1/2))*b)*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/( 
cos(d*x+c)+1))^(1/2)/(b+a*cos(d*x+c))*(cos(d*x+c)^2+cos(d*x+c))+B/d/((a-b) 
/(a+b))^(1/2)*((1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x 
+c)+1))^(1/2)*a*EllipticPi(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(a+ 
b)/(a-b),I/((a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^3+4*cos(d*x+c)^2+2*cos(d*x+c 
))+(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2 
)*a*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^( 
1/2))*(-cos(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c))+(1/(a+b)*(b+a*cos(d*x+c))/ 
(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*b*EllipticE(((a-b)/(a+b))^( 
1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x 
+c)^2+cos(d*x+c))+((a-b)/(a+b))^(1/2)*a*cos(d*x+c)*sin(d*x+c)+((a-b)/(a+b) 
)^(1/2)*b*sin(d*x+c))*cos(d*x+c)*sec(d*x+c)^(3/2)*(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)+1/4*C/d/b/((a-b)/(a+b))^(1/2)* 
sec(d*x+c)^(5/2)*(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)*((1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c) 
+1))^(1/2)*a^2*EllipticPi(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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