\(\int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [580]

Optimal result

Integrand size = 45, antiderivative size = 131 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a} (8 A+4 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a (4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

1/4*a^(1/2)*(8*A+4*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2 
))/d+1/4*a*(4*B+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/ 
2*C*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.17 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left ((4 B+3 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-8 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+3 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a*((4*B + 3*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 8*A*ArcSin[Sqrt[Sec[c + d 
*x]]] + 2*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 4*B*Sqrt[-((-1 + S 
ec[c + d*x])*Sec[c + d*x])] + 3*C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x]) 
])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 4576, 27, 3042, 4504, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4504

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {a^{3/2} (8 A+4 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (4 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\)

Input:

Int[Sqrt[Sec[c + d*x]]*Sqrt[a + a*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 + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((a^( 
3/2)*(8*A + 4*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d 
*x]]])/d + (a^2*(4*B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*S 
ec[c + d*x]]))/(4*a)
 

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

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

rule 4288

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

rule 4504

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

rule 4576

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(111)=222\).

Time = 7.19 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.79

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (8 A \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+4 B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 C \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+8 A \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+4 B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 C \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-4 B \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \sqrt {2}-\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (3 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(366\)
parts	\(-\frac {A \left (\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {B \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}-\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-3 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(471\)

Input:

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

Output:

-1/8/d*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d 
*x+c)+1))^(1/2)*(8*A*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/ 
(cos(d*x+c)+1))^(1/2))+4*B*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1 
)/(-1/(cos(d*x+c)+1))^(1/2))+3*C*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d* 
x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+8*A*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+ 
csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+4*B*cos(d*x+c)*arctan(1/2*(-cot(d 
*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+3*C*cos(d*x+c)*arctan(1/2*( 
-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))-4*B*sin(d*x+c)*(-2/(c 
os(d*x+c)+1))^(1/2)*2^(1/2)-2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*(3*sin(d*x 
+c)+2*tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 411, normalized size of antiderivative = 3.14 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/16*(((8*A + 4*B + 3*C)*cos(d*x + c)^2 + (8*A + 4*B + 3*C)*cos(d*x + c)) 
*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 
2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + 
c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((4*B 
+ 3*C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x 
 + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/8*(((8*A 
+ 4*B + 3*C)*cos(d*x + c)^2 + (8*A + 4*B + 3*C)*cos(d*x + c))*sqrt(-a)*arc 
tan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + 
a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*((4*B + 3*C)*cos 
(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt 
(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2167 vs. \(2 (111) = 222\).

Time = 0.38 (sec) , antiderivative size = 2167, normalized size of antiderivative = 16.54 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(8*A*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 
 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/ 
2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 
 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2* 
sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1 
/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x 
 + 1/2*c) + 2)) - 4*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)) 
)*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)) 
)*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d* 
x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin 
(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(si 
n(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d 
*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 
2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2 
*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d* 
x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + 
 c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) 
 + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arc 
tan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (111) = 222\).

Time = 2.25 (sec) , antiderivative size = 583, normalized size of antiderivative = 4.45 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*((8*A*sqrt(a)*sgn(cos(d*x + c)) + 4*B*sqrt(a)*sgn(cos(d*x + c)) + 3*C* 
sqrt(a)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a* 
tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - (8*A*sqrt(a)*sgn(co 
s(d*x + c)) + 4*B*sqrt(a)*sgn(cos(d*x + c)) + 3*C*sqrt(a)*sgn(cos(d*x + c) 
))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + 
 a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(12*(sqrt(a)*tan(1/2*d*x + 1/2*c) 
 - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(3/2)*sgn(cos(d*x + c)) - 5*( 
sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^( 
3/2)*sgn(cos(d*x + c)) - 76*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2 
*d*x + 1/2*c)^2 + a))^4*B*a^(5/2)*sgn(cos(d*x + c)) - 19*(sqrt(a)*tan(1/2* 
d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a^(5/2)*sgn(cos(d*x 
 + c)) + 36*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 
+ a))^2*B*a^(7/2)*sgn(cos(d*x + c)) + 17*(sqrt(a)*tan(1/2*d*x + 1/2*c) - s 
qrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(7/2)*sgn(cos(d*x + c)) - 4*B*a^( 
9/2)*sgn(cos(d*x + c)) - C*a^(9/2)*sgn(cos(d*x + c)))/((sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2)/d
 

Mupad [F(-1)]

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

int((a + a/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 + a/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]

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

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

Output:

sqrt(a)*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)* 
c + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*b + int( 
sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*a)