\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [513]

Optimal result

Integrand size = 41, antiderivative size = 118 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (3 B-2 C) \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 a d} \] Output:

2^(1/2)*(A-B+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/ 
2))/a^(1/2)/d+2/3*(3*B-2*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*C*(a+a 
*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (3 \sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} (3 B-C+C \sec (c+d x))\right ) \tan (c+d x)}{3 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3042, 4570, 27, 3042, 4489, 3042, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {2 a (3 B-2 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a (A-B+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {3 \sqrt {2} \sqrt {a} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a (3 B-2 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}+\frac {2 C \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 a d}\)

Input:

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

Output:

(2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d) + ((3*Sqrt[2]*Sqrt[a]* 
(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x] 
])])/d + (2*a*(3*B - 2*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*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 216

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

rule 3042

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

rule 4282

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

rule 4489

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(269\) vs. \(2(101)=202\).

Time = 1.45 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.29

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (C \left (-2 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )+6 B \sin \left (d x +c \right )+3 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) A -3 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) B +3 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) C \right )}{3 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(270\)
parts	\(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{d a}+\frac {B \left (-\ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-2 \cot \left (d x +c \right )+2 \csc \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d a}+\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-2 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )+3 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{3 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(290\)

Input:

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

Output:

1/3/d/a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(C*(-2*sin(d*x+c)+2*tan(d* 
x+c))+6*B*sin(d*x+c)+3*2^(1/2)*(cos(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*A-3* 
2^(1/2)*(cos(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*B+3*2^(1/2)*(cos(d*x+c)+1)* 
(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2) 
-cot(d*x+c)+csc(d*x+c))*C)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 356, normalized size of antiderivative = 3.02 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left ({\left (3 \, B - C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/6*(3*sqrt(2)*((A - B + C)*a*cos(d*x + c)^2 + (A - B + C)*a*cos(d*x + c) 
)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt( 
-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/( 
cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*B - C)*cos(d*x + c) + C)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^2 + a 
*d*cos(d*x + c)), 1/3*(2*((3*B - C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) 
 + a)/cos(d*x + c))*sin(d*x + c) - 3*sqrt(2)*((A - B + C)*a*cos(d*x + c)^2 
 + (A - B + C)*a*cos(d*x + c))*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/co 
s(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c 
)^2 + a*d*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.41 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (A - B + C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\frac {\sqrt {2} {\left (3 \, B a - 2 \, C a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {3 \, \sqrt {2} B a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{3 \, d} \] Input:

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

Output:

-1/3*(3*sqrt(2)*(A - B + 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)))/(sqrt(-a)*sgn(cos(d*x + c))) - 2*(sqrt(2) 
*(3*B*a - 2*C*a)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 3*sqrt(2)*B*a/ 
sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sq 
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:

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

Output:

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