\(\int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [488]

Optimal result

Integrand size = 43, antiderivative size = 117 \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a} (3 A+4 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a (A+4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98 \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left ((3 A+4 B+8 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+\cos (c+d x) (3 A+4 B+2 A \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \] Input:

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

Output:

(((3*A + 4*B + 8*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*(3*A + 
4*B + 2*A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c + d*x])] 
*Sin[c + d*x])/(4*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {3042, 4574, 27, 3042, 4503, 3042, 4261, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4503

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(A*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((a^(3/2)*( 
3*A + 4*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/ 
d + (a^2*(A + 4*B)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[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 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 4261

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

rule 4503

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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 4574

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(101)=202\).

Time = 7.00 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.79

method	result	size
default	\(\frac {\left (\left (3 \cos \left (d x +c \right )+3\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (4 \cos \left (d x +c \right )+4\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (8 \cos \left (d x +c \right )+8\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (3+2 \cos \left (d x +c \right )\right ) A +4 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \left (\cos \left (d x +c \right )+1\right )}\)	\(326\)

Input:

int(cos(d*x+c)^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:

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

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.74 \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (3 \, A + 4 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 4 \, B + 8 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A + 4 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (3 \, A + 4 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 4 \, B + 8 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A + 4 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^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:

[1/8*(((3*A + 4*B + 8*C)*cos(d*x + c) + 3*A + 4*B + 8*C)*sqrt(-a)*log((2*a 
*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d 
*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*A*co 
s(d*x + c)^2 + (3*A + 4*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/4*(((3*A + 4*B + 8*C)*cos(d* 
x + c) + 3*A + 4*B + 8*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*A*cos(d*x + c)^2 + (3*A + 
 4*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/ 
(d*cos(d*x + c) + d)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1996 vs. \(2 (101) = 202\).

Time = 0.38 (sec) , antiderivative size = 1996, normalized size of antiderivative = 17.06 \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^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*(16*C*sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co 
s(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* 
c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos( 
2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
 + 1)) + cos(d*x + c)) + (2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c 
os(2*d*x + 2*c) + 1)^(1/4)*((cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c)))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x 
 + 2*c), cos(2*d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c))) - cos(2*d*x + 2*c) + 2)*sin(1/2*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*sqrt(a)*(arctan2((c 
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(co 
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d 
*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(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)^(1/4)*( 
cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(s 
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (101) = 202\).

Time = 0.85 (sec) , antiderivative size = 660, normalized size of antiderivative = 5.64 \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^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*((3*A*sqrt(-a)*sgn(cos(d*x + c)) + 4*B*sqrt(-a)*sgn(cos(d*x + c)) + 8 
*C*sqrt(-a)*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq 
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - (3*A*sqrt(-a) 
*sgn(cos(d*x + c)) + 4*B*sqrt(-a)*sgn(cos(d*x + c)) + 8*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)*(5*(sqrt(-a)*tan(1/2*d 
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a*sgn(cos( 
d*x + c)) - 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*sqrt(-a)*a*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*A*sqrt(-a)*a^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*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 17*(sqrt(-a)*tan(1/2*d*x + 1/2 
*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^3*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*sqrt(-a)*a^3*sgn(cos(d*x + c)) + A*sqrt(-a)*a^4*sgn(cos(d*x + c)) 
 + 4*B*sqrt(-a)*a^4*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - s 
qrt(-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. \[ \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^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) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)*c + 
 int(sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x),x)*b + int(sqrt(s 
ec(c + d*x) + 1)*cos(c + d*x)**2,x)*a)