\(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [189]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {(7 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {A \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(7*A+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/d- 
2^(1/2)*(A+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-1/4*A*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*A*cos(d*x+c)*sin 
(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((7 A+8 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+A (1-\cos (c+d x)+\cos (2 (c+d x))) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(((7*A + 8*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*(A + C)*ArcTanh[ 
Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + A*(1 - Cos[c + d*x] + Cos[2*(c + d*x)])* 
Sqrt[1 - 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 = 1.01 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4575, 27, 3042, 4510, 27, 3042, 4408, 3042, 4261, 216, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {a^2 (7 A+8 C)-a^2 A \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (7 A+8 C)-a^2 A \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (7 A+8 C)-a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a (7 A+8 C) \int \sqrt {\sec (c+d x) a+a}dx-8 a^2 (A+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a (7 A+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-8 a^2 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-8 a^2 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^2 (7 A+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-8 a^2 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {16 a^2 (A+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (7 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{3/2} (7 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

Input:

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

Output:

(A*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((2*a 
^(3/2)*(7*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] 
)/d - (8*Sqrt[2]*a^(3/2)*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sq 
rt[a + a*Sec[c + d*x]])])/d)/a + (a*A*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 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 4408
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c/a   Int[Sqrt[a + b*Csc[e + f*x]], x], x] - 
Simp[(b*c - a*d)/a   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
 

rule 4510
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), 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 - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

rule 4575
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co 
t[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 
*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 
 C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || 
 EqQ[m + n + 1, 0])
 
Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(134)=268\).

Time = 0.43 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.30

method result size
default \(\frac {\left (-7 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right ) A -8 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right ) C -4 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) A -4 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) C +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )-1\right ) A \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d a \left (\cos \left (d x +c \right )+1\right )}\) \(366\)

Input:

int(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/4/d/a*(-7*(cos(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1 
/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot(d*x 
+c)^2-1)^(1/2))*A-8*(cos(d*x+c)+1)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct 
anh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c) 
+cot(d*x+c)^2-1)^(1/2))*C-4*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-4*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+sin(d*x+c)*cos(d*x+c 
)*(2*cos(d*x+c)-1)*A)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 489, normalized size of antiderivative = 3.08 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\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 ) - {\left ({\left (7 \, A + 8 \, C\right )} \cos \left (d x + c\right ) + 7 \, A + 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} - A \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 (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {{\left ({\left (7 \, A + 8 \, C\right )} \cos \left (d x + c\right ) + 7 \, A + 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} - A \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 ) - \frac {4 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\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}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:

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

Output:

[1/8*(4*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*sqrt(-1/a)*log((2*sqr 
t(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)) - ((7*A + 8*C)*cos(d*x + c) + 7*A + 8*C)*sqrt(-a)*log((2*a*c 
os(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*cos( 
d*x + c)^2 - A*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d 
*x + c))/(a*d*cos(d*x + c) + a*d), -1/4*(((7*A + 8*C)*cos(d*x + c) + 7*A + 
 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 - A*cos(d*x + c))*sqrt((a*co 
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*((A + C)*a*cos(d*x 
+ c) + (A + C)*a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*c 
os(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (134) = 268\).

Time = 1.15 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.93 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {4 \, \sqrt {2} {\left (A + 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 )}^{2}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (7 \, A + 8 \, C\right )} \log \left ({\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 )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left (7 \, A + 8 \, C\right )} \log \left ({\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 )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, \sqrt {2} {\left (17 \, {\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 )}^{6} A \sqrt {-a} - 57 \, {\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 )}^{4} A \sqrt {-a} a + 19 \, {\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 )}^{2} A \sqrt {-a} a^{2} - 3 \, A \sqrt {-a} a^{3}\right )}}{{\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 )}^{4} - 6 \, {\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 )}^{2} a + a^{2}\right )}^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \] Input:

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

Output:

1/8*(4*sqrt(2)*(A + C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/ 
2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*sgn(cos(d*x + c))) + (7*A + 8*C)*log(a 
bs((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)))/(sqrt(-a)*sgn(cos(d*x + c))) - (7*A + 8*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)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt(2)*(17*(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) 
- 57*(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 + 19*(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^2 - 3*A*sqrt(-a)*a^3)/(((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*sgn(cos 
(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(cos(d*x+c)^2*(A+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)*cos(c + d*x)**2*sec(c + d*x)**2)/(se 
c(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2)/(sec(c 
 + d*x) + 1),x)*a))/a