\(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [198]

Optimal result

Integrand size = 35, antiderivative size = 217 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A+5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 A+2 C) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 A+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(19*A+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d 
-1/4*(13*A+5*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/ 
2))*2^(1/2)/a^(3/2)/d-1/2*(A+C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^( 
3/2)-1/4*(7*A+2*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/2*(2*A+C)*cos(d 
*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (\frac {\left (-2 (13 A+5 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+\sqrt {2} (19 A+8 C) \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {1}{2} (6 A+2 C+3 A \cos (c+d x)-A \cos (2 (c+d x))) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{2 d (a (1+\sec (c+d x)))^{3/2}} \] Input:

Integrate[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/ 
2),x]
 

Output:

(Cos[(c + d*x)/2]^2*Sec[c + d*x]^(3/2)*(((-2*(13*A + 5*C)*ArcSin[Tan[(c + 
d*x)/2]] + Sqrt[2]*(19*A + 8*C)*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/ 
(1 + Cos[c + d*x])]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[1 + Sec[c 
 + d*x]])/Sqrt[Sec[(c + d*x)/2]^2] + ((6*A + 2*C + 3*A*Cos[c + d*x] - A*Co 
s[2*(c + d*x)])*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(Sin[(c + d*x)/2] - Si 
n[(3*(c + d*x))/2]))/2))/(2*d*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4573, 27, 3042, 4510, 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.

\(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\cos ^2(c+d x) (4 a (2 A+C)-a (5 A+C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (4 a (2 A+C)-a (5 A+C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {4 a (2 A+C)-a (5 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {a^3 (19 A+8 C)-a^3 (7 A+2 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (19 A+8 C)-a^3 (7 A+2 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (19 A+8 C)-a^3 (7 A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (19 A+8 C) \int \sqrt {\sec (c+d x) a+a}dx-2 a^3 (13 A+5 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (19 A+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-2 a^3 (13 A+5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-2 a^3 (13 A+5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^3 (19 A+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-2 a^3 (13 A+5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^3 (13 A+5 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (2 A+C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A+2 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} a^{5/2} (13 A+5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

Int[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

-1/2*((A + C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + 
((2*a*(2*A + C)*Cos[c + d*x]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) - 
(-1/2*((2*a^(5/2)*(19*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Se 
c[c + d*x]]])/d - (2*Sqrt[2]*a^(5/2)*(13*A + 5*C)*ArcTan[(Sqrt[a]*Tan[c + 
d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a^2*(7*A + 2*C)*Sin[c + 
 d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*a^2)
 

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 4573

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(444\) vs. \(2(186)=372\).

Time = 0.71 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.05

method	result	size
default	\(\frac {\left (\left (-19 \cos \left (d x +c \right )^{2}-38 \cos \left (d x +c \right )-19\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-8 \cos \left (d x +c \right )^{2}-16 \cos \left (d x +c \right )-8\right ) C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-13 \cos \left (d x +c \right )^{2}-26 \cos \left (d x +c \right )-13\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )-7\right ) A -2 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(445\)

Input:

int(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/4/d/a^2*((-19*cos(d*x+c)^2-38*cos(d*x+c)-19)*A*(-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))+(-8*cos(d*x+c)^2-16*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)+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))+(-13*cos(d* 
x+c)^2-26*cos(d*x+c)-13)*2^(1/2)*A*(-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))+(-5*cos(d*x+c)^ 
2-10*cos(d*x+c)-5)*2^(1/2)*C*(-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))+sin(d*x+c)*cos(d*x+c) 
*(2*cos(d*x+c)^2-3*cos(d*x+c)-7)*A-2*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d* 
x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 2.99 (sec) , antiderivative size = 631, normalized size of antiderivative = 2.91 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algori 
thm="fricas")
 

Output:

[-1/8*(sqrt(2)*((13*A + 5*C)*cos(d*x + c)^2 + 2*(13*A + 5*C)*cos(d*x + c) 
+ 13*A + 5*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x 
 + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A + 8*C)*cos(d*x 
+ c)^2 + 2*(19*A + 8*C)*cos(d*x + c) + 19*A + 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*cos(d*x 
+ c)^3 - 3*A*cos(d*x + c)^2 - (7*A + 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d* 
x + c) + a^2*d), 1/4*(sqrt(2)*((13*A + 5*C)*cos(d*x + c)^2 + 2*(13*A + 5*C 
)*cos(d*x + c) + 13*A + 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((19*A + 8*C)*cos 
(d*x + c)^2 + 2*(19*A + 8*C)*cos(d*x + c) + 19*A + 8*C)*sqrt(a)*arctan(sqr 
t((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)^3 - 3*A*cos(d*x + c)^2 - (7*A + 2*C)*cos(d*x + c))*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 
 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F]

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

integrate(cos(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a*(sec(c + d*x) + 1))**( 
3/2), x)
 

Maxima [F]

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

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algori 
thm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^2/(a*sec(d*x + c) + a)^(3/2) 
, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 504 vs. \(2 (186) = 372\).

Time = 1.18 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.32 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algori 
thm="giac")
 

Output:

1/8*(sqrt(2)*(13*A + 5*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)*a*sgn(cos(d*x + c))) + (19*A + 8*C) 
*log(abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^2 - 295147905179352825856*sqrt(2)*abs(a) - 4427 
21857769029238784*a)/abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1/2 
*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 295147905179352825856*sqrt( 
2)*abs(a) - 442721857769029238784*a))/(sqrt(-a)*abs(a)*sgn(cos(d*x + c))) 
- 2*(sqrt(2)*A*a*sgn(cos(d*x + c)) + sqrt(2)*C*a*sgn(cos(d*x + c)))*sqrt(- 
a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3 - 4*sqrt(2)*(29*(sq 
rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A - 1 
33*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 
*A*a + 55*(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 - 7*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*sqrt(-a)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((cos(c + d*x)^2*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(3/2),x)
 

Output:

int((cos(c + d*x)^2*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),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)**2 + 2*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*cos 
(c + d*x)**2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*a))/a**2