\(\int \frac {\cos ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [399]

Optimal result

Integrand size = 42, antiderivative size = 170 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(3 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 B-5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(B-C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 B-C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-8 (3 B-2 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+2 \sqrt {2} (9 B-5 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+2 \cos (c+d x) (3 B-C+2 B \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \sec (c+d x) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

((-8*(3*B - 2*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2 + 2*Sq 
rt[2]*(9*B - 5*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2] 
^2 + 2*Cos[c + d*x]*(3*B - C + 2*B*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*S 
ec[c + d*x]*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x] 
))^(3/2))
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4560, 3042, 4508, 27, 3042, 4510, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 B-2 C) \int \sqrt {\sec (c+d x) a+a}dx-a^2 (9 B-5 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 B-2 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^2 (9 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^2 (9 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {4 a^2 (3 B-2 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^2 (9 B-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (9 B-5 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {4 a^{3/2} (3 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (3 B-C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (9 B-5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((B - C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-(((4*a^(3/2 
)*(3*B - 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - 
 (Sqrt[2]*a^(3/2)*(9*B - 5*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[ 
a + a*Sec[c + d*x]])])/d)/a) + (2*a*(3*B - C)*Sin[c + d*x])/(d*Sqrt[a + a* 
Sec[c + d*x]]))/(4*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4508

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B 
, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 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 4560

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(434\) vs. \(2(145)=290\).

Time = 3.37 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.56

method	result	size
default	\(\frac {\left (\left (6 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+6\right ) \sqrt {2}\, B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-4 \cos \left (d x +c \right )^{2}-8 \cos \left (d x +c \right )-4\right ) \sqrt {2}\, C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (9 \cos \left (d x +c \right )^{2}+18 \cos \left (d x +c \right )+9\right ) B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (4 \cos \left (d x +c \right )+6\right ) B -2 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(435\)

Input:

int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4/d/a^2*((6*cos(d*x+c)^2+12*cos(d*x+c)+6)*2^(1/2)*B*(-2*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))+(-4*cos(d*x+c)^2-8*cos(d*x+c) 
-4)*2^(1/2)*C*(-2*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))+(9*cos(d*x+c)^2+18*cos(d*x+c)+9)*B*(-2*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)*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*co 
s(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 
)*(4*cos(d*x+c)+6)*B-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 = 2.00 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.58 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/8*(sqrt(2)*((9*B - 5*C)*cos(d*x + c)^2 + 2*(9*B - 5*C)*cos(d*x + c) + 9 
*B - 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)) + 4*((3*B - 2*C)*cos(d*x + c 
)^2 + 2*(3*B - 2*C)*cos(d*x + c) + 3*B - 2*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)) + 4*(2*B*cos(d*x + c)^ 
2 + (3*B - 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)*((9*B - 5*C)*cos(d*x + c)^2 + 2*(9*B - 5*C)*cos(d*x + c) + 9*B - 5*C)*s 
qrt(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))) - 4*((3*B - 2*C)*cos(d*x + c)^2 + 2*(3*B - 2*C)*c 
os(d*x + c) + 3*B - 2*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*(2*B*cos(d*x + c)^2 + (3*B 
- 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)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*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. 418 vs. \(2 (145) = 290\).

Time = 1.44 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (9 \, B - 5 \, 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} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {4 \, {\left (3 \, B - 2 \, C\right )} \log \left (\frac {{\left | -17179869184 \, {\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} - 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}{{\left | -17179869184 \, {\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} + 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}\right )}{\sqrt {-a} {\left | a \right |} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {16 \, \sqrt {2} {\left (3 \, {\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} B - B a\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 )} \sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \] Input:

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

Output:

-1/8*(sqrt(2)*(9*B - 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))) - 4*(3*B - 2*C 
)*log(abs(-17179869184*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d* 
x + 1/2*c)^2 + a))^2 - 34359738368*sqrt(2)*abs(a) + 51539607552*a)/abs(-17 
179869184*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 
+ a))^2 + 34359738368*sqrt(2)*abs(a) + 51539607552*a))/(sqrt(-a)*abs(a)*sg 
n(cos(d*x + c))) - 2*(sqrt(2)*B*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 - 
16*sqrt(2)*(3*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c 
)^2 + a))^2*B - B*a)/(((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)*sqrt(-a)*sgn(cos(d*x + c))))/d
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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