\(\int (a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [782]

Optimal result

Integrand size = 40, antiderivative size = 60 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=b (b B+2 a C) x+\frac {a (2 b B+a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 C \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d} \] Output:

b*(B*b+2*C*a)*x+a*(2*B*b+C*a)*arctanh(sin(d*x+c))/d+b^2*C*sin(d*x+c)/d+a^2 
*B*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.88 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.82 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {b (b B+2 a C) (c+d x)-a (2 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a (2 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 C \sin (c+d x)+a^2 B \tan (c+d x)}{d} \] Input:

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

Output:

(b*(b*B + 2*a*C)*(c + d*x) - a*(2*b*B + a*C)*Log[Cos[(c + d*x)/2] - Sin[(c 
 + d*x)/2]] + a*(2*b*B + a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b 
^2*C*Sin[c + d*x] + a^2*B*Tan[c + d*x])/d
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {3042, 3508, 3042, 3467, 25, 3042, 3502, 3042, 3214, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^2 (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

\(\displaystyle \frac {a^2 B \tan (c+d x)}{d}-\int -\left (\left (b^2 C \cos ^2(c+d x)+b (b B+2 a C) \cos (c+d x)+a (2 b B+a C)\right ) \sec (c+d x)\right )dx\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (b B+2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+a (2 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 B \tan (c+d x)}{d}\)

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a (2 b B+a C)+b (b B+2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 B \tan (c+d x)}{d}+\frac {b^2 C \sin (c+d x)}{d}\)

\(\Big \downarrow \) 3214

\(\displaystyle a (a C+2 b B) \int \sec (c+d x)dx+\frac {a^2 B \tan (c+d x)}{d}+b x (2 a C+b B)+\frac {b^2 C \sin (c+d x)}{d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {a^2 B \tan (c+d x)}{d}+\frac {a (a C+2 b B) \text {arctanh}(\sin (c+d x))}{d}+b x (2 a C+b B)+\frac {b^2 C \sin (c+d x)}{d}\)

Input:

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

Output:

b*(b*B + 2*a*C)*x + (a*(2*b*B + a*C)*ArcTanh[Sin[c + d*x]])/d + (b^2*C*Sin 
[c + d*x])/d + (a^2*B*Tan[c + d*x])/d
 

Defintions of rubi rules used

rule 25

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

rule 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3467

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ 
(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 
*(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2))   Int[(c + d* 
Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c 
- 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* 
a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 
1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ 
n, -1]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3508

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32

method	result	size
parts	\(\frac {\left (B \,b^{2}+2 a b C \right ) \left (d x +c \right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {b^{2} C \sin \left (d x +c \right )}{d}\)	\(79\)
derivativedivides	\(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 a b C \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (d x +c \right )}{d}\)	\(86\)
default	\(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 a b C \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (d x +c \right )}{d}\)	\(86\)
parallelrisch	\(\frac {\left (-4 B a b -2 a^{2} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (4 B a b +2 a^{2} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+b^{2} C \sin \left (2 d x +2 c \right )+2 b d x \left (B b +2 C a \right ) \cos \left (d x +c \right )+2 B \,a^{2} \sin \left (d x +c \right )}{2 d \cos \left (d x +c \right )}\)	\(122\)
risch	\(x B \,b^{2}+2 a b C x -\frac {i b^{2} C \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i b^{2} C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i B \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(160\)
norman	\(\frac {\left (B \,b^{2}+2 a b C \right ) x +\left (-4 B \,b^{2}-8 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-B \,b^{2}-2 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-B \,b^{2}-2 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (B \,b^{2}+2 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (2 B \,b^{2}+4 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (2 B \,b^{2}+4 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {4 \left (B \,a^{2}-b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 \left (B \,a^{2}-b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 \left (B \,a^{2}+b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (B \,a^{2}+b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 \left (3 B \,a^{2}-b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 \left (3 B \,a^{2}+b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {a \left (2 B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (2 B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(412\)

Input:

int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method 
=_RETURNVERBOSE)
 

Output:

(B*b^2+2*C*a*b)/d*(d*x+c)+(2*B*a*b+C*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+a^2* 
B*tan(d*x+c)/d+b^2*C*sin(d*x+c)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.95 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (2 \, C a b + B b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

1/2*(2*(2*C*a*b + B*b^2)*d*x*cos(d*x + c) + (C*a^2 + 2*B*a*b)*cos(d*x + c) 
*log(sin(d*x + c) + 1) - (C*a^2 + 2*B*a*b)*cos(d*x + c)*log(-sin(d*x + c) 
+ 1) + 2*(C*b^2*cos(d*x + c) + B*a^2)*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.72 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a b + 2 \, {\left (d x + c\right )} B b^{2} + C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C b^{2} \sin \left (d x + c\right ) + 2 \, B a^{2} \tan \left (d x + c\right )}{2 \, d} \] Input:

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

Output:

1/2*(4*(d*x + c)*C*a*b + 2*(d*x + c)*B*b^2 + C*a^2*(log(sin(d*x + c) + 1) 
- log(sin(d*x + c) - 1)) + 2*B*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + 
c) - 1)) + 2*C*b^2*sin(d*x + c) + 2*B*a^2*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (60) = 120\).

Time = 0.15 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.53 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (2 \, C a b + B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \] Input:

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

Output:

((2*C*a*b + B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b)*log(abs(tan(1/2*d*x + 1/2 
*c) + 1)) - (C*a^2 + 2*B*a*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(B*a^ 
2*tan(1/2*d*x + 1/2*c)^3 - C*b^2*tan(1/2*d*x + 1/2*c)^3 + B*a^2*tan(1/2*d* 
x + 1/2*c) + C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1))/d
 

Mupad [B] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.82 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )}+\frac {4\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \] Input:

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

Output:

(B*a^2*tan(c + d*x))/d + (2*B*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 
2)))/d - (C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + ( 
C*b^2*sin(2*c + 2*d*x))/(2*d*cos(c + d*x)) - (B*a*b*atan((sin(c/2 + (d*x)/ 
2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (4*C*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/ 
2 + (d*x)/2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 185, normalized size of antiderivative = 3.08 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} c -2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} c +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2} c +2 \cos \left (d x +c \right ) a b \,c^{2}+2 \cos \left (d x +c \right ) a b c d x +\cos \left (d x +c \right ) b^{3} c +\cos \left (d x +c \right ) b^{3} d x +\sin \left (d x +c \right ) a^{2} b}{\cos \left (d x +c \right ) d} \] Input:

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

Output:

( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*c - 2*cos(c + d*x)*log(tan 
((c + d*x)/2) - 1)*a*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*c 
+ 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 + cos(c + d*x)*sin(c + d 
*x)*b**2*c + 2*cos(c + d*x)*a*b*c**2 + 2*cos(c + d*x)*a*b*c*d*x + cos(c + 
d*x)*b**3*c + cos(c + d*x)*b**3*d*x + sin(c + d*x)*a**2*b)/(cos(c + d*x)*d 
)