\(\int (a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [781]

Optimal result

Integrand size = 40, antiderivative size = 86 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) x+\frac {a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {b (2 b B+3 a C) \sin (c+d x)}{2 d}+\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d} \] Output:

1/2*(4*B*a*b+2*C*a^2+C*b^2)*x+a^2*B*arctanh(sin(d*x+c))/d+1/2*b*(2*B*b+3*C 
*a)*sin(d*x+c)/d+1/2*b*C*(a+b*cos(d*x+c))*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.40 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \left (4 a b B+2 a^2 C+b^2 C\right ) (c+d x)-4 a^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b (b B+2 a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 d} \] Input:

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

Output:

(2*(4*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) - 4*a^2*B*Log[Cos[(c + d*x)/2] - 
Sin[(c + d*x)/2]] + 4*a^2*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*b 
*(b*B + 2*a*C)*Sin[c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3508, 3042, 3469, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{2} \left (\frac {2 a^2 B \text {arctanh}(\sin (c+d x))}{d}+x \left (2 a^2 C+4 a b B+b^2 C\right )+\frac {b (3 a C+2 b B) \sin (c+d x)}{d}\right )+\frac {b C \sin (c+d x) (a+b \cos (c+d x))}{2 d}\)

Input:

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

Output:

(b*C*(a + b*Cos[c + d*x])*Sin[c + d*x])/(2*d) + ((4*a*b*B + 2*a^2*C + b^2* 
C)*x + (2*a^2*B*ArcTanh[Sin[c + d*x]])/d + (b*(2*b*B + 3*a*C)*Sin[c + d*x] 
)/d)/2
 

Defintions of rubi rules used

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 3469

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

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 = 0.48 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b C \sin \left (d x +c \right )+2 B a b \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{2} \sin \left (d x +c \right )}{d}\)	\(94\)
default	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b C \sin \left (d x +c \right )+2 B a b \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{2} \sin \left (d x +c \right )}{d}\)	\(94\)
parts	\(\frac {\left (B \,b^{2}+2 a b C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(95\)
parallelrisch	\(\frac {-4 B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+b^{2} C \sin \left (2 d x +2 c \right )+\left (4 B \,b^{2}+8 a b C \right ) \sin \left (d x +c \right )+8 x \left (B a b +\frac {1}{2} a^{2} C +\frac {1}{4} b^{2} C \right ) d}{4 d}\)	\(97\)
risch	\(2 x B a b +a^{2} C x +\frac {b^{2} C x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a b C}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a b C}{d}+\frac {B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) b^{2} C}{4 d}\)	\(156\)
norman	\(\frac {\left (-2 B a b -a^{2} C -\frac {1}{2} b^{2} C \right ) x +\left (-6 B a b -3 a^{2} C -\frac {3}{2} b^{2} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (2 B a b +a^{2} C +\frac {1}{2} b^{2} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (6 B a b +3 a^{2} C +\frac {3}{2} b^{2} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-4 B a b -2 a^{2} C -b^{2} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (4 B a b +2 a^{2} C +b^{2} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {b \left (2 B b +4 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {4 b \left (B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {4 b \left (B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {b \left (2 B b +4 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 b^{2} C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(374\)

Input:

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

Output:

1/d*(a^2*C*(d*x+c)+B*a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b*C*sin(d*x+c)+2*B* 
a*b*(d*x+c)+b^2*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+B*b^2*sin(d*x+ 
c))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} d x + {\left (C b^{2} \cos \left (d x + c\right ) + 4 \, C a b + 2 \, B b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

1/4*(4*(d*x + c)*C*a^2 + 8*(d*x + c)*B*a*b + (2*d*x + 2*c + sin(2*d*x + 2* 
c))*C*b^2 + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 8*C* 
a*b*sin(d*x + c) + 4*B*b^2*sin(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (80) = 160\).

Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.07 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.97 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \] Input:

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

Output:

(B*b^2*sin(c + d*x))/d + (2*B*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) 
/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C*b^2 
*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C*b^2*sin(2*c + 2*d*x)) 
/(4*d) + (2*C*a*b*sin(c + d*x))/d + (4*B*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 = 126, normalized size of antiderivative = 1.47 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +4 \sin \left (d x +c \right ) a b c +2 \sin \left (d x +c \right ) b^{3}+2 a^{2} c^{2}+2 a^{2} c d x +4 a \,b^{2} c +4 a \,b^{2} d x +b^{2} c^{2}+b^{2} c d x}{2 d} \] Input:

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

Output:

(cos(c + d*x)*sin(c + d*x)*b**2*c - 2*log(tan((c + d*x)/2) - 1)*a**2*b + 2 
*log(tan((c + d*x)/2) + 1)*a**2*b + 4*sin(c + d*x)*a*b*c + 2*sin(c + d*x)* 
b**3 + 2*a**2*c**2 + 2*a**2*c*d*x + 4*a*b**2*c + 4*a*b**2*d*x + b**2*c**2 
+ b**2*c*d*x)/(2*d)