\(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [949]

Optimal result

Integrand size = 39, antiderivative size = 134 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (3 A b^2+6 a b B+2 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 d}+\frac {b (3 b B+2 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \] Output:

1/2*(2*B*a^2+B*b^2+2*a*b*(2*A+C))*x+a^2*A*arctanh(sin(d*x+c))/d+1/3*(3*A*b 
^2+6*B*a*b+2*C*a^2+2*C*b^2)*sin(d*x+c)/d+1/6*b*(3*B*b+2*C*a)*cos(d*x+c)*si 
n(d*x+c)/d+1/3*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.44 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.18 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {6 \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) (c+d x)-12 a^2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \left (4 A b^2+8 a b B+4 a^2 C+3 b^2 C\right ) \sin (c+d x)+3 b (b B+2 a C) \sin (2 (c+d x))+b^2 C \sin (3 (c+d x))}{12 d} \] Input:

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

Output:

(6*(2*a^2*B + b^2*B + 2*a*b*(2*A + C))*(c + d*x) - 12*a^2*A*Log[Cos[(c + d 
*x)/2] - Sin[(c + d*x)/2]] + 12*a^2*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x) 
/2]] + 3*(4*A*b^2 + 8*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b*(b*B + 
 2*a*C)*Sin[2*(c + d*x)] + b^2*C*Sin[3*(c + d*x)])/(12*d)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3042, 3528, 3042, 3512, 3042, 3502, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (6 A a^2+2 \left (2 C a^2+6 b B a+3 A b^2+2 b^2 C\right ) \cos ^2(c+d x)+3 \left (2 B a^2+2 b (2 A+C) a+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {6 A a^2+2 \left (2 C a^2+6 b B a+3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 \left (2 B a^2+2 b (2 A+C) a+b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (2 A a^2+\left (2 B a^2+2 b (2 A+C) a+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (2 A a^2+\left (2 B a^2+2 b (2 A+C) a+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {2 A a^2+\left (2 B a^2+2 b (2 A+C) a+b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (2 a^2 A \int \sec (c+d x)dx+x \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )\right )+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (2 a^2 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+x \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )\right )+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {2 a^2 A \text {arctanh}(\sin (c+d x))}{d}+x \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )\right )+\frac {2 \sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{d}\right )+\frac {b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\)

Input:

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

Output:

(C*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((b*(3*b*B + 2*a*C)*Cos[c 
+ d*x]*Sin[c + d*x])/(2*d) + (3*((2*a^2*B + b^2*B + 2*a*b*(2*A + C))*x + ( 
2*a^2*A*ArcTanh[Sin[c + d*x]])/d) + (2*(3*A*b^2 + 6*a*b*B + 2*a^2*C + 2*b^ 
2*C)*Sin[c + d*x])/d)/2)/3
 

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 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 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 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3528

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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[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 = 0.62 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99

method	result	size
parts	\(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 a A b +B \,a^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (B \,b^{2}+2 a b C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,b^{2}+2 B a b +a^{2} C \right ) \sin \left (d x +c \right )}{d}+\frac {C \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}\)	\(132\)
parallelrisch	\(\frac {-12 A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (3 B \,b^{2}+6 a b C \right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{2}+\left (\left (12 A +9 C \right ) b^{2}+24 B a b +12 a^{2} C \right ) \sin \left (d x +c \right )+24 \left (\frac {B \,b^{2}}{4}+a \left (A +\frac {C}{2}\right ) b +\frac {B \,a^{2}}{2}\right ) x d}{12 d}\)	\(134\)
derivativedivides	\(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+C \sin \left (d x +c \right ) a^{2}+2 a A b \left (d x +c \right )+2 B \sin \left (d x +c \right ) a b +2 a b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {C \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(154\)
default	\(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+C \sin \left (d x +c \right ) a^{2}+2 a A b \left (d x +c \right )+2 B \sin \left (d x +c \right ) a b +2 a b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{2}+B \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {C \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(154\)
risch	\(2 x a A b +x B \,a^{2}+\frac {x B \,b^{2}}{2}+a b C x +\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{2}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a b}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B a b}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{2}}{8 d}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) C \,b^{2}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a b C}{2 d}\)	\(272\)
norman	\(\frac {\left (2 a A b +B \,a^{2}+\frac {1}{2} B \,b^{2}+a b C \right ) x +\left (2 a A b +B \,a^{2}+\frac {1}{2} B \,b^{2}+a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (8 a A b +4 B \,a^{2}+2 B \,b^{2}+4 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (8 a A b +4 B \,a^{2}+2 B \,b^{2}+4 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (12 a A b +6 B \,a^{2}+3 B \,b^{2}+6 a b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (2 A \,b^{2}+4 B a b -B \,b^{2}+2 a^{2} C -2 a b C +2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (2 A \,b^{2}+4 B a b +B \,b^{2}+2 a^{2} C +2 a b C +2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (18 A \,b^{2}+36 B a b -3 B \,b^{2}+18 a^{2} C -6 a b C +10 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {\left (18 A \,b^{2}+36 B a b +3 B \,b^{2}+18 a^{2} C +6 a b C +10 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(424\)

Input:

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

Output:

A*a^2/d*ln(sec(d*x+c)+tan(d*x+c))+(2*A*a*b+B*a^2)/d*(d*x+c)+(B*b^2+2*C*a*b 
)/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+(A*b^2+2*B*a*b+C*a^2)/d*sin( 
d*x+c)+1/3*C*b^2/d*(2+cos(d*x+c)^2)*sin(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 \, A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, B a^{2} + 2 \, {\left (2 \, A + C\right )} a b + B b^{2}\right )} d x + {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 6 \, C a^{2} + 12 \, B a b + 2 \, {\left (3 \, A + 2 \, C\right )} b^{2} + 3 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

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

Output:

1/6*(3*A*a^2*log(sin(d*x + c) + 1) - 3*A*a^2*log(-sin(d*x + c) + 1) + 3*(2 
*B*a^2 + 2*(2*A + C)*a*b + B*b^2)*d*x + (2*C*b^2*cos(d*x + c)^2 + 6*C*a^2 
+ 12*B*a*b + 2*(3*A + 2*C)*b^2 + 3*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*x 
 + c))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} B a^{2} + 24 \, {\left (d x + c\right )} A a b + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{2} + 12 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, C a^{2} \sin \left (d x + c\right ) + 24 \, B a b \sin \left (d x + c\right ) + 12 \, A b^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

1/12*(12*(d*x + c)*B*a^2 + 24*(d*x + c)*A*a*b + 6*(2*d*x + 2*c + sin(2*d*x 
 + 2*c))*C*a*b + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^2 - 4*(sin(d*x + c 
)^3 - 3*sin(d*x + c))*C*b^2 + 12*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 
12*C*a^2*sin(d*x + c) + 24*B*a*b*sin(d*x + c) + 12*A*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. 346 vs. \(2 (126) = 252\).

Time = 0.15 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.58 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (2 \, B a^{2} + 4 \, A a b + 2 \, C a b + B b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^2*log(abs(tan(1/2* 
d*x + 1/2*c) - 1)) + 3*(2*B*a^2 + 4*A*a*b + 2*C*a*b + B*b^2)*(d*x + c) + 2 
*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a 
*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1 
/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x 
+ 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2* 
c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) + 12* 
B*a*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/ 
2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2* 
c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.96 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,B\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \] Input:

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

Output:

(A*b^2*sin(c + d*x))/d + (C*a^2*sin(c + d*x))/d + (3*C*b^2*sin(c + d*x))/( 
4*d) + (2*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^2 
*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (B*b^2*atan(sin(c/2 + (d 
*x)/2)/cos(c/2 + (d*x)/2)))/d + (B*b^2*sin(2*c + 2*d*x))/(4*d) + (C*b^2*si 
n(3*c + 3*d*x))/(12*d) + (2*B*a*b*sin(c + d*x))/d + (4*A*a*b*atan(sin(c/2 
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a*b*atan(sin(c/2 + (d*x)/2)/cos(c 
/2 + (d*x)/2)))/d + (C*a*b*sin(2*c + 2*d*x))/(2*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-2 \sin \left (d x +c \right )^{3} b^{2} c +6 \sin \left (d x +c \right ) a^{2} c +18 \sin \left (d x +c \right ) a \,b^{2}+6 \sin \left (d x +c \right ) b^{2} c +18 a^{2} b d x +6 a b c d x +3 b^{3} d x}{6 d} \] Input:

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

Output:

(6*cos(c + d*x)*sin(c + d*x)*a*b*c + 3*cos(c + d*x)*sin(c + d*x)*b**3 - 6* 
log(tan((c + d*x)/2) - 1)*a**3 + 6*log(tan((c + d*x)/2) + 1)*a**3 - 2*sin( 
c + d*x)**3*b**2*c + 6*sin(c + d*x)*a**2*c + 18*sin(c + d*x)*a*b**2 + 6*si 
n(c + d*x)*b**2*c + 18*a**2*b*d*x + 6*a*b*c*d*x + 3*b**3*d*x)/(6*d)