\(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [945]

Optimal result

Integrand size = 39, antiderivative size = 137 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {(3 a A+4 b B+4 a C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(2 A b+2 a B+3 b C) \tan (c+d x)}{3 d}+\frac {(3 a A+4 b B+4 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

1/8*(3*A*a+4*B*b+4*C*a)*arctanh(sin(d*x+c))/d+1/3*(2*A*b+2*B*a+3*C*b)*tan( 
d*x+c)/d+1/8*(3*A*a+4*B*b+4*C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*(A*b+B*a)*sec 
(d*x+c)^2*tan(d*x+c)/d+1/4*a*A*sec(d*x+c)^3*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.73 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 (3 a A+4 b B+4 a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 (A b+a B+b C)+3 (3 a A+4 b B+4 a C) \sec (c+d x)+6 a A \sec ^3(c+d x)+8 (A b+a B) \tan ^2(c+d x)\right )}{24 d} \] Input:

Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec 
[c + d*x]^5,x]
 

Output:

(3*(3*a*A + 4*b*B + 4*a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(A*b + 
 a*B + b*C) + 3*(3*a*A + 4*b*B + 4*a*C)*Sec[c + d*x] + 6*a*A*Sec[c + d*x]^ 
3 + 8*(A*b + a*B)*Tan[c + d*x]^2))/(24*d)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 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.

\(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d 
*x]^5,x]
 

Output:

(a*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*(A*b + a*B)*Sec[c + d*x]^2*T 
an[c + d*x])/(3*d) + ((4*(2*A*b + 2*a*B + 3*b*C)*Tan[c + d*x])/d + 3*(3*a* 
A + 4*b*B + 4*a*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d* 
x])/(2*d)))/3)/4
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500

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

rule 3510

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[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

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 = 0.83 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99

method	result	size
parts	\(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (A b +B a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B b +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C b \tan \left (d x +c \right )}{d}\)	\(136\)
derivativedivides	\(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \tan \left (d x +c \right )}{d}\)	\(174\)
default	\(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \tan \left (d x +c \right )}{d}\)	\(174\)
parallelrisch	\(\frac {-9 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \left (\left (A +\frac {4 C}{3}\right ) a +\frac {4 B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \left (\left (A +\frac {4 C}{3}\right ) a +\frac {4 B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+64 \left (\left (A +\frac {3 C}{4}\right ) b +B a \right ) \sin \left (2 d x +2 c \right )+6 \left (\left (3 A +4 C \right ) a +4 B b \right ) \sin \left (3 d x +3 c \right )+16 \left (B a +\left (A +\frac {3 C}{2}\right ) b \right ) \sin \left (4 d x +4 c \right )+66 \sin \left (d x +c \right ) \left (\left (A +\frac {4 C}{11}\right ) a +\frac {4 B b}{11}\right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(216\)
norman	\(\frac {\frac {\left (7 a A -4 B b -4 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (27 a A -16 A b -16 B a +12 B b +12 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {\left (27 a A +16 A b +16 B a +12 B b +12 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}+\frac {\left (5 a A -8 A b -8 B a +4 B b +4 C a -8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (5 a A +8 A b +8 B a +4 B b +4 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (81 a A -8 A b -8 B a -12 B b -12 C a -72 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {\left (81 a A +8 A b +8 B a -12 B b -12 C a +72 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (3 a A +4 B b +4 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a A +4 B b +4 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(358\)
risch	\(-\frac {i \left (9 A a \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 C a \,{\mathrm e}^{7 i \left (d x +c \right )}-24 C b \,{\mathrm e}^{6 i \left (d x +c \right )}+33 A a \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B a \,{\mathrm e}^{4 i \left (d x +c \right )}-72 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A a \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-12 C a \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-72 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a A \,{\mathrm e}^{i \left (d x +c \right )}-12 B b \,{\mathrm e}^{i \left (d x +c \right )}-12 C a \,{\mathrm e}^{i \left (d x +c \right )}-16 A b -16 B a -24 C b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C a}{2 d}+\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C a}{2 d}\)	\(401\)

Input:

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x,method 
=_RETURNVERBOSE)
 

Output:

a*A/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+ta 
n(d*x+c)))-(A*b+B*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(B*b+C*a)/d*(1/2 
*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*b/d*tan(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, B a + {\left (2 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{2} + 6 \, A a + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="fricas")
 

Output:

1/48*(3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*( 
(3*A + 4*C)*a + 4*B*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(2*B*a 
 + (2*A + 3*C)*b)*cos(d*x + c)^3 + 3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c)^ 
2 + 6*A*a + 8*(B*a + A*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5, 
x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.59 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C b \tan \left (d x + c\right )}{48 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="maxima")
 

Output:

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*ta 
n(d*x + c))*A*b - 3*A*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + 
c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c 
) - 1)) - 12*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 
 1) + log(sin(d*x + c) - 1)) - 12*B*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) 
 - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*C*b*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (127) = 254\).

Time = 0.16 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.12 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="giac")
 

Output:

1/24*(3*(3*A*a + 4*C*a + 4*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3* 
A*a + 4*C*a + 4*B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*A*a*tan(1/ 
2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1/ 
2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b*tan(1/2*d*x + 1/2*c)^7 - 2 
4*C*b*tan(1/2*d*x + 1/2*c)^7 + 9*A*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*a*tan(1 
/2*d*x + 1/2*c)^5 - 12*C*a*tan(1/2*d*x + 1/2*c)^5 + 40*A*b*tan(1/2*d*x + 1 
/2*c)^5 - 12*B*b*tan(1/2*d*x + 1/2*c)^5 + 72*C*b*tan(1/2*d*x + 1/2*c)^5 + 
9*A*a*tan(1/2*d*x + 1/2*c)^3 - 40*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan( 
1/2*d*x + 1/2*c)^3 - 40*A*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*d*x + 
1/2*c)^3 - 72*C*b*tan(1/2*d*x + 1/2*c)^3 + 15*A*a*tan(1/2*d*x + 1/2*c) + 2 
4*B*a*tan(1/2*d*x + 1/2*c) + 12*C*a*tan(1/2*d*x + 1/2*c) + 24*A*b*tan(1/2* 
d*x + 1/2*c) + 12*B*b*tan(1/2*d*x + 1/2*c) + 24*C*b*tan(1/2*d*x + 1/2*c))/ 
(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 2.82 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.87 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a}{8}+\frac {B\,b}{2}+\frac {C\,a}{2}\right )}{\frac {3\,A\,a}{2}+2\,B\,b+2\,C\,a}\right )\,\left (\frac {3\,A\,a}{4}+B\,b+C\,a\right )}{d}+\frac {\left (\frac {5\,A\,a}{4}-2\,A\,b-2\,B\,a+B\,b+C\,a-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a}{4}+\frac {10\,A\,b}{3}+\frac {10\,B\,a}{3}-B\,b-C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a}{4}-\frac {10\,A\,b}{3}-\frac {10\,B\,a}{3}-B\,b-C\,a-6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a}{4}+2\,A\,b+2\,B\,a+B\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int(((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + 
 d*x)^5,x)
 

Output:

(atanh((4*tan(c/2 + (d*x)/2)*((3*A*a)/8 + (B*b)/2 + (C*a)/2))/((3*A*a)/2 + 
 2*B*b + 2*C*a))*((3*A*a)/4 + B*b + C*a))/d + (tan(c/2 + (d*x)/2)^7*((5*A* 
a)/4 - 2*A*b - 2*B*a + B*b + C*a - 2*C*b) - tan(c/2 + (d*x)/2)^3*((10*A*b) 
/3 - (3*A*a)/4 + (10*B*a)/3 + B*b + C*a + 6*C*b) + tan(c/2 + (d*x)/2)^5*(( 
3*A*a)/4 + (10*A*b)/3 + (10*B*a)/3 - B*b - C*a + 6*C*b) + tan(c/2 + (d*x)/ 
2)*((5*A*a)/4 + 2*A*b + 2*B*a + B*b + C*a + 2*C*b))/(d*(6*tan(c/2 + (d*x)/ 
2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2 
)^8 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 714, normalized size of antiderivative = 5.21 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)
 

Output:

( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 12*cos 
(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*c - 12*cos(c + d*x)* 
log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**2 + 18*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 24*cos(c + d*x)*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**2*a*c + 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*si 
n(c + d*x)**2*b**2 - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2 - 12*co 
s(c + d*x)*log(tan((c + d*x)/2) - 1)*a*c - 12*cos(c + d*x)*log(tan((c + d* 
x)/2) - 1)*b**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 
*a**2 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*c + 12 
*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 - 18*cos(c + 
d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 24*cos(c + d*x)*log( 
tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*c - 24*cos(c + d*x)*log(tan((c + d 
*x)/2) + 1)*sin(c + d*x)**2*b**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1 
)*a**2 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*c + 12*cos(c + d*x)*l 
og(tan((c + d*x)/2) + 1)*b**2 - 9*cos(c + d*x)*sin(c + d*x)**3*a**2 - 12*c 
os(c + d*x)*sin(c + d*x)**3*a*c - 12*cos(c + d*x)*sin(c + d*x)**3*b**2 + 1 
5*cos(c + d*x)*sin(c + d*x)*a**2 + 12*cos(c + d*x)*sin(c + d*x)*a*c + 12*c 
os(c + d*x)*sin(c + d*x)*b**2 + 32*sin(c + d*x)**5*a*b + 24*sin(c + d*x)** 
5*b*c - 80*sin(c + d*x)**3*a*b - 48*sin(c + d*x)**3*b*c + 48*sin(c + d*x)* 
a*b + 24*sin(c + d*x)*b*c)/(24*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(...