\(\int (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\) [792]

Optimal result

Integrand size = 40, antiderivative size = 188 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \tan (c+d x)}{3 d}+\frac {a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

1/8*(3*B*a^3+12*B*a*b^2+12*C*a^2*b+8*C*b^3)*arctanh(sin(d*x+c))/d+1/3*(6*B 
*a^2*b+3*B*b^3+2*C*a^3+9*C*a*b^2)*tan(d*x+c)/d+1/8*a*(3*B*a^2+10*B*b^2+12* 
C*a*b)*sec(d*x+c)*tan(d*x+c)/d+1/6*a^2*(3*B*b+2*C*a)*sec(d*x+c)^2*tan(d*x+ 
c)/d+1/4*a*B*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {24 b^3 C \coth ^{-1}(\sin (c+d x))+9 a \left (a^2 B+4 b^2 B+4 a b C\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 \left (3 a^2 b B+b^3 B+a^3 C+3 a b^2 C\right )+9 a \left (a^2 B+4 b^2 B+4 a b C\right ) \sec (c+d x)+6 a^3 B \sec ^3(c+d x)+8 a^2 (3 b B+a C) \tan ^2(c+d x)\right )}{24 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c 
 + d*x]^6,x]
 

Output:

(24*b^3*C*ArcCoth[Sin[c + d*x]] + 9*a*(a^2*B + 4*b^2*B + 4*a*b*C)*ArcTanh[ 
Sin[c + d*x]] + Tan[c + d*x]*(24*(3*a^2*b*B + b^3*B + a^3*C + 3*a*b^2*C) + 
 9*a*(a^2*B + 4*b^2*B + 4*a*b*C)*Sec[c + d*x] + 6*a^3*B*Sec[c + d*x]^3 + 8 
*a^2*(3*b*B + a*C)*Tan[c + d*x]^2))/(24*d)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3508, 3042, 3468, 3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

\(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^3 (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 3468

\(\displaystyle \frac {1}{4} \int (a+b \cos (c+d x)) \left (b (a B+4 b C) \cos ^2(c+d x)+\left (3 B a^2+8 b C a+4 b^2 B\right ) \cos (c+d x)+2 a (3 b B+2 a C)\right ) \sec ^4(c+d x)dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (a B+4 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B a^2+8 b C a+4 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (3 b B+2 a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{4} \left (\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 b^2 (a B+4 b C) \cos ^2(c+d x)+4 \left (2 C a^3+6 b B a^2+9 b^2 C a+3 b^3 B\right ) \cos (c+d x)+3 a \left (3 B a^2+12 b C a+10 b^2 B\right )\right ) \sec ^3(c+d x)\right )dx\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (3 b^2 (a B+4 b C) \cos ^2(c+d x)+4 \left (2 C a^3+6 b B a^2+9 b^2 C a+3 b^3 B\right ) \cos (c+d x)+3 a \left (3 B a^2+12 b C a+10 b^2 B\right )\right ) \sec ^3(c+d x)dx+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 b^2 (a B+4 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 \left (2 C a^3+6 b B a^2+9 b^2 C a+3 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (3 B a^2+12 b C a+10 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 \left (2 C a^3+6 b B a^2+9 b^2 C a+3 b^3 B\right )+3 \left (3 B a^3+12 b C a^2+12 b^2 B a+8 b^3 C\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 \left (2 C a^3+6 b B a^2+9 b^2 C a+3 b^3 B\right )+3 \left (3 B a^3+12 b C a^2+12 b^2 B a+8 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 \left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \int \sec ^2(c+d x)dx+3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \int \sec (c+d x)dx\right )+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 \left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 \left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 \left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d}\right )+\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{4} \left (\frac {2 a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {3 a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 \left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d}\right )\right )\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

Input:

Int[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x 
]^6,x]
 

Output:

(a*B*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*a^2*( 
3*b*B + 2*a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a*(3*a^2*B + 10*b^ 
2*B + 12*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*(3*a^3*B + 12*a*b^2 
*B + 12*a^2*b*C + 8*b^3*C)*ArcTanh[Sin[c + d*x]])/d + (8*(6*a^2*b*B + 3*b^ 
3*B + 2*a^3*C + 9*a*b^2*C)*Tan[c + d*x])/d)/2)/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 3468

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*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c 
 + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 
1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1]
 

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 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 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 4257

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

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.98

method	result	size
parts	\(\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B a \,b^{2}+3 a^{2} b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 B \,a^{2} b +a^{3} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(185\)
derivativedivides	\(\frac {-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(223\)
default	\(\frac {-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(223\)
parallelrisch	\(\frac {-18 \left (B \,a^{3}+4 B a \,b^{2}+4 a^{2} b C +\frac {8}{3} C \,b^{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (B \,a^{3}+4 B a \,b^{2}+4 a^{2} b C +\frac {8}{3} C \,b^{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \left (12 B \,a^{2} b +3 B \,b^{3}+4 a^{3} C +9 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+4 \left (6 B \,a^{2} b +3 B \,b^{3}+2 a^{3} C +9 C a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+33 \left (\frac {3 \left (B \,a^{2}+4 B \,b^{2}+4 a b C \right ) \sin \left (3 d x +3 c \right )}{11}+\sin \left (d x +c \right ) \left (B \,a^{2}+\frac {12}{11} B \,b^{2}+\frac {12}{11} a b C \right )\right ) a}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(287\)
risch	\(-\frac {i \left (-24 B \,b^{3}-16 a^{3} C -48 B \,a^{2} b -72 C a \,b^{2}-36 B \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}-36 C \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b -192 B \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +36 C \,a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-144 B \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-216 C a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-36 B a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-36 C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-216 C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+36 B a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+36 C \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-72 C a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+36 B a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-9 B \,{\mathrm e}^{i \left (d x +c \right )} a^{3}-33 B \,{\mathrm e}^{3 i \left (d x +c \right )} a^{3}-64 C \,{\mathrm e}^{2 i \left (d x +c \right )} a^{3}-72 B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+9 B \,a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-24 B \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+33 B \,a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-72 B \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-48 C \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{2 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{3}}{d}-\frac {3 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{2 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{3}}{d}\)	\(570\)

Input:

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method 
=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.12 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, B a^{3} + 8 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (B a^{3} + 4 \, C a^{2} b + 4 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, 
 algorithm="fricas")
 

Output:

1/48*(3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*cos(d*x + c)^4*log(s 
in(d*x + c) + 1) - 3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*cos(d*x 
 + c)^4*log(-sin(d*x + c) + 1) + 2*(6*B*a^3 + 8*(2*C*a^3 + 6*B*a^2*b + 9*C 
*a*b^2 + 3*B*b^3)*cos(d*x + c)^3 + 9*(B*a^3 + 4*C*a^2*b + 4*B*a*b^2)*cos(d 
*x + c)^2 + 8*(C*a^3 + 3*B*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + 
 c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6 
,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.45 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b - 3 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, C a b^{2} \tan \left (d x + c\right ) + 48 \, B b^{3} \tan \left (d x + c\right )}{48 \, d} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, 
 algorithm="maxima")
 

Output:

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 48*(tan(d*x + c)^3 + 3* 
tan(d*x + c))*B*a^2*b - 3*B*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(si 
n(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)) - 36*C*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si 
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 36*B*a*b^2*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*b 
^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 144*C*a*b^2*tan(d*x + 
 c) + 48*B*b^3*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (178) = 356\).

Time = 0.18 (sec) , antiderivative size = 586, normalized size of antiderivative = 3.12 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, 
 algorithm="giac")
 

Output:

1/24*(3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*log(abs(tan(1/2*d*x 
+ 1/2*c) + 1)) - 3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*log(abs(t 
an(1/2*d*x + 1/2*c) - 1)) + 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^3* 
tan(1/2*d*x + 1/2*c)^7 - 72*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^2*b*ta 
n(1/2*d*x + 1/2*c)^7 + 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 72*C*a*b^2*tan( 
1/2*d*x + 1/2*c)^7 - 24*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*B*a^3*tan(1/2*d*x 
 + 1/2*c)^5 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 120*B*a^2*b*tan(1/2*d*x + 
1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 36*B*a*b^2*tan(1/2*d*x + 1/ 
2*c)^5 + 216*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*b^3*tan(1/2*d*x + 1/2*c 
)^5 + 9*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1 
20*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36 
*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 216*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 72* 
B*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c) + 24*C*a^3*ta 
n(1/2*d*x + 1/2*c) + 72*B*a^2*b*tan(1/2*d*x + 1/2*c) + 36*C*a^2*b*tan(1/2* 
d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 72*C*a*b^2*tan(1/2*d*x + 
1/2*c) + 24*B*b^3*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 = 2.78 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.10 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^3}{8}+\frac {3\,C\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}+C\,b^3\right )}{\frac {3\,B\,a^3}{2}+6\,C\,a^2\,b+6\,B\,a\,b^2+4\,C\,b^3}\right )\,\left (\frac {3\,B\,a^3}{4}+3\,C\,a^2\,b+3\,B\,a\,b^2+2\,C\,b^3\right )}{d}-\frac {\left (2\,B\,b^3-\frac {5\,B\,a^3}{4}+2\,C\,a^3-3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (3\,B\,a\,b^2-6\,B\,b^3-\frac {10\,C\,a^3}{3}-\frac {3\,B\,a^3}{4}-10\,B\,a^2\,b-18\,C\,a\,b^2+3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,B\,b^3-\frac {3\,B\,a^3}{4}+\frac {10\,C\,a^3}{3}+3\,B\,a\,b^2+10\,B\,a^2\,b+18\,C\,a\,b^2+3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {5\,B\,a^3}{4}-2\,B\,b^3-2\,C\,a^3-3\,B\,a\,b^2-6\,B\,a^2\,b-6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d 
*x)^6,x)
 

Output:

(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^3)/8 + C*b^3 + (3*B*a*b^2)/2 + (3*C*a 
^2*b)/2))/((3*B*a^3)/2 + 4*C*b^3 + 6*B*a*b^2 + 6*C*a^2*b))*((3*B*a^3)/4 + 
2*C*b^3 + 3*B*a*b^2 + 3*C*a^2*b))/d - (tan(c/2 + (d*x)/2)^7*(2*B*b^3 - (5* 
B*a^3)/4 + 2*C*a^3 - 3*B*a*b^2 + 6*B*a^2*b + 6*C*a*b^2 - 3*C*a^2*b) + tan( 
c/2 + (d*x)/2)^3*(6*B*b^3 - (3*B*a^3)/4 + (10*C*a^3)/3 + 3*B*a*b^2 + 10*B* 
a^2*b + 18*C*a*b^2 + 3*C*a^2*b) - tan(c/2 + (d*x)/2)^5*((3*B*a^3)/4 + 6*B* 
b^3 + (10*C*a^3)/3 - 3*B*a*b^2 + 10*B*a^2*b + 18*C*a*b^2 - 3*C*a^2*b) - ta 
n(c/2 + (d*x)/2)*((5*B*a^3)/4 + 2*B*b^3 + 2*C*a^3 + 3*B*a*b^2 + 6*B*a^2*b 
+ 6*C*a*b^2 + 3*C*a^2*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 = 0.16 (sec) , antiderivative size = 830, normalized size of antiderivative = 4.41 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
 

Output:

( - 16*cos(c + d*x)*sin(c + d*x)**3*a**3*c - 48*cos(c + d*x)*sin(c + d*x)* 
*3*a**2*b**2 - 72*cos(c + d*x)*sin(c + d*x)**3*a*b**2*c - 24*cos(c + d*x)* 
sin(c + d*x)**3*b**4 + 24*cos(c + d*x)*sin(c + d*x)*a**3*c + 72*cos(c + d* 
x)*sin(c + d*x)*a**2*b**2 + 72*cos(c + d*x)*sin(c + d*x)*a*b**2*c + 24*cos 
(c + d*x)*sin(c + d*x)*b**4 - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4* 
a**3*b - 36*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b*c - 36*log(ta 
n((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**3 - 24*log(tan((c + d*x)/2) - 1)* 
sin(c + d*x)**4*b**3*c + 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 
*b + 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b*c + 72*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + 48*log(tan((c + d*x)/2) - 1)*sin( 
c + d*x)**2*b**3*c - 9*log(tan((c + d*x)/2) - 1)*a**3*b - 36*log(tan((c + 
d*x)/2) - 1)*a**2*b*c - 36*log(tan((c + d*x)/2) - 1)*a*b**3 - 24*log(tan(( 
c + d*x)/2) - 1)*b**3*c + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 
*b + 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b*c + 36*log(tan((c 
 + d*x)/2) + 1)*sin(c + d*x)**4*a*b**3 + 24*log(tan((c + d*x)/2) + 1)*sin( 
c + d*x)**4*b**3*c - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b - 
 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b*c - 72*log(tan((c + d 
*x)/2) + 1)*sin(c + d*x)**2*a*b**3 - 48*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**2*b**3*c + 9*log(tan((c + d*x)/2) + 1)*a**3*b + 36*log(tan((c + d*x) 
/2) + 1)*a**2*b*c + 36*log(tan((c + d*x)/2) + 1)*a*b**3 + 24*log(tan((c...