\(\int (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\) [791]

Optimal result

Integrand size = 40, antiderivative size = 145 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=b^3 C x+\frac {\left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \tan (c+d x)}{3 d}+\frac {a^2 (5 b B+3 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:

b^3*C*x+1/2*(3*B*a^2*b+2*B*b^3+C*a^3+6*C*a*b^2)*arctanh(sin(d*x+c))/d+1/3* 
a*(2*B*a^2+8*B*b^2+9*C*a*b)*tan(d*x+c)/d+1/6*a^2*(5*B*b+3*C*a)*sec(d*x+c)* 
tan(d*x+c)/d+1/3*a*B*(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {6 b^3 C d x+6 b^2 (b B+3 a C) \coth ^{-1}(\sin (c+d x))+3 a^2 (3 b B+a C) \text {arctanh}(\sin (c+d x))+18 a b (b B+a C) \tan (c+d x)+3 a^2 (3 b B+a C) \sec (c+d x) \tan (c+d x)+2 a^3 B \tan (c+d x) \left (3+\tan ^2(c+d x)\right )}{6 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c 
 + d*x]^5,x]
 

Output:

(6*b^3*C*d*x + 6*b^2*(b*B + 3*a*C)*ArcCoth[Sin[c + d*x]] + 3*a^2*(3*b*B + 
a*C)*ArcTanh[Sin[c + d*x]] + 18*a*b*(b*B + a*C)*Tan[c + d*x] + 3*a^2*(3*b* 
B + a*C)*Sec[c + d*x]*Tan[c + d*x] + 2*a^3*B*Tan[c + d*x]*(3 + Tan[c + d*x 
]^2))/(6*d)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3468, 3042, 3510, 25, 3042, 3500, 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.

\(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^3 (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

\(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (3 b^2 C \cos ^2(c+d x)+\left (2 B a^2+6 b C a+3 b^2 B\right ) \cos (c+d x)+a (5 b B+3 a C)\right ) \sec ^3(c+d x)dx+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 B a^2+6 b C a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+3 a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{3} \left (\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int -\left (\left (6 C \cos ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x)\right )dx\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (6 C \cos ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {6 C \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B+2 b^3 C \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B+2 b^3 C \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {C a^3+3 b B a^2+6 b^2 C a+2 b^3 B+2 b^3 C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right ) \int \sec (c+d x)dx+2 b^3 C x\right )+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 b^3 C x\right )+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{d}+3 \left (\frac {\left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right ) \text {arctanh}(\sin (c+d x))}{d}+2 b^3 C x\right )\right )\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

Input:

Int[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x 
]^5,x]
 

Output:

(a*B*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^2*(5* 
b*B + 3*a*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (3*(2*b^3*C*x + ((3*a^2*b* 
B + 2*b^3*B + a^3*C + 6*a*b^2*C)*ArcTanh[Sin[c + d*x]])/d) + (2*a*(2*a^2*B 
 + 8*b^2*B + 9*a*b*C)*Tan[c + d*x])/d)/2)/3
 

Defintions of rubi rules used

rule 25

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

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 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 4257

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

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01

method	result	size
parts	\(\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 B a \,b^{2}+3 a^{2} b C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{2} b +a^{3} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (d x +c \right )}{d}\)	\(146\)
derivativedivides	\(\frac {a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a^{2} b +3 B \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a \,b^{2}+C \,b^{3} \left (d x +c \right )+B \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(180\)
default	\(\frac {a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a^{2} b +3 B \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a \,b^{2}+C \,b^{3} \left (d x +c \right )+B \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(180\)
parallelrisch	\(\frac {-27 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B \,a^{2} b +\frac {2}{3} B \,b^{3}+\frac {1}{3} a^{3} C +2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+27 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B \,a^{2} b +\frac {2}{3} B \,b^{3}+\frac {1}{3} a^{3} C +2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 C \,b^{3} d x \cos \left (3 d x +3 c \right )+\left (4 B \,a^{3}+18 B a \,b^{2}+18 a^{2} b C \right ) \sin \left (3 d x +3 c \right )+\left (18 B \,a^{2} b +6 a^{3} C \right ) \sin \left (2 d x +2 c \right )+18 C \,b^{3} d x \cos \left (d x +c \right )+12 \left (B \,a^{2}+\frac {3}{2} B \,b^{2}+\frac {3}{2} a b C \right ) \sin \left (d x +c \right ) a}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(255\)
risch	\(b^{3} C x -\frac {i a \left (9 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-18 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-36 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-36 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 B a b \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 B \,a^{2}-18 B \,b^{2}-18 a b C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C a \,b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{3}}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C a \,b^{2}}{d}\)	\(356\)

Input:

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x,method 
=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.30 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {12 \, C b^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{3} + 2 \, {\left (2 \, B a^{3} + 9 \, C a^{2} b + 9 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5 
,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.49 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 12 \, {\left (d x + c\right )} C b^{3} - 3 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 9 \, B a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, C a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="maxima")
 

Output:

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 12*(d*x + c)*C*b^3 - 3*C 
*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si 
n(d*x + c) - 1)) - 9*B*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)) + 18*C*a*b^2*(log(sin(d*x + c) + 
1) - log(sin(d*x + c) - 1)) + 6*B*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x 
 + c) - 1)) + 36*C*a^2*b*tan(d*x + c) + 36*B*a*b^2*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (137) = 274\).

Time = 0.17 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.32 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {6 \, {\left (d x + c\right )} C b^{3} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, 
 algorithm="giac")
 

Output:

1/6*(6*(d*x + c)*C*b^3 + 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*log(a 
bs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3 
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 
 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*C* 
a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*B*a^3 
*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*t 
an(1/2*d*x + 1/2*c)^3 + 6*B*a^3*tan(1/2*d*x + 1/2*c) + 3*C*a^3*tan(1/2*d*x 
 + 1/2*c) + 9*B*a^2*b*tan(1/2*d*x + 1/2*c) + 18*C*a^2*b*tan(1/2*d*x + 1/2* 
c) + 18*B*a*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 = 1.48 (sec) , antiderivative size = 526, normalized size of antiderivative = 3.63 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d 
*x)^5,x)
 

Output:

((B*a^3*sin(3*c + 3*d*x))/6 + (C*a^3*sin(2*c + 2*d*x))/4 + (B*a^3*sin(c + 
d*x))/2 + (3*B*a*b^2*sin(c + d*x))/4 + (3*C*a^2*b*sin(c + d*x))/4 - (B*b^3 
*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 - (C* 
a^3*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/4 + 
(3*C*b^3*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (3* 
B*a^2*b*sin(2*c + 2*d*x))/4 + (3*B*a*b^2*sin(3*c + 3*d*x))/4 + (3*C*a^2*b* 
sin(3*c + 3*d*x))/4 - (B*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/ 
2))*cos(3*c + 3*d*x)*1i)/2 - (C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + 
 (d*x)/2))*cos(3*c + 3*d*x)*1i)/4 + (C*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 
 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (B*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/c 
os(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/4 - (C*a*b^2*atan((sin(c/2 + (d*x) 
/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/2 - (B*a^2*b*cos(c + d*x) 
*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/4 - (C*a*b^2*cos(c + 
 d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/2)/(d*((3*cos(c 
 + d*x))/4 + cos(3*c + 3*d*x)/4))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 636, normalized size of antiderivative = 4.39 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)
 

Output:

( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*c - 9*co 
s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 - 18*cos(c 
+ d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2*c - 6*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + 3*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*a**3*c + 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b* 
*2 + 18*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2*c + 6*cos(c + d*x)*l 
og(tan((c + d*x)/2) - 1)*b**4 + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**2*a**3*c + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d 
*x)**2*a**2*b**2 + 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* 
*2*a*b**2*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b** 
4 - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*c - 9*cos(c + d*x)*log(t 
an((c + d*x)/2) + 1)*a**2*b**2 - 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1) 
*a*b**2*c - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**4 + 6*cos(c + d*x) 
*sin(c + d*x)**2*b**3*c*d*x - 3*cos(c + d*x)*sin(c + d*x)*a**3*c - 9*cos(c 
 + d*x)*sin(c + d*x)*a**2*b**2 - 6*cos(c + d*x)*b**3*c*d*x + 4*sin(c + d*x 
)**3*a**3*b + 18*sin(c + d*x)**3*a**2*b*c + 18*sin(c + d*x)**3*a*b**3 - 6* 
sin(c + d*x)*a**3*b - 18*sin(c + d*x)*a**2*b*c - 18*sin(c + d*x)*a*b**3)/( 
6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))