\(\int (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [789]

Optimal result

Integrand size = 40, antiderivative size = 131 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {1}{2} b \left (6 a b B+6 a^2 C+b^2 C\right ) x+\frac {a^2 (3 b B+a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (2 a^2 B-b^2 B-3 a b C\right ) \sin (c+d x)}{d}-\frac {b^2 (2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B (a+b \cos (c+d x))^2 \tan (c+d x)}{d} \] Output:

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

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.66 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 b \left (6 a b B+6 a^2 C+b^2 C\right ) (c+d x)-4 a^2 (3 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 (3 b B+a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 a^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^2 (b B+3 a C) \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3508, 3042, 3468, 3042, 3512, 3042, 3502, 3042, 3214, 3042, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

-1/2*(b^2*(2*a*B - b*C)*Cos[c + d*x]*Sin[c + d*x])/d + (b*(6*a*b*B + 6*a^2 
*C + b^2*C)*x + (2*a^2*(3*b*B + a*C)*ArcTanh[Sin[c + d*x]])/d - (2*b*(2*a^ 
2*B - b^2*B - 3*a*b*C)*Sin[c + d*x])/d)/2 + (a*B*(a + b*Cos[c + d*x])^2*Ta 
n[c + d*x])/d
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 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 3502

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

rule 3508

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

rule 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 4257

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

Maple [A] (verified)

Time = 1.77 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95

method	result	size
parts	\(\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 B a \,b^{2}+3 a^{2} b C \right ) \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{2} b +a^{3} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,a^{3} \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(124\)
derivativedivides	\(\frac {a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+3 a^{2} b C \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (d x +c \right )+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )}{d}\)	\(132\)
default	\(\frac {a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+3 a^{2} b C \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (d x +c \right )+C \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )}{d}\)	\(132\)
parallelrisch	\(\frac {8 \left (-3 B \,a^{2} b -a^{3} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \left (3 B \,a^{2} b +a^{3} C \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{3}+24 x \left (B a b +a^{2} C +\frac {1}{6} b^{2} C \right ) b d \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (8 B \,a^{3}+C \,b^{3}\right )}{8 d \cos \left (d x +c \right )}\)	\(166\)
risch	\(3 x B a \,b^{2}+3 x \,a^{2} b C +\frac {b^{3} C x}{2}-\frac {i C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{2}}{2 d}+\frac {i C \,b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i B \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(253\)
norman	\(\frac {\left (3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x +\left (-15 B a \,b^{2}-15 a^{2} b C -\frac {5}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-15 B a \,b^{2}-15 a^{2} b C -\frac {5}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (3 B a \,b^{2}+3 a^{2} b C +\frac {1}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (9 B a \,b^{2}+9 a^{2} b C +\frac {3}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (9 B a \,b^{2}+9 a^{2} b C +\frac {3}{2} C \,b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {\left (2 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (10 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}-3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 \left (2 B \,a^{3}-B \,b^{3}-3 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {4 \left (2 B \,a^{3}+B \,b^{3}+3 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (2 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {\left (10 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}-3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {8 b^{2} \left (B b +3 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(610\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C b^{3} \cos \left (d x + c\right )^{2} + 2 \, B a^{3} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.10 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a^{2} b + 12 \, {\left (d x + c\right )} B a b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 4 \, B b^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.79 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (6 \, C a^{2} b + 6 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.80 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {C\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+B\,a^3\,\sin \left (c+d\,x\right )+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{8}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \] Input:

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

Output:

(C*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - C*a^3*atan((sin(c/2 + 
 (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i + 6*B*a*b^2*atan(sin(c/2 + (d*x)/2)/c 
os(c/2 + (d*x)/2)) - B*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/ 
2))*6i + 6*C*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + ((B*b^ 
3*sin(2*c + 2*d*x))/2 + (C*b^3*sin(3*c + 3*d*x))/8 + B*a^3*sin(c + d*x) + 
(C*b^3*sin(c + d*x))/8 + (3*C*a*b^2*sin(2*c + 2*d*x))/2)/(d*cos(c + d*x))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.06 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} c +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2} c +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}+6 \cos \left (d x +c \right ) a^{2} b \,c^{2}+6 \cos \left (d x +c \right ) a^{2} b c d x +6 \cos \left (d x +c \right ) a \,b^{3} c +6 \cos \left (d x +c \right ) a \,b^{3} d x +\cos \left (d x +c \right ) b^{3} c^{2}+\cos \left (d x +c \right ) b^{3} c d x -\sin \left (d x +c \right )^{3} b^{3} c +2 \sin \left (d x +c \right ) a^{3} b +\sin \left (d x +c \right ) b^{3} c}{2 \cos \left (d x +c \right ) d} \] Input:

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

Output:

( - 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*c - 6*cos(c + d*x)*log(t 
an((c + d*x)/2) - 1)*a**2*b**2 + 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)* 
a**3*c + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**2 + 6*cos(c + d* 
x)*sin(c + d*x)*a*b**2*c + 2*cos(c + d*x)*sin(c + d*x)*b**4 + 6*cos(c + d* 
x)*a**2*b*c**2 + 6*cos(c + d*x)*a**2*b*c*d*x + 6*cos(c + d*x)*a*b**3*c + 6 
*cos(c + d*x)*a*b**3*d*x + cos(c + d*x)*b**3*c**2 + cos(c + d*x)*b**3*c*d* 
x - sin(c + d*x)**3*b**3*c + 2*sin(c + d*x)*a**3*b + sin(c + d*x)*b**3*c)/ 
(2*cos(c + d*x)*d)