\(\int (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [788]

Optimal result

Integrand size = 40, antiderivative size = 137 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) x+\frac {a^3 B \text {arctanh}(\sin (c+d x))}{d}+\frac {b \left (9 a b B+8 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 b B+5 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.98 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {6 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 C\right ) (c+d x)-12 a^3 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^3 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 b \left (4 a b B+4 a^2 C+b^2 C\right ) \sin (c+d x)+3 b^2 (b B+3 a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \] Input:

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

Output:

(6*(6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*C)*(c + d*x) - 12*a^3*B*Log[Cos[ 
(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^3*B*Log[Cos[(c + d*x)/2] + Sin[(c 
+ d*x)/2]] + 9*b*(4*a*b*B + 4*a^2*C + b^2*C)*Sin[c + d*x] + 3*b^2*(b*B + 3 
*a*C)*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)])/(12*d)
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 3508, 3042, 3469, 3042, 3512, 3042, 3502, 27, 3042, 3214, 3042, 4257}

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

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3214

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

rule 3469

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( 
n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
 f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( 
m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m 
+ n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !(IGt 
Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

rule 3502

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

rule 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 = 1.65 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99

method	result	size
parts	\(\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 B a \,b^{2}+3 a^{2} b C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{2} b +a^{3} C \right ) \left (d x +c \right )}{d}+\frac {B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}\)	\(135\)
parallelrisch	\(\frac {-12 B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \left (B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{3}+9 \left (4 B a \,b^{2}+4 a^{2} b C +C \,b^{3}\right ) \sin \left (d x +c \right )+36 x \left (B \,a^{2} b +\frac {1}{6} B \,b^{3}+\frac {1}{3} a^{3} C +\frac {1}{2} C a \,b^{2}\right ) d}{12 d}\)	\(139\)
derivativedivides	\(\frac {a^{3} C \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a^{2} b +3 B \,a^{2} b \left (d x +c \right )+3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \sin \left (d x +c \right ) a \,b^{2}+\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(151\)
default	\(\frac {a^{3} C \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a^{2} b +3 B \,a^{2} b \left (d x +c \right )+3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \sin \left (d x +c \right ) a \,b^{2}+\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(151\)
risch	\(3 x B \,a^{2} b +\frac {x B \,b^{3}}{2}+a^{3} C x +\frac {3 a \,b^{2} C x}{2}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B a \,b^{2}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b C}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{3}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B a \,b^{2}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b C}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{3}}{8 d}+\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) C \,b^{3}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) C a \,b^{2}}{4 d}\)	\(247\)
norman	\(\frac {\left (-3 B \,a^{2} b -\frac {1}{2} B \,b^{3}-a^{3} C -\frac {3}{2} C a \,b^{2}\right ) x +\left (-15 B \,a^{2} b -\frac {5}{2} B \,b^{3}-5 a^{3} C -\frac {15}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (3 B \,a^{2} b +\frac {1}{2} B \,b^{3}+a^{3} C +\frac {3}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (15 B \,a^{2} b +\frac {5}{2} B \,b^{3}+5 a^{3} C +\frac {15}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-12 B \,a^{2} b -2 B \,b^{3}-4 a^{3} C -6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (12 B \,a^{2} b +2 B \,b^{3}+4 a^{3} C +6 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {b \left (6 B a b -B \,b^{2}+6 a^{2} C -3 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {b \left (6 B a b +B \,b^{2}+6 a^{2} C +3 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b \left (18 B a b -3 B \,b^{2}+18 a^{2} C -9 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {2 b \left (18 B a b +3 B \,b^{2}+18 a^{2} C +9 a b C +2 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {b \left (54 B a b -3 B \,b^{2}+54 a^{2} C -9 a b C +10 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {b \left (54 B a b +3 B \,b^{2}+54 a^{2} C +9 a b C +10 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(575\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 \, B a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} d x + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{2} + 18 \, C a^{2} b + 18 \, B a b^{2} + 4 \, C b^{3} + 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

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

Output:

1/6*(3*B*a^3*log(sin(d*x + c) + 1) - 3*B*a^3*log(-sin(d*x + c) + 1) + 3*(2 
*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*d*x + (2*C*b^3*cos(d*x + c)^2 + 18 
*C*a^2*b + 18*B*a*b^2 + 4*C*b^3 + 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin( 
d*x + c))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a^{3} + 36 \, {\left (d x + c\right )} B a^{2} b + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \sin \left (d x + c\right ) + 36 \, B a b^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (129) = 258\).

Time = 0.16 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.29 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(6*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*B*a^3*log(abs(tan(1/2* 
d*x + 1/2*c) - 1)) + 3*(2*C*a^3 + 6*B*a^2*b + 3*C*a*b^2 + B*b^3)*(d*x + c) 
 + 2*(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 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 6* 
C*b^3*tan(1/2*d*x + 1/2*c)^5 + 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 + 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*t 
an(1/2*d*x + 1/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2* 
d*x + 1/2*c) + 3*B*b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^3*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 = 1924, normalized size of antiderivative = 14.04 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)^5*(2*C*b^3 - B*b^3 + 6*B*a*b^2 - 3*C*a*b^2 + 6*C*a^2*b 
) + tan(c/2 + (d*x)/2)*(B*b^3 + 2*C*b^3 + 6*B*a*b^2 + 3*C*a*b^2 + 6*C*a^2* 
b) + tan(c/2 + (d*x)/2)^3*((4*C*b^3)/3 + 12*B*a*b^2 + 12*C*a^2*b))/(d*(3*t 
an(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) 
+ (atan(((((B*b^3*1i)/2 + C*a^3*1i + B*a^2*b*3i + (C*a*b^2*3i)/2)*(32*B*a^ 
3 + 16*B*b^3 + 32*C*a^3 + 96*B*a^2*b + 48*C*a*b^2) + tan(c/2 + (d*x)/2)*(3 
2*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72 
*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 48*B*C*a*b^5 + 192*B*C*a^5*b + 320*B*C*a^3 
*b^3))*((B*b^3*1i)/2 + C*a^3*1i + B*a^2*b*3i + (C*a*b^2*3i)/2)*1i - (((B*b 
^3*1i)/2 + C*a^3*1i + B*a^2*b*3i + (C*a*b^2*3i)/2)*(32*B*a^3 + 16*B*b^3 + 
32*C*a^3 + 96*B*a^2*b + 48*C*a*b^2) - tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B 
^2*b^6 + 32*C^2*a^6 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 
96*C^2*a^4*b^2 + 48*B*C*a*b^5 + 192*B*C*a^5*b + 320*B*C*a^3*b^3))*((B*b^3* 
1i)/2 + C*a^3*1i + B*a^2*b*3i + (C*a*b^2*3i)/2)*1i)/((((B*b^3*1i)/2 + C*a^ 
3*1i + B*a^2*b*3i + (C*a*b^2*3i)/2)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*B 
*a^2*b + 48*C*a*b^2) + tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2 
*a^6 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 
+ 48*B*C*a*b^5 + 192*B*C*a^5*b + 320*B*C*a^3*b^3))*((B*b^3*1i)/2 + C*a^3*1 
i + B*a^2*b*3i + (C*a*b^2*3i)/2) + (((B*b^3*1i)/2 + C*a^3*1i + B*a^2*b*3i 
+ (C*a*b^2*3i)/2)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*B*a^2*b + 48*C*a...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.42 \[ \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} b -2 \sin \left (d x +c \right )^{3} b^{3} c +18 \sin \left (d x +c \right ) a^{2} b c +18 \sin \left (d x +c \right ) a \,b^{3}+6 \sin \left (d x +c \right ) b^{3} c +6 a^{3} c^{2}+6 a^{3} c d x +18 a^{2} b^{2} c +18 a^{2} b^{2} d x +9 a \,b^{2} c^{2}+9 a \,b^{2} c d x +3 b^{4} c +3 b^{4} d x}{6 d} \] Input:

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

Output:

(9*cos(c + d*x)*sin(c + d*x)*a*b**2*c + 3*cos(c + d*x)*sin(c + d*x)*b**4 - 
 6*log(tan((c + d*x)/2) - 1)*a**3*b + 6*log(tan((c + d*x)/2) + 1)*a**3*b - 
 2*sin(c + d*x)**3*b**3*c + 18*sin(c + d*x)*a**2*b*c + 18*sin(c + d*x)*a*b 
**3 + 6*sin(c + d*x)*b**3*c + 6*a**3*c**2 + 6*a**3*c*d*x + 18*a**2*b**2*c 
+ 18*a**2*b**2*d*x + 9*a*b**2*c**2 + 9*a*b**2*c*d*x + 3*b**4*c + 3*b**4*d* 
x)/(6*d)