\(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [330]

Optimal result

Integrand size = 33, antiderivative size = 200 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {7}{16} a^4 (10 A+8 B+7 C) x+\frac {4 a^4 (10 A+8 B+7 C) \sin (c+d x)}{5 d}+\frac {27 a^4 (10 A+8 B+7 C) \cos (c+d x) \sin (c+d x)}{80 d}+\frac {a^4 (10 A+8 B+7 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {(6 B-C) (a+a \cos (c+d x))^4 \sin (c+d x)}{30 d}+\frac {C (a+a \cos (c+d x))^5 \sin (c+d x)}{6 a d}-\frac {2 a^4 (10 A+8 B+7 C) \sin ^3(c+d x)}{15 d} \] Output:

7/16*a^4*(10*A+8*B+7*C)*x+4/5*a^4*(10*A+8*B+7*C)*sin(d*x+c)/d+27/80*a^4*(1 
0*A+8*B+7*C)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^4*(10*A+8*B+7*C)*cos(d*x+c)^3* 
sin(d*x+c)/d+1/30*(6*B-C)*(a+a*cos(d*x+c))^4*sin(d*x+c)/d+1/6*C*(a+a*cos(d 
*x+c))^5*sin(d*x+c)/a/d-2/15*a^4*(10*A+8*B+7*C)*sin(d*x+c)^3/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.84 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a^4 \sin (c+d x) \left (210 (10 A+8 B+7 C) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )+\left (16 (100 A+83 B+72 C)+15 (54 A+56 B+49 C) \cos (c+d x)+32 (10 A+17 B+18 C) \cos ^2(c+d x)+10 (6 A+24 B+41 C) \cos ^3(c+d x)+48 (B+4 C) \cos ^4(c+d x)+40 C \cos ^5(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{240 d \sqrt {\sin ^2(c+d x)}} \] Input:

Integrate[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x 
]
 

Output:

(a^4*Sin[c + d*x]*(210*(10*A + 8*B + 7*C)*ArcSin[Sqrt[Sin[(c + d*x)/2]^2]] 
 + (16*(100*A + 83*B + 72*C) + 15*(54*A + 56*B + 49*C)*Cos[c + d*x] + 32*( 
10*A + 17*B + 18*C)*Cos[c + d*x]^2 + 10*(6*A + 24*B + 41*C)*Cos[c + d*x]^3 
 + 48*(B + 4*C)*Cos[c + d*x]^4 + 40*C*Cos[c + d*x]^5)*Sqrt[Sin[c + d*x]^2] 
))/(240*d*Sqrt[Sin[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 3502, 3042, 3230, 3042, 3124, 2009}

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.

\(\displaystyle \int (a \cos (c+d x)+a)^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^4 (a (6 A+5 C)+a (6 B-C) \cos (c+d x))dx}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (6 A+5 C)+a (6 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {3}{5} a (10 A+8 B+7 C) \int (\cos (c+d x) a+a)^4dx+\frac {a (6 B-C) \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{5} a (10 A+8 B+7 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4dx+\frac {a (6 B-C) \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

\(\Big \downarrow \) 3124

\(\displaystyle \frac {\frac {3}{5} a (10 A+8 B+7 C) \int \left (\cos ^4(c+d x) a^4+4 \cos ^3(c+d x) a^4+6 \cos ^2(c+d x) a^4+4 \cos (c+d x) a^4+a^4\right )dx+\frac {a (6 B-C) \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3}{5} a (10 A+8 B+7 C) \left (-\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8}\right )+\frac {a (6 B-C) \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^5}{6 a d}\)

Input:

Int[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 

Output:

(C*(a + a*Cos[c + d*x])^5*Sin[c + d*x])/(6*a*d) + ((a*(6*B - C)*(a + a*Cos 
[c + d*x])^4*Sin[c + d*x])/(5*d) + (3*a*(10*A + 8*B + 7*C)*((35*a^4*x)/8 + 
 (8*a^4*Sin[c + d*x])/d + (27*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4* 
Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (4*a^4*Sin[c + d*x]^3)/(3*d)))/5)/(6* 
a)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3124

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTri 
g[(a + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - 
b^2, 0] && IGtQ[n, 0]
 

rule 3230

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(415\) vs. \(2(186)=372\).

Time = 1.14 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.08

\[\frac {a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{4} B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{4} B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a^{4} C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{4} B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 a^{4} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} A \sin \left (d x +c \right )+4 a^{4} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} A \left (d x +c \right )+a^{4} B \sin \left (d x +c \right )+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\]

Input:

int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

1/d*(a^4*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/ 
5*a^4*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^4*C*(1/6*(cos(d*x 
+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+4/3*a^ 
4*A*(2+cos(d*x+c)^2)*sin(d*x+c)+4*a^4*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c)) 
*sin(d*x+c)+3/8*d*x+3/8*c)+4/5*a^4*C*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*s 
in(d*x+c)+6*a^4*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*a^4*B*(2+cos 
(d*x+c)^2)*sin(d*x+c)+6*a^4*C*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c 
)+3/8*d*x+3/8*c)+4*a^4*A*sin(d*x+c)+4*a^4*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2 
*d*x+1/2*c)+4/3*a^4*C*(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*A*(d*x+c)+a^4*B*sin( 
d*x+c)+a^4*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.72 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {105 \, {\left (10 \, A + 8 \, B + 7 \, C\right )} a^{4} d x + {\left (40 \, C a^{4} \cos \left (d x + c\right )^{5} + 48 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 10 \, {\left (6 \, A + 24 \, B + 41 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 32 \, {\left (10 \, A + 17 \, B + 18 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (54 \, A + 56 \, B + 49 \, C\right )} a^{4} \cos \left (d x + c\right ) + 16 \, {\left (100 \, A + 83 \, B + 72 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"fricas")
 

Output:

1/240*(105*(10*A + 8*B + 7*C)*a^4*d*x + (40*C*a^4*cos(d*x + c)^5 + 48*(B + 
 4*C)*a^4*cos(d*x + c)^4 + 10*(6*A + 24*B + 41*C)*a^4*cos(d*x + c)^3 + 32* 
(10*A + 17*B + 18*C)*a^4*cos(d*x + c)^2 + 15*(54*A + 56*B + 49*C)*a^4*cos( 
d*x + c) + 16*(100*A + 83*B + 72*C)*a^4)*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1005 vs. \(2 (187) = 374\).

Time = 0.45 (sec) , antiderivative size = 1005, normalized size of antiderivative = 5.02 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate((a+a*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
 

Output:

Piecewise((3*A*a**4*x*sin(c + d*x)**4/8 + 3*A*a**4*x*sin(c + d*x)**2*cos(c 
 + d*x)**2/4 + 3*A*a**4*x*sin(c + d*x)**2 + 3*A*a**4*x*cos(c + d*x)**4/8 + 
 3*A*a**4*x*cos(c + d*x)**2 + A*a**4*x + 3*A*a**4*sin(c + d*x)**3*cos(c + 
d*x)/(8*d) + 8*A*a**4*sin(c + d*x)**3/(3*d) + 5*A*a**4*sin(c + d*x)*cos(c 
+ d*x)**3/(8*d) + 4*A*a**4*sin(c + d*x)*cos(c + d*x)**2/d + 3*A*a**4*sin(c 
 + d*x)*cos(c + d*x)/d + 4*A*a**4*sin(c + d*x)/d + 3*B*a**4*x*sin(c + d*x) 
**4/2 + 3*B*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2 + 2*B*a**4*x*sin(c + d* 
x)**2 + 3*B*a**4*x*cos(c + d*x)**4/2 + 2*B*a**4*x*cos(c + d*x)**2 + 8*B*a* 
*4*sin(c + d*x)**5/(15*d) + 4*B*a**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) 
 + 3*B*a**4*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 4*B*a**4*sin(c + d*x)**3/ 
d + B*a**4*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*a**4*sin(c + d*x)*cos(c + 
d*x)**3/(2*d) + 6*B*a**4*sin(c + d*x)*cos(c + d*x)**2/d + 2*B*a**4*sin(c + 
 d*x)*cos(c + d*x)/d + B*a**4*sin(c + d*x)/d + 5*C*a**4*x*sin(c + d*x)**6/ 
16 + 15*C*a**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*C*a**4*x*sin(c + d 
*x)**4/4 + 15*C*a**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 9*C*a**4*x*sin 
(c + d*x)**2*cos(c + d*x)**2/2 + C*a**4*x*sin(c + d*x)**2/2 + 5*C*a**4*x*c 
os(c + d*x)**6/16 + 9*C*a**4*x*cos(c + d*x)**4/4 + C*a**4*x*cos(c + d*x)** 
2/2 + 5*C*a**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 32*C*a**4*sin(c + d*x 
)**5/(15*d) + 5*C*a**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 16*C*a**4*s 
in(c + d*x)**3*cos(c + d*x)**2/(3*d) + 9*C*a**4*sin(c + d*x)**3*cos(c +...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (186) = 372\).

Time = 0.05 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.00 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 1440 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 960 \, {\left (d x + c\right )} A a^{4} - 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{4} + 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{4} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 3840 \, A a^{4} \sin \left (d x + c\right ) - 960 \, B a^{4} \sin \left (d x + c\right )}{960 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"maxima")
 

Output:

-1/960*(1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 30*(12*d*x + 12*c + 
 sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 1440*(2*d*x + 2*c + sin(2* 
d*x + 2*c))*A*a^4 - 960*(d*x + c)*A*a^4 - 64*(3*sin(d*x + c)^5 - 10*sin(d* 
x + c)^3 + 15*sin(d*x + c))*B*a^4 + 1920*(sin(d*x + c)^3 - 3*sin(d*x + c)) 
*B*a^4 - 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 
 - 960*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 256*(3*sin(d*x + c)^5 - 10 
*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^4 + 5*(4*sin(2*d*x + 2*c)^3 - 60*d* 
x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*C*a^4 + 1280*(sin(d*x 
 + c)^3 - 3*sin(d*x + c))*C*a^4 - 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 
8*sin(2*d*x + 2*c))*C*a^4 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 3 
840*A*a^4*sin(d*x + c) - 960*B*a^4*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {7}{16} \, {\left (10 \, A a^{4} + 8 \, B a^{4} + 7 \, C a^{4}\right )} x + \frac {{\left (B a^{4} + 4 \, C a^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (2 \, A a^{4} + 8 \, B a^{4} + 15 \, C a^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, A a^{4} + 29 \, B a^{4} + 36 \, C a^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (112 \, A a^{4} + 128 \, B a^{4} + 127 \, C a^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (56 \, A a^{4} + 49 \, B a^{4} + 44 \, C a^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"giac")
 

Output:

1/192*C*a^4*sin(6*d*x + 6*c)/d + 7/16*(10*A*a^4 + 8*B*a^4 + 7*C*a^4)*x + 1 
/80*(B*a^4 + 4*C*a^4)*sin(5*d*x + 5*c)/d + 1/64*(2*A*a^4 + 8*B*a^4 + 15*C* 
a^4)*sin(4*d*x + 4*c)/d + 1/48*(16*A*a^4 + 29*B*a^4 + 36*C*a^4)*sin(3*d*x 
+ 3*c)/d + 1/64*(112*A*a^4 + 128*B*a^4 + 127*C*a^4)*sin(2*d*x + 2*c)/d + 1 
/8*(56*A*a^4 + 49*B*a^4 + 44*C*a^4)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 2.14 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4+\frac {49\,C\,a^4}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {595\,A\,a^4}{12}+\frac {119\,B\,a^4}{3}+\frac {833\,C\,a^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {231\,A\,a^4}{2}+\frac {462\,B\,a^4}{5}+\frac {1617\,C\,a^4}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {281\,A\,a^4}{2}+\frac {562\,B\,a^4}{5}+\frac {1967\,C\,a^4}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {1069\,A\,a^4}{12}+\frac {233\,B\,a^4}{3}+\frac {1471\,C\,a^4}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+25\,B\,a^4+\frac {207\,C\,a^4}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (10\,A+8\,B+7\,C\right )}{8\,\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4+\frac {49\,C\,a^4}{8}\right )}\right )\,\left (10\,A+8\,B+7\,C\right )}{8\,d} \] Input:

int((a + a*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
 

Output:

(tan(c/2 + (d*x)/2)^11*((35*A*a^4)/4 + 7*B*a^4 + (49*C*a^4)/8) + tan(c/2 + 
 (d*x)/2)^9*((595*A*a^4)/12 + (119*B*a^4)/3 + (833*C*a^4)/24) + tan(c/2 + 
(d*x)/2)^7*((231*A*a^4)/2 + (462*B*a^4)/5 + (1617*C*a^4)/20) + tan(c/2 + ( 
d*x)/2)^3*((1069*A*a^4)/12 + (233*B*a^4)/3 + (1471*C*a^4)/24) + tan(c/2 + 
(d*x)/2)^5*((281*A*a^4)/2 + (562*B*a^4)/5 + (1967*C*a^4)/20) + tan(c/2 + ( 
d*x)/2)*((93*A*a^4)/4 + 25*B*a^4 + (207*C*a^4)/8))/(d*(6*tan(c/2 + (d*x)/2 
)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d* 
x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (7*a^4*a 
tan((7*a^4*tan(c/2 + (d*x)/2)*(10*A + 8*B + 7*C))/(8*((35*A*a^4)/4 + 7*B*a 
^4 + (49*C*a^4)/8)))*(10*A + 8*B + 7*C))/(8*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.10 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a^{4} \left (40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} c -60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b -490 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +870 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +1080 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +1185 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +48 \sin \left (d x +c \right )^{5} b +192 \sin \left (d x +c \right )^{5} c -320 \sin \left (d x +c \right )^{3} a -640 \sin \left (d x +c \right )^{3} b -960 \sin \left (d x +c \right )^{3} c +1920 \sin \left (d x +c \right ) a +1920 \sin \left (d x +c \right ) b +1920 \sin \left (d x +c \right ) c +1050 a d x +840 b d x +735 c d x \right )}{240 d} \] Input:

int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

(a**4*(40*cos(c + d*x)*sin(c + d*x)**5*c - 60*cos(c + d*x)*sin(c + d*x)**3 
*a - 240*cos(c + d*x)*sin(c + d*x)**3*b - 490*cos(c + d*x)*sin(c + d*x)**3 
*c + 870*cos(c + d*x)*sin(c + d*x)*a + 1080*cos(c + d*x)*sin(c + d*x)*b + 
1185*cos(c + d*x)*sin(c + d*x)*c + 48*sin(c + d*x)**5*b + 192*sin(c + d*x) 
**5*c - 320*sin(c + d*x)**3*a - 640*sin(c + d*x)**3*b - 960*sin(c + d*x)** 
3*c + 1920*sin(c + d*x)*a + 1920*sin(c + d*x)*b + 1920*sin(c + d*x)*c + 10 
50*a*d*x + 840*b*d*x + 735*c*d*x))/(240*d)