\(\int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\) [392]

Optimal result

Integrand size = 29, antiderivative size = 203 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {15 a^3 x}{256}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}+\frac {9 a^3 \cos ^7(c+d x)}{7 d}-\frac {2 a^3 \cos ^9(c+d x)}{3 d}+\frac {a^3 \cos ^{11}(c+d x)}{11 d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{256 d}+\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{128 d}-\frac {5 a^3 \cos ^5(c+d x) \sin (c+d x)}{32 d}-\frac {5 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{16 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^5(c+d x)}{10 d} \] Output:

15/256*a^3*x-4/5*a^3*cos(d*x+c)^5/d+9/7*a^3*cos(d*x+c)^7/d-2/3*a^3*cos(d*x 
+c)^9/d+1/11*a^3*cos(d*x+c)^11/d+15/256*a^3*cos(d*x+c)*sin(d*x+c)/d+5/128* 
a^3*cos(d*x+c)^3*sin(d*x+c)/d-5/32*a^3*cos(d*x+c)^5*sin(d*x+c)/d-5/16*a^3* 
cos(d*x+c)^5*sin(d*x+c)^3/d-3/10*a^3*cos(d*x+c)^5*sin(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 12.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.62 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 (138600 c+138600 d x-198660 \cos (c+d x)-41580 \cos (3 (c+d x))+27258 \cos (5 (c+d x))+3630 \cos (7 (c+d x))-3850 \cos (9 (c+d x))+210 \cos (11 (c+d x))-13860 \sin (2 (c+d x))-46200 \sin (4 (c+d x))+6930 \sin (6 (c+d x))+5775 \sin (8 (c+d x))-1386 \sin (10 (c+d x)))}{2365440 d} \] Input:

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]
 

Output:

(a^3*(138600*c + 138600*d*x - 198660*Cos[c + d*x] - 41580*Cos[3*(c + d*x)] 
 + 27258*Cos[5*(c + d*x)] + 3630*Cos[7*(c + d*x)] - 3850*Cos[9*(c + d*x)] 
+ 210*Cos[11*(c + d*x)] - 13860*Sin[2*(c + d*x)] - 46200*Sin[4*(c + d*x)] 
+ 6930*Sin[6*(c + d*x)] + 5775*Sin[8*(c + d*x)] - 1386*Sin[10*(c + d*x)])) 
/(2365440*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 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.

\(\displaystyle \int \sin ^4(c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x)^4 (a \sin (c+d x)+a)^3dx\)

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (a^3 \sin ^7(c+d x) \cos ^4(c+d x)+3 a^3 \sin ^6(c+d x) \cos ^4(c+d x)+3 a^3 \sin ^5(c+d x) \cos ^4(c+d x)+a^3 \sin ^4(c+d x) \cos ^4(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \cos ^{11}(c+d x)}{11 d}-\frac {2 a^3 \cos ^9(c+d x)}{3 d}+\frac {9 a^3 \cos ^7(c+d x)}{7 d}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a^3 \sin ^5(c+d x) \cos ^5(c+d x)}{10 d}-\frac {5 a^3 \sin ^3(c+d x) \cos ^5(c+d x)}{16 d}-\frac {5 a^3 \sin (c+d x) \cos ^5(c+d x)}{32 d}+\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{128 d}+\frac {15 a^3 \sin (c+d x) \cos (c+d x)}{256 d}+\frac {15 a^3 x}{256}\)

Input:

Int[Cos[c + d*x]^4*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]
 

Output:

(15*a^3*x)/256 - (4*a^3*Cos[c + d*x]^5)/(5*d) + (9*a^3*Cos[c + d*x]^7)/(7* 
d) - (2*a^3*Cos[c + d*x]^9)/(3*d) + (a^3*Cos[c + d*x]^11)/(11*d) + (15*a^3 
*Cos[c + d*x]*Sin[c + d*x])/(256*d) + (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/ 
(128*d) - (5*a^3*Cos[c + d*x]^5*Sin[c + d*x])/(32*d) - (5*a^3*Cos[c + d*x] 
^5*Sin[c + d*x]^3)/(16*d) - (3*a^3*Cos[c + d*x]^5*Sin[c + d*x]^5)/(10*d)
 

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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.42

\[\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{6} \cos \left (d x +c \right )^{5}}{11}-\frac {2 \sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{5}}{33}-\frac {8 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{231}-\frac {16 \cos \left (d x +c \right )^{5}}{1155}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{5}}{10}-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{16}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{32}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{128}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{5}}{9}-\frac {4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{63}-\frac {8 \cos \left (d x +c \right )^{5}}{315}\right )+a^{3} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{8}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{16}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )}{d}\]

Input:

int(cos(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x)
                                                                                    
                                                                                    
 

Output:

1/d*(a^3*(-1/11*sin(d*x+c)^6*cos(d*x+c)^5-2/33*sin(d*x+c)^4*cos(d*x+c)^5-8 
/231*sin(d*x+c)^2*cos(d*x+c)^5-16/1155*cos(d*x+c)^5)+3*a^3*(-1/10*sin(d*x+ 
c)^5*cos(d*x+c)^5-1/16*sin(d*x+c)^3*cos(d*x+c)^5-1/32*sin(d*x+c)*cos(d*x+c 
)^5+1/128*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/256*c)+3*a^ 
3*(-1/9*sin(d*x+c)^4*cos(d*x+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos 
(d*x+c)^5)+a^3*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*cos(d*x+c)^ 
5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128*c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.67 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {26880 \, a^{3} \cos \left (d x + c\right )^{11} - 197120 \, a^{3} \cos \left (d x + c\right )^{9} + 380160 \, a^{3} \cos \left (d x + c\right )^{7} - 236544 \, a^{3} \cos \left (d x + c\right )^{5} + 17325 \, a^{3} d x - 231 \, {\left (384 \, a^{3} \cos \left (d x + c\right )^{9} - 1168 \, a^{3} \cos \left (d x + c\right )^{7} + 984 \, a^{3} \cos \left (d x + c\right )^{5} - 50 \, a^{3} \cos \left (d x + c\right )^{3} - 75 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{295680 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

1/295680*(26880*a^3*cos(d*x + c)^11 - 197120*a^3*cos(d*x + c)^9 + 380160*a 
^3*cos(d*x + c)^7 - 236544*a^3*cos(d*x + c)^5 + 17325*a^3*d*x - 231*(384*a 
^3*cos(d*x + c)^9 - 1168*a^3*cos(d*x + c)^7 + 984*a^3*cos(d*x + c)^5 - 50* 
a^3*cos(d*x + c)^3 - 75*a^3*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 648 vs. \(2 (194) = 388\).

Time = 2.05 (sec) , antiderivative size = 648, normalized size of antiderivative = 3.19 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**4*(a+a*sin(d*x+c))**3,x)
 

Output:

Piecewise((9*a**3*x*sin(c + d*x)**10/256 + 45*a**3*x*sin(c + d*x)**8*cos(c 
 + d*x)**2/256 + 3*a**3*x*sin(c + d*x)**8/128 + 45*a**3*x*sin(c + d*x)**6* 
cos(c + d*x)**4/128 + 3*a**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 45*a** 
3*x*sin(c + d*x)**4*cos(c + d*x)**6/128 + 9*a**3*x*sin(c + d*x)**4*cos(c + 
 d*x)**4/64 + 45*a**3*x*sin(c + d*x)**2*cos(c + d*x)**8/256 + 3*a**3*x*sin 
(c + d*x)**2*cos(c + d*x)**6/32 + 9*a**3*x*cos(c + d*x)**10/256 + 3*a**3*x 
*cos(c + d*x)**8/128 + 9*a**3*sin(c + d*x)**9*cos(c + d*x)/(256*d) + 21*a* 
*3*sin(c + d*x)**7*cos(c + d*x)**3/(128*d) + 3*a**3*sin(c + d*x)**7*cos(c 
+ d*x)/(128*d) - a**3*sin(c + d*x)**6*cos(c + d*x)**5/(5*d) - 3*a**3*sin(c 
 + d*x)**5*cos(c + d*x)**5/(10*d) + 11*a**3*sin(c + d*x)**5*cos(c + d*x)** 
3/(128*d) - 6*a**3*sin(c + d*x)**4*cos(c + d*x)**7/(35*d) - 3*a**3*sin(c + 
 d*x)**4*cos(c + d*x)**5/(5*d) - 21*a**3*sin(c + d*x)**3*cos(c + d*x)**7/( 
128*d) - 11*a**3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) - 8*a**3*sin(c + 
d*x)**2*cos(c + d*x)**9/(105*d) - 12*a**3*sin(c + d*x)**2*cos(c + d*x)**7/ 
(35*d) - 9*a**3*sin(c + d*x)*cos(c + d*x)**9/(256*d) - 3*a**3*sin(c + d*x) 
*cos(c + d*x)**7/(128*d) - 16*a**3*cos(c + d*x)**11/(1155*d) - 8*a**3*cos( 
c + d*x)**9/(105*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**4*cos(c)**4, 
True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.83 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2048 \, {\left (105 \, \cos \left (d x + c\right )^{11} - 385 \, \cos \left (d x + c\right )^{9} + 495 \, \cos \left (d x + c\right )^{7} - 231 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 22528 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 693 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} - 120 \, d x - 120 \, c - 5 \, \sin \left (8 \, d x + 8 \, c\right ) + 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} + 2310 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{2365440 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

1/2365440*(2048*(105*cos(d*x + c)^11 - 385*cos(d*x + c)^9 + 495*cos(d*x + 
c)^7 - 231*cos(d*x + c)^5)*a^3 - 22528*(35*cos(d*x + c)^9 - 90*cos(d*x + c 
)^7 + 63*cos(d*x + c)^5)*a^3 - 693*(32*sin(2*d*x + 2*c)^5 - 120*d*x - 120* 
c - 5*sin(8*d*x + 8*c) + 40*sin(4*d*x + 4*c))*a^3 + 2310*(24*d*x + 24*c + 
sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a^3)/d
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.94 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {15}{256} \, a^{3} x + \frac {a^{3} \cos \left (11 \, d x + 11 \, c\right )}{11264 \, d} - \frac {5 \, a^{3} \cos \left (9 \, d x + 9 \, c\right )}{3072 \, d} + \frac {11 \, a^{3} \cos \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac {59 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{5120 \, d} - \frac {9 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{512 \, d} - \frac {43 \, a^{3} \cos \left (d x + c\right )}{512 \, d} - \frac {3 \, a^{3} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {5 \, a^{3} \sin \left (8 \, d x + 8 \, c\right )}{2048 \, d} + \frac {3 \, a^{3} \sin \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

15/256*a^3*x + 1/11264*a^3*cos(11*d*x + 11*c)/d - 5/3072*a^3*cos(9*d*x + 9 
*c)/d + 11/7168*a^3*cos(7*d*x + 7*c)/d + 59/5120*a^3*cos(5*d*x + 5*c)/d - 
9/512*a^3*cos(3*d*x + 3*c)/d - 43/512*a^3*cos(d*x + c)/d - 3/5120*a^3*sin( 
10*d*x + 10*c)/d + 5/2048*a^3*sin(8*d*x + 8*c)/d + 3/1024*a^3*sin(6*d*x + 
6*c)/d - 5/256*a^3*sin(4*d*x + 4*c)/d - 3/512*a^3*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 19.74 (sec) , antiderivative size = 506, normalized size of antiderivative = 2.49 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^4*sin(c + d*x)^4*(a + a*sin(c + d*x))^3,x)
 

Output:

(15*a^3*x)/256 - ((15*a^3*(c + d*x))/256 + (5*a^3*tan(c/2 + (d*x)/2)^3)/4 
- (231*a^3*tan(c/2 + (d*x)/2)^5)/640 - (242*a^3*tan(c/2 + (d*x)/2)^7)/5 + 
(3987*a^3*tan(c/2 + (d*x)/2)^9)/64 - (3987*a^3*tan(c/2 + (d*x)/2)^13)/64 + 
 (242*a^3*tan(c/2 + (d*x)/2)^15)/5 + (231*a^3*tan(c/2 + (d*x)/2)^17)/640 - 
 (5*a^3*tan(c/2 + (d*x)/2)^19)/4 - (15*a^3*tan(c/2 + (d*x)/2)^21)/128 - (a 
^3*(17325*c + 17325*d*x - 53248))/295680 + tan(c/2 + (d*x)/2)^2*((165*a^3* 
(c + d*x))/256 - (a^3*(190575*c + 190575*d*x - 585728))/295680) + tan(c/2 
+ (d*x)/2)^4*((825*a^3*(c + d*x))/256 - (a^3*(952875*c + 952875*d*x - 2928 
640))/295680) + tan(c/2 + (d*x)/2)^6*((2475*a^3*(c + d*x))/256 - (a^3*(285 
8625*c + 2858625*d*x + 675840))/295680) + tan(c/2 + (d*x)/2)^8*((2475*a^3* 
(c + d*x))/128 - (a^3*(5717250*c + 5717250*d*x - 3379200))/295680) + tan(c 
/2 + (d*x)/2)^16*((2475*a^3*(c + d*x))/256 - (a^3*(2858625*c + 2858625*d*x 
 - 9461760))/295680) + tan(c/2 + (d*x)/2)^14*((2475*a^3*(c + d*x))/128 - ( 
a^3*(5717250*c + 5717250*d*x - 14192640))/295680) + tan(c/2 + (d*x)/2)^12* 
((3465*a^3*(c + d*x))/128 - (a^3*(8004150*c + 8004150*d*x + 16084992))/295 
680) + tan(c/2 + (d*x)/2)^10*((3465*a^3*(c + d*x))/128 - (a^3*(8004150*c + 
 8004150*d*x - 40685568))/295680) + (15*a^3*tan(c/2 + (d*x)/2))/128)/(d*(t 
an(c/2 + (d*x)/2)^2 + 1)^11)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.89 \[ \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (-26880 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{10}-88704 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}-62720 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}+85008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+139520 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+49896 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-9984 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-11550 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-13312 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-17325 \cos \left (d x +c \right ) \sin \left (d x +c \right )-26624 \cos \left (d x +c \right )+17325 d x +26624\right )}{295680 d} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x)
 

Output:

(a**3*( - 26880*cos(c + d*x)*sin(c + d*x)**10 - 88704*cos(c + d*x)*sin(c + 
 d*x)**9 - 62720*cos(c + d*x)*sin(c + d*x)**8 + 85008*cos(c + d*x)*sin(c + 
 d*x)**7 + 139520*cos(c + d*x)*sin(c + d*x)**6 + 49896*cos(c + d*x)*sin(c 
+ d*x)**5 - 9984*cos(c + d*x)*sin(c + d*x)**4 - 11550*cos(c + d*x)*sin(c + 
 d*x)**3 - 13312*cos(c + d*x)*sin(c + d*x)**2 - 17325*cos(c + d*x)*sin(c + 
 d*x) - 26624*cos(c + d*x) + 17325*d*x + 26624))/(295680*d)