\(\int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [588]

Optimal result

Integrand size = 29, antiderivative size = 209 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {17 a^2 x}{1024}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {4 a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \cos ^{11}(c+d x)}{11 d}+\frac {17 a^2 \cos (c+d x) \sin (c+d x)}{1024 d}+\frac {17 a^2 \cos ^3(c+d x) \sin (c+d x)}{1536 d}+\frac {17 a^2 \cos ^5(c+d x) \sin (c+d x)}{1920 d}-\frac {17 a^2 \cos ^7(c+d x) \sin (c+d x)}{320 d}-\frac {17 a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{120 d}-\frac {a^2 \cos ^7(c+d x) \sin ^5(c+d x)}{12 d} \] Output:

17/1024*a^2*x-2/7*a^2*cos(d*x+c)^7/d+4/9*a^2*cos(d*x+c)^9/d-2/11*a^2*cos(d 
*x+c)^11/d+17/1024*a^2*cos(d*x+c)*sin(d*x+c)/d+17/1536*a^2*cos(d*x+c)^3*si 
n(d*x+c)/d+17/1920*a^2*cos(d*x+c)^5*sin(d*x+c)/d-17/320*a^2*cos(d*x+c)^7*s 
in(d*x+c)/d-17/120*a^2*cos(d*x+c)^7*sin(d*x+c)^3/d-1/12*a^2*cos(d*x+c)^7*s 
in(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.65 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (166320 c+471240 d x-554400 \cos (c+d x)-184800 \cos (3 (c+d x))+55440 \cos (5 (c+d x))+39600 \cos (7 (c+d x))-6160 \cos (9 (c+d x))-5040 \cos (11 (c+d x))+55440 \sin (2 (c+d x))-162855 \sin (4 (c+d x))-27720 \sin (6 (c+d x))+24255 \sin (8 (c+d x))+5544 \sin (10 (c+d x))-1155 \sin (12 (c+d x)))}{28385280 d} \] Input:

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
 

Output:

(a^2*(166320*c + 471240*d*x - 554400*Cos[c + d*x] - 184800*Cos[3*(c + d*x) 
] + 55440*Cos[5*(c + d*x)] + 39600*Cos[7*(c + d*x)] - 6160*Cos[9*(c + d*x) 
] - 5040*Cos[11*(c + d*x)] + 55440*Sin[2*(c + d*x)] - 162855*Sin[4*(c + d* 
x)] - 27720*Sin[6*(c + d*x)] + 24255*Sin[8*(c + d*x)] + 5544*Sin[10*(c + d 
*x)] - 1155*Sin[12*(c + d*x)]))/(28385280*d)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 209, 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.

\(\displaystyle \int \sin ^4(c+d x) \cos ^6(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 \cos (c+d x)^6 (a \sin (c+d x)+a)^2dx\)

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (a^2 \sin ^6(c+d x) \cos ^6(c+d x)+2 a^2 \sin ^5(c+d x) \cos ^6(c+d x)+a^2 \sin ^4(c+d x) \cos ^6(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2 \cos ^{11}(c+d x)}{11 d}+\frac {4 a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^5(c+d x) \cos ^7(c+d x)}{12 d}-\frac {17 a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{120 d}-\frac {17 a^2 \sin (c+d x) \cos ^7(c+d x)}{320 d}+\frac {17 a^2 \sin (c+d x) \cos ^5(c+d x)}{1920 d}+\frac {17 a^2 \sin (c+d x) \cos ^3(c+d x)}{1536 d}+\frac {17 a^2 \sin (c+d x) \cos (c+d x)}{1024 d}+\frac {17 a^2 x}{1024}\)

Input:

Int[Cos[c + d*x]^6*Sin[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]
 

Output:

(17*a^2*x)/1024 - (2*a^2*Cos[c + d*x]^7)/(7*d) + (4*a^2*Cos[c + d*x]^9)/(9 
*d) - (2*a^2*Cos[c + d*x]^11)/(11*d) + (17*a^2*Cos[c + d*x]*Sin[c + d*x])/ 
(1024*d) + (17*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(1536*d) + (17*a^2*Cos[c + 
 d*x]^5*Sin[c + d*x])/(1920*d) - (17*a^2*Cos[c + d*x]^7*Sin[c + d*x])/(320 
*d) - (17*a^2*Cos[c + d*x]^7*Sin[c + d*x]^3)/(120*d) - (a^2*Cos[c + d*x]^7 
*Sin[c + d*x]^5)/(12*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 = 2.71 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.14

\[\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{5}}{12}-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{24}-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{64}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{384}+\frac {5 d x}{1024}+\frac {5 c}{1024}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{7}}{11}-\frac {4 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{99}-\frac {8 \cos \left (d x +c \right )^{7}}{693}\right )+a^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{80}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )}{d}\]

Input:

int(cos(d*x+c)^6*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x)
                                                                                    
                                                                                    
 

Output:

1/d*(a^2*(-1/12*cos(d*x+c)^7*sin(d*x+c)^5-1/24*sin(d*x+c)^3*cos(d*x+c)^7-1 
/64*cos(d*x+c)^7*sin(d*x+c)+1/384*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos( 
d*x+c))*sin(d*x+c)+5/1024*d*x+5/1024*c)+2*a^2*(-1/11*sin(d*x+c)^4*cos(d*x+ 
c)^7-4/99*cos(d*x+c)^7*sin(d*x+c)^2-8/693*cos(d*x+c)^7)+a^2*(-1/10*sin(d*x 
+c)^3*cos(d*x+c)^7-3/80*cos(d*x+c)^7*sin(d*x+c)+1/160*(cos(d*x+c)^5+5/4*co 
s(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/256*c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.66 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {645120 \, a^{2} \cos \left (d x + c\right )^{11} - 1576960 \, a^{2} \cos \left (d x + c\right )^{9} + 1013760 \, a^{2} \cos \left (d x + c\right )^{7} - 58905 \, a^{2} d x + 231 \, {\left (1280 \, a^{2} \cos \left (d x + c\right )^{11} - 4736 \, a^{2} \cos \left (d x + c\right )^{9} + 4272 \, a^{2} \cos \left (d x + c\right )^{7} - 136 \, a^{2} \cos \left (d x + c\right )^{5} - 170 \, a^{2} \cos \left (d x + c\right )^{3} - 255 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3548160 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-1/3548160*(645120*a^2*cos(d*x + c)^11 - 1576960*a^2*cos(d*x + c)^9 + 1013 
760*a^2*cos(d*x + c)^7 - 58905*a^2*d*x + 231*(1280*a^2*cos(d*x + c)^11 - 4 
736*a^2*cos(d*x + c)^9 + 4272*a^2*cos(d*x + c)^7 - 136*a^2*cos(d*x + c)^5 
- 170*a^2*cos(d*x + c)^3 - 255*a^2*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. 656 vs. \(2 (201) = 402\).

Time = 2.60 (sec) , antiderivative size = 656, normalized size of antiderivative = 3.14 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)**6*sin(d*x+c)**4*(a+a*sin(d*x+c))**2,x)
 

Output:

Piecewise((5*a**2*x*sin(c + d*x)**12/1024 + 15*a**2*x*sin(c + d*x)**10*cos 
(c + d*x)**2/512 + 3*a**2*x*sin(c + d*x)**10/256 + 75*a**2*x*sin(c + d*x)* 
*8*cos(c + d*x)**4/1024 + 15*a**2*x*sin(c + d*x)**8*cos(c + d*x)**2/256 + 
25*a**2*x*sin(c + d*x)**6*cos(c + d*x)**6/256 + 15*a**2*x*sin(c + d*x)**6* 
cos(c + d*x)**4/128 + 75*a**2*x*sin(c + d*x)**4*cos(c + d*x)**8/1024 + 15* 
a**2*x*sin(c + d*x)**4*cos(c + d*x)**6/128 + 15*a**2*x*sin(c + d*x)**2*cos 
(c + d*x)**10/512 + 15*a**2*x*sin(c + d*x)**2*cos(c + d*x)**8/256 + 5*a**2 
*x*cos(c + d*x)**12/1024 + 3*a**2*x*cos(c + d*x)**10/256 + 5*a**2*sin(c + 
d*x)**11*cos(c + d*x)/(1024*d) + 85*a**2*sin(c + d*x)**9*cos(c + d*x)**3/( 
3072*d) + 3*a**2*sin(c + d*x)**9*cos(c + d*x)/(256*d) + 33*a**2*sin(c + d* 
x)**7*cos(c + d*x)**5/(512*d) + 7*a**2*sin(c + d*x)**7*cos(c + d*x)**3/(12 
8*d) - 33*a**2*sin(c + d*x)**5*cos(c + d*x)**7/(512*d) + a**2*sin(c + d*x) 
**5*cos(c + d*x)**5/(10*d) - 2*a**2*sin(c + d*x)**4*cos(c + d*x)**7/(7*d) 
- 85*a**2*sin(c + d*x)**3*cos(c + d*x)**9/(3072*d) - 7*a**2*sin(c + d*x)** 
3*cos(c + d*x)**7/(128*d) - 8*a**2*sin(c + d*x)**2*cos(c + d*x)**9/(63*d) 
- 5*a**2*sin(c + d*x)*cos(c + d*x)**11/(1024*d) - 3*a**2*sin(c + d*x)*cos( 
c + d*x)**9/(256*d) - 16*a**2*cos(c + d*x)**11/(693*d), Ne(d, 0)), (x*(a*s 
in(c) + a)**2*sin(c)**4*cos(c)**6, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.66 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {81920 \, {\left (63 \, \cos \left (d x + c\right )^{11} - 154 \, \cos \left (d x + c\right )^{9} + 99 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 2772 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 1155 \, {\left (4 \, \sin \left (4 \, d x + 4 \, c\right )^{3} + 120 \, d x + 120 \, c + 9 \, \sin \left (8 \, d x + 8 \, c\right ) - 48 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{28385280 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

-1/28385280*(81920*(63*cos(d*x + c)^11 - 154*cos(d*x + c)^9 + 99*cos(d*x + 
 c)^7)*a^2 - 2772*(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^2 - 1155*(4*sin(4*d*x + 4*c)^3 + 120*d*x + 
120*c + 9*sin(8*d*x + 8*c) - 48*sin(4*d*x + 4*c))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {17}{1024} \, a^{2} x - \frac {a^{2} \cos \left (11 \, d x + 11 \, c\right )}{5632 \, d} - \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{4608 \, d} + \frac {5 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{3584 \, d} + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{512 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{768 \, d} - \frac {5 \, a^{2} \cos \left (d x + c\right )}{256 \, d} - \frac {a^{2} \sin \left (12 \, d x + 12 \, c\right )}{24576 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {7 \, a^{2} \sin \left (8 \, d x + 8 \, c\right )}{8192 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {47 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{8192 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

17/1024*a^2*x - 1/5632*a^2*cos(11*d*x + 11*c)/d - 1/4608*a^2*cos(9*d*x + 9 
*c)/d + 5/3584*a^2*cos(7*d*x + 7*c)/d + 1/512*a^2*cos(5*d*x + 5*c)/d - 5/7 
68*a^2*cos(3*d*x + 3*c)/d - 5/256*a^2*cos(d*x + c)/d - 1/24576*a^2*sin(12* 
d*x + 12*c)/d + 1/5120*a^2*sin(10*d*x + 10*c)/d + 7/8192*a^2*sin(8*d*x + 8 
*c)/d - 1/1024*a^2*sin(6*d*x + 6*c)/d - 47/8192*a^2*sin(4*d*x + 4*c)/d + 1 
/512*a^2*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 35.99 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.48 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^6*sin(c + d*x)^4*(a + a*sin(c + d*x))^2,x)
 

Output:

(a^2*((17*c)/1024 - (17*tan(c/2 + (d*x)/2))/512 - (128*tan(c/2 + (d*x)/2)^ 
2)/231 - (595*tan(c/2 + (d*x)/2)^3)/1536 - (64*tan(c/2 + (d*x)/2)^4)/21 + 
(11097*tan(c/2 + (d*x)/2)^5)/2560 + (704*tan(c/2 + (d*x)/2)^6)/63 + (27449 
*tan(c/2 + (d*x)/2)^7)/2560 - (384*tan(c/2 + (d*x)/2)^8)/7 - (202307*tan(c 
/2 + (d*x)/2)^9)/3840 + (192*tan(c/2 + (d*x)/2)^10)/7 + (28659*tan(c/2 + ( 
d*x)/2)^11)/256 - (64*tan(c/2 + (d*x)/2)^12)/3 - (28659*tan(c/2 + (d*x)/2) 
^13)/256 - 64*tan(c/2 + (d*x)/2)^14 + (202307*tan(c/2 + (d*x)/2)^15)/3840 
+ 32*tan(c/2 + (d*x)/2)^16 - (27449*tan(c/2 + (d*x)/2)^17)/2560 - (64*tan( 
c/2 + (d*x)/2)^18)/3 - (11097*tan(c/2 + (d*x)/2)^19)/2560 + (595*tan(c/2 + 
 (d*x)/2)^21)/1536 + (17*tan(c/2 + (d*x)/2)^23)/512 + (17*d*x)/1024 + (51* 
tan(c/2 + (d*x)/2)^2*(c + d*x))/256 + (561*tan(c/2 + (d*x)/2)^4*(c + d*x)) 
/512 + (935*tan(c/2 + (d*x)/2)^6*(c + d*x))/256 + (8415*tan(c/2 + (d*x)/2) 
^8*(c + d*x))/1024 + (1683*tan(c/2 + (d*x)/2)^10*(c + d*x))/128 + (3927*ta 
n(c/2 + (d*x)/2)^12*(c + d*x))/256 + (1683*tan(c/2 + (d*x)/2)^14*(c + d*x) 
)/128 + (8415*tan(c/2 + (d*x)/2)^16*(c + d*x))/1024 + (935*tan(c/2 + (d*x) 
/2)^18*(c + d*x))/256 + (561*tan(c/2 + (d*x)/2)^20*(c + d*x))/512 + (51*ta 
n(c/2 + (d*x)/2)^22*(c + d*x))/256 + (17*tan(c/2 + (d*x)/2)^24*(c + d*x))/ 
1024 - 32/693))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^12)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.94 \[ \int \cos ^6(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (295680 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{11}+645120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{10}-384384 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}-1648640 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}-432432 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+1157120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+678216 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-30720 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-39270 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-40960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-58905 \cos \left (d x +c \right ) \sin \left (d x +c \right )-81920 \cos \left (d x +c \right )+58905 d x +81920\right )}{3548160 d} \] Input:

int(cos(d*x+c)^6*sin(d*x+c)^4*(a+a*sin(d*x+c))^2,x)
 

Output:

(a**2*(295680*cos(c + d*x)*sin(c + d*x)**11 + 645120*cos(c + d*x)*sin(c + 
d*x)**10 - 384384*cos(c + d*x)*sin(c + d*x)**9 - 1648640*cos(c + d*x)*sin( 
c + d*x)**8 - 432432*cos(c + d*x)*sin(c + d*x)**7 + 1157120*cos(c + d*x)*s 
in(c + d*x)**6 + 678216*cos(c + d*x)*sin(c + d*x)**5 - 30720*cos(c + d*x)* 
sin(c + d*x)**4 - 39270*cos(c + d*x)*sin(c + d*x)**3 - 40960*cos(c + d*x)* 
sin(c + d*x)**2 - 58905*cos(c + d*x)*sin(c + d*x) - 81920*cos(c + d*x) + 5 
8905*d*x + 81920))/(3548160*d)