\(\int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\) [589]

Optimal result

Integrand size = 29, antiderivative size = 183 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{128}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{80 d}-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d} \] Output:

3/128*a^2*x-2/7*a^2*cos(d*x+c)^7/d+1/3*a^2*cos(d*x+c)^9/d-1/11*a^2*cos(d*x 
+c)^11/d+3/128*a^2*cos(d*x+c)*sin(d*x+c)/d+1/64*a^2*cos(d*x+c)^3*sin(d*x+c 
)/d+1/80*a^2*cos(d*x+c)^5*sin(d*x+c)/d-3/40*a^2*cos(d*x+c)^7*sin(d*x+c)/d- 
1/5*a^2*cos(d*x+c)^7*sin(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.69 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (27720 c+27720 d x-39270 \cos (c+d x)-16170 \cos (3 (c+d x))+1155 \cos (5 (c+d x))+2805 \cos (7 (c+d x))+385 \cos (9 (c+d x))-105 \cos (11 (c+d x))+4620 \sin (2 (c+d x))-9240 \sin (4 (c+d x))-2310 \sin (6 (c+d x))+1155 \sin (8 (c+d x))+462 \sin (10 (c+d x)))}{1182720 d} \] Input:

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
 

Output:

(a^2*(27720*c + 27720*d*x - 39270*Cos[c + d*x] - 16170*Cos[3*(c + d*x)] + 
1155*Cos[5*(c + d*x)] + 2805*Cos[7*(c + d*x)] + 385*Cos[9*(c + d*x)] - 105 
*Cos[11*(c + d*x)] + 4620*Sin[2*(c + d*x)] - 9240*Sin[4*(c + d*x)] - 2310* 
Sin[6*(c + d*x)] + 1155*Sin[8*(c + d*x)] + 462*Sin[10*(c + d*x)]))/(118272 
0*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 183, 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.

\(\displaystyle \int \sin ^3(c+d x) \cos ^6(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^3 \cos (c+d x)^6 (a \sin (c+d x)+a)^2dx\)

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (a^2 \sin ^5(c+d x) \cos ^6(c+d x)+2 a^2 \sin ^4(c+d x) \cos ^6(c+d x)+a^2 \sin ^3(c+d x) \cos ^6(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{5 d}-\frac {3 a^2 \sin (c+d x) \cos ^7(c+d x)}{40 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{80 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 a^2 x}{128}\)

Input:

Int[Cos[c + d*x]^6*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]
 

Output:

(3*a^2*x)/128 - (2*a^2*Cos[c + d*x]^7)/(7*d) + (a^2*Cos[c + d*x]^9)/(3*d) 
- (a^2*Cos[c + d*x]^11)/(11*d) + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/(128*d) 
 + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) + (a^2*Cos[c + d*x]^5*Sin[c + 
d*x])/(80*d) - (3*a^2*Cos[c + d*x]^7*Sin[c + d*x])/(40*d) - (a^2*Cos[c + d 
*x]^7*Sin[c + d*x]^3)/(5*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 = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.94

\[\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{7}}{11}-\frac {4 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{99}-\frac {8 \cos \left (d x +c \right )^{7}}{693}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{80}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )}{d}\]

Input:

int(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)
                                                                                    
                                                                                    
 

Output:

1/d*(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)+2*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*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x 
+c)+3/256*d*x+3/256*c)+a^2*(-1/9*cos(d*x+c)^7*sin(d*x+c)^2-2/63*cos(d*x+c) 
^7))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.68 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {13440 \, a^{2} \cos \left (d x + c\right )^{11} - 49280 \, a^{2} \cos \left (d x + c\right )^{9} + 42240 \, a^{2} \cos \left (d x + c\right )^{7} - 3465 \, a^{2} d x - 231 \, {\left (128 \, a^{2} \cos \left (d x + c\right )^{9} - 176 \, a^{2} \cos \left (d x + c\right )^{7} + 8 \, a^{2} \cos \left (d x + c\right )^{5} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{147840 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-1/147840*(13440*a^2*cos(d*x + c)^11 - 49280*a^2*cos(d*x + c)^9 + 42240*a^ 
2*cos(d*x + c)^7 - 3465*a^2*d*x - 231*(128*a^2*cos(d*x + c)^9 - 176*a^2*co 
s(d*x + c)^7 + 8*a^2*cos(d*x + c)^5 + 10*a^2*cos(d*x + c)^3 + 15*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. 384 vs. \(2 (168) = 336\).

Time = 1.85 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.10 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{10}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 a^{2} x \cos ^{10}{\left (c + d x \right )}}{128} + \frac {3 a^{2} \sin ^{9}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {7 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {7 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{63 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {3 a^{2} \sin {\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{128 d} - \frac {8 a^{2} \cos ^{11}{\left (c + d x \right )}}{693 d} - \frac {2 a^{2} \cos ^{9}{\left (c + d x \right )}}{63 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{6}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**6*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)
 

Output:

Piecewise((3*a**2*x*sin(c + d*x)**10/128 + 15*a**2*x*sin(c + d*x)**8*cos(c 
 + d*x)**2/128 + 15*a**2*x*sin(c + d*x)**6*cos(c + d*x)**4/64 + 15*a**2*x* 
sin(c + d*x)**4*cos(c + d*x)**6/64 + 15*a**2*x*sin(c + d*x)**2*cos(c + d*x 
)**8/128 + 3*a**2*x*cos(c + d*x)**10/128 + 3*a**2*sin(c + d*x)**9*cos(c + 
d*x)/(128*d) + 7*a**2*sin(c + d*x)**7*cos(c + d*x)**3/(64*d) + a**2*sin(c 
+ d*x)**5*cos(c + d*x)**5/(5*d) - a**2*sin(c + d*x)**4*cos(c + d*x)**7/(7* 
d) - 7*a**2*sin(c + d*x)**3*cos(c + d*x)**7/(64*d) - 4*a**2*sin(c + d*x)** 
2*cos(c + d*x)**9/(63*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 3* 
a**2*sin(c + d*x)*cos(c + d*x)**9/(128*d) - 8*a**2*cos(c + d*x)**11/(693*d 
) - 2*a**2*cos(c + d*x)**9/(63*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)* 
*3*cos(c)**6, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.63 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {5120 \, {\left (63 \, \cos \left (d x + c\right )^{11} - 154 \, \cos \left (d x + c\right )^{9} + 99 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 56320 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 693 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{3548160 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

-1/3548160*(5120*(63*cos(d*x + c)^11 - 154*cos(d*x + c)^9 + 99*cos(d*x + c 
)^7)*a^2 - 56320*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^2 - 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^2)/d
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3}{128} \, a^{2} x - \frac {a^{2} \cos \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{3072 \, d} + \frac {17 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{1024 \, d} - \frac {7 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{512 \, d} - \frac {17 \, a^{2} \cos \left (d x + c\right )}{512 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{2560 \, d} + \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{512 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{256 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

3/128*a^2*x - 1/11264*a^2*cos(11*d*x + 11*c)/d + 1/3072*a^2*cos(9*d*x + 9* 
c)/d + 17/7168*a^2*cos(7*d*x + 7*c)/d + 1/1024*a^2*cos(5*d*x + 5*c)/d - 7/ 
512*a^2*cos(3*d*x + 3*c)/d - 17/512*a^2*cos(d*x + c)/d + 1/2560*a^2*sin(10 
*d*x + 10*c)/d + 1/1024*a^2*sin(8*d*x + 8*c)/d - 1/512*a^2*sin(6*d*x + 6*c 
)/d - 1/128*a^2*sin(4*d*x + 4*c)/d + 1/256*a^2*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 36.52 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.97 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^6*sin(c + d*x)^3*(a + a*sin(c + d*x))^2,x)
 

Output:

(3*a^2*x)/128 - ((3*a^2*(c + d*x))/128 + (a^2*tan(c/2 + (d*x)/2)^3)/2 - (3 
323*a^2*tan(c/2 + (d*x)/2)^5)/320 + (108*a^2*tan(c/2 + (d*x)/2)^7)/5 - (84 
1*a^2*tan(c/2 + (d*x)/2)^9)/32 + (841*a^2*tan(c/2 + (d*x)/2)^13)/32 - (108 
*a^2*tan(c/2 + (d*x)/2)^15)/5 + (3323*a^2*tan(c/2 + (d*x)/2)^17)/320 - (a^ 
2*tan(c/2 + (d*x)/2)^19)/2 - (3*a^2*tan(c/2 + (d*x)/2)^21)/64 - a^2*((3*c) 
/128 + (3*d*x)/128 - 20/231) + tan(c/2 + (d*x)/2)^2*((33*a^2*(c + d*x))/12 
8 - a^2*((33*c)/128 + (33*d*x)/128 - 20/21)) + tan(c/2 + (d*x)/2)^18*((165 
*a^2*(c + d*x))/128 - a^2*((165*c)/128 + (165*d*x)/128 - 4)) + tan(c/2 + ( 
d*x)/2)^4*((165*a^2*(c + d*x))/128 - a^2*((165*c)/128 + (165*d*x)/128 - 16 
/21)) + tan(c/2 + (d*x)/2)^14*((495*a^2*(c + d*x))/64 - a^2*((495*c)/64 + 
(495*d*x)/64 + 16)) + tan(c/2 + (d*x)/2)^16*((495*a^2*(c + d*x))/128 - a^2 
*((495*c)/128 + (495*d*x)/128 - 12)) + tan(c/2 + (d*x)/2)^6*((495*a^2*(c + 
 d*x))/128 - a^2*((495*c)/128 + (495*d*x)/128 - 16/7)) + tan(c/2 + (d*x)/2 
)^8*((495*a^2*(c + d*x))/64 - a^2*((495*c)/64 + (495*d*x)/64 - 312/7)) + t 
an(c/2 + (d*x)/2)^10*((693*a^2*(c + d*x))/64 - a^2*((693*c)/64 + (693*d*x) 
/64 + 40)) + tan(c/2 + (d*x)/2)^12*((693*a^2*(c + d*x))/64 - a^2*((693*c)/ 
64 + (693*d*x)/64 - 80)) + (3*a^2*tan(c/2 + (d*x)/2))/64)/(d*(tan(c/2 + (d 
*x)/2)^2 + 1)^11)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.98 \[ \int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (13440 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{10}+29568 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}-17920 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}-77616 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-20480 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+57288 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+34560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-2310 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-3200 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-3465 \cos \left (d x +c \right ) \sin \left (d x +c \right )-6400 \cos \left (d x +c \right )+3465 d x +6400\right )}{147840 d} \] Input:

int(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)
 

Output:

(a**2*(13440*cos(c + d*x)*sin(c + d*x)**10 + 29568*cos(c + d*x)*sin(c + d* 
x)**9 - 17920*cos(c + d*x)*sin(c + d*x)**8 - 77616*cos(c + d*x)*sin(c + d* 
x)**7 - 20480*cos(c + d*x)*sin(c + d*x)**6 + 57288*cos(c + d*x)*sin(c + d* 
x)**5 + 34560*cos(c + d*x)*sin(c + d*x)**4 - 2310*cos(c + d*x)*sin(c + d*x 
)**3 - 3200*cos(c + d*x)*sin(c + d*x)**2 - 3465*cos(c + d*x)*sin(c + d*x) 
- 6400*cos(c + d*x) + 3465*d*x + 6400))/(147840*d)