\(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\) [394]

Optimal result

Integrand size = 29, antiderivative size = 159 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {17 a^3 x}{128}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {17 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {17 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {3 a^3 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d} \] Output:

17/128*a^3*x-4/5*a^3*cos(d*x+c)^5/d+5/7*a^3*cos(d*x+c)^7/d-1/9*a^3*cos(d*x 
+c)^9/d+17/128*a^3*cos(d*x+c)*sin(d*x+c)/d+17/192*a^3*cos(d*x+c)^3*sin(d*x 
+c)/d-17/48*a^3*cos(d*x+c)^5*sin(d*x+c)/d-3/8*a^3*cos(d*x+c)^5*sin(d*x+c)^ 
3/d
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.67 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 (30240 c+42840 d x-52920 \cos (c+d x)-16800 \cos (3 (c+d x))+4032 \cos (5 (c+d x))+2340 \cos (7 (c+d x))-140 \cos (9 (c+d x))+5040 \sin (2 (c+d x))-12600 \sin (4 (c+d x))-1680 \sin (6 (c+d x))+945 \sin (8 (c+d x)))}{322560 d} \] Input:

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
 

Output:

(a^3*(30240*c + 42840*d*x - 52920*Cos[c + d*x] - 16800*Cos[3*(c + d*x)] + 
4032*Cos[5*(c + d*x)] + 2340*Cos[7*(c + d*x)] - 140*Cos[9*(c + d*x)] + 504 
0*Sin[2*(c + d*x)] - 12600*Sin[4*(c + d*x)] - 1680*Sin[6*(c + d*x)] + 945* 
Sin[8*(c + d*x)]))/(322560*d)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 159, 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.

\(\displaystyle \int \sin ^2(c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3352

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3 \cos ^9(c+d x)}{9 d}+\frac {5 a^3 \cos ^7(c+d x)}{7 d}-\frac {4 a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a^3 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {17 a^3 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {17 a^3 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {17 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {17 a^3 x}{128}\)

Input:

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]
 

Output:

(17*a^3*x)/128 - (4*a^3*Cos[c + d*x]^5)/(5*d) + (5*a^3*Cos[c + d*x]^7)/(7* 
d) - (a^3*Cos[c + d*x]^9)/(9*d) + (17*a^3*Cos[c + d*x]*Sin[c + d*x])/(128* 
d) + (17*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(192*d) - (17*a^3*Cos[c + d*x]^5 
*Sin[c + d*x])/(48*d) - (3*a^3*Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*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 = 216, normalized size of antiderivative = 1.36

\[\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{5}}{9}-\frac {4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{63}-\frac {8 \cos \left (d x +c \right )^{5}}{315}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{8}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{16}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\]

Input:

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
                                                                                    
                                                                                    
 

Output:

1/d*(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)+3*a^3*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*co 
s(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 
)+3*a^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+a^3*(-1/6*sin(d 
*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+ 
1/16*c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {4480 \, a^{3} \cos \left (d x + c\right )^{9} - 28800 \, a^{3} \cos \left (d x + c\right )^{7} + 32256 \, a^{3} \cos \left (d x + c\right )^{5} - 5355 \, a^{3} d x - 105 \, {\left (144 \, a^{3} \cos \left (d x + c\right )^{7} - 280 \, a^{3} \cos \left (d x + c\right )^{5} + 34 \, a^{3} \cos \left (d x + c\right )^{3} + 51 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40320 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

-1/40320*(4480*a^3*cos(d*x + c)^9 - 28800*a^3*cos(d*x + c)^7 + 32256*a^3*c 
os(d*x + c)^5 - 5355*a^3*d*x - 105*(144*a^3*cos(d*x + c)^7 - 280*a^3*cos(d 
*x + c)^5 + 34*a^3*cos(d*x + c)^3 + 51*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. 486 vs. \(2 (151) = 302\).

Time = 1.22 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.06 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\begin {cases} \frac {9 a^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {9 a^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {27 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {33 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {33 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {4 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {3 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {9 a^{3} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {8 a^{3} \cos ^{9}{\left (c + d x \right )}}{315 d} - \frac {6 a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)
 

Output:

Piecewise((9*a**3*x*sin(c + d*x)**8/128 + 9*a**3*x*sin(c + d*x)**6*cos(c + 
 d*x)**2/32 + a**3*x*sin(c + d*x)**6/16 + 27*a**3*x*sin(c + d*x)**4*cos(c 
+ d*x)**4/64 + 3*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a**3*x*sin( 
c + d*x)**2*cos(c + d*x)**6/32 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/ 
16 + 9*a**3*x*cos(c + d*x)**8/128 + a**3*x*cos(c + d*x)**6/16 + 9*a**3*sin 
(c + d*x)**7*cos(c + d*x)/(128*d) + 33*a**3*sin(c + d*x)**5*cos(c + d*x)** 
3/(128*d) + a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**3*sin(c + d*x)** 
4*cos(c + d*x)**5/(5*d) - 33*a**3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) 
+ a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 4*a**3*sin(c + d*x)**2*cos( 
c + d*x)**7/(35*d) - 3*a**3*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 9*a**3 
*sin(c + d*x)*cos(c + d*x)**7/(128*d) - a**3*sin(c + d*x)*cos(c + d*x)**5/ 
(16*d) - 8*a**3*cos(c + d*x)**9/(315*d) - 6*a**3*cos(c + d*x)**7/(35*d), N 
e(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.87 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {1024 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 27648 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 1680 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 945 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{322560 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

-1/322560*(1024*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5 
)*a^3 - 27648*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^3 - 1680*(4*sin(2*d* 
x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3 - 945*(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 = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {17}{128} \, a^{3} x - \frac {a^{3} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {13 \, a^{3} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{96 \, d} - \frac {21 \, a^{3} \cos \left (d x + c\right )}{128 \, d} + \frac {3 \, a^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

17/128*a^3*x - 1/2304*a^3*cos(9*d*x + 9*c)/d + 13/1792*a^3*cos(7*d*x + 7*c 
)/d + 1/80*a^3*cos(5*d*x + 5*c)/d - 5/96*a^3*cos(3*d*x + 3*c)/d - 21/128*a 
^3*cos(d*x + c)/d + 3/1024*a^3*sin(8*d*x + 8*c)/d - 1/192*a^3*sin(6*d*x + 
6*c)/d - 5/128*a^3*sin(4*d*x + 4*c)/d + 1/64*a^3*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 20.31 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.75 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx =\text {Too large to display} \] Input:

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^3,x)
 

Output:

(17*a^3*x)/128 - ((17*a^3*(c + d*x))/128 - (35*a^3*tan(c/2 + (d*x)/2)^3)/9 
6 - (537*a^3*tan(c/2 + (d*x)/2)^5)/32 + (531*a^3*tan(c/2 + (d*x)/2)^7)/32 
- (531*a^3*tan(c/2 + (d*x)/2)^11)/32 + (537*a^3*tan(c/2 + (d*x)/2)^13)/32 
+ (35*a^3*tan(c/2 + (d*x)/2)^15)/96 - (17*a^3*tan(c/2 + (d*x)/2)^17)/64 - 
(a^3*(5355*c + 5355*d*x - 15872))/40320 + tan(c/2 + (d*x)/2)^2*((153*a^3*( 
c + d*x))/128 - (a^3*(48195*c + 48195*d*x - 142848))/40320) + tan(c/2 + (d 
*x)/2)^4*((153*a^3*(c + d*x))/32 - (a^3*(192780*c + 192780*d*x - 87552))/4 
0320) + tan(c/2 + (d*x)/2)^14*((153*a^3*(c + d*x))/32 - (a^3*(192780*c + 1 
92780*d*x - 483840))/40320) + tan(c/2 + (d*x)/2)^6*((357*a^3*(c + d*x))/32 
 - (a^3*(449820*c + 449820*d*x - 419328))/40320) + tan(c/2 + (d*x)/2)^10*( 
(1071*a^3*(c + d*x))/64 - (a^3*(674730*c + 674730*d*x + 161280))/40320) + 
tan(c/2 + (d*x)/2)^12*((357*a^3*(c + d*x))/32 - (a^3*(449820*c + 449820*d* 
x - 913920))/40320) + tan(c/2 + (d*x)/2)^8*((1071*a^3*(c + d*x))/64 - (a^3 
*(674730*c + 674730*d*x - 2161152))/40320) + (17*a^3*tan(c/2 + (d*x)/2))/6 
4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^9)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.93 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (-4480 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}-15120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-10880 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+15960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+27264 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+9870 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-3968 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-5355 \cos \left (d x +c \right ) \sin \left (d x +c \right )-7936 \cos \left (d x +c \right )+5355 d x +7936\right )}{40320 d} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)
 

Output:

(a**3*( - 4480*cos(c + d*x)*sin(c + d*x)**8 - 15120*cos(c + d*x)*sin(c + d 
*x)**7 - 10880*cos(c + d*x)*sin(c + d*x)**6 + 15960*cos(c + d*x)*sin(c + d 
*x)**5 + 27264*cos(c + d*x)*sin(c + d*x)**4 + 9870*cos(c + d*x)*sin(c + d* 
x)**3 - 3968*cos(c + d*x)*sin(c + d*x)**2 - 5355*cos(c + d*x)*sin(c + d*x) 
 - 7936*cos(c + d*x) + 5355*d*x + 7936))/(40320*d)