\(\int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1103]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 129 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b x}{8}-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{105 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{21 d}-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d} \] Output:

1/8*a*b*x-1/105*(a^2+6*b^2)*cos(d*x+c)^5/d+1/8*a*b*cos(d*x+c)*sin(d*x+c)/d 
+1/12*a*b*cos(d*x+c)^3*sin(d*x+c)/d-1/21*a*cos(d*x+c)^5*(a+b*sin(d*x+c))/d 
-1/7*cos(d*x+c)^5*(a+b*sin(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {840 a b c+840 a b d x-105 \left (8 a^2+3 b^2\right ) \cos (c+d x)-105 \left (4 a^2+b^2\right ) \cos (3 (c+d x))-84 a^2 \cos (5 (c+d x))+21 b^2 \cos (5 (c+d x))+15 b^2 \cos (7 (c+d x))+210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))-70 a b \sin (6 (c+d x))}{6720 d} \] Input:

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

Output:

(840*a*b*c + 840*a*b*d*x - 105*(8*a^2 + 3*b^2)*Cos[c + d*x] - 105*(4*a^2 + 
 b^2)*Cos[3*(c + d*x)] - 84*a^2*Cos[5*(c + d*x)] + 21*b^2*Cos[5*(c + d*x)] 
 + 15*b^2*Cos[7*(c + d*x)] + 210*a*b*Sin[2*(c + d*x)] - 210*a*b*Sin[4*(c + 
 d*x)] - 70*a*b*Sin[6*(c + d*x)])/(6720*d)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3341, 27, 3042, 3341, 3042, 3148, 3042, 3115, 3042, 3115, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3341

\(\displaystyle \frac {1}{7} \int 2 \cos ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{7} \int \cos ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \int \cos (c+d x)^4 (b+a \sin (c+d x)) (a+b \sin (c+d x))dx-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3341

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \int \cos ^4(c+d x) \left (7 a b+\left (a^2+6 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \int \cos (c+d x)^4 \left (7 a b+\left (a^2+6 b^2\right ) \sin (c+d x)\right )dx-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \int \cos ^4(c+d x)dx-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {2}{7} \left (\frac {1}{6} \left (7 a b \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\left (a^2+6 b^2\right ) \cos ^5(c+d x)}{5 d}\right )-\frac {a \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\right )-\frac {\cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\)

Input:

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

Output:

-1/7*(Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2)/d + (2*(-1/6*(a*Cos[c + d*x]^ 
5*(a + b*Sin[c + d*x]))/d + (-1/5*((a^2 + 6*b^2)*Cos[c + d*x]^5)/d + 7*a*b 
*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d* 
x])/(2*d)))/4))/6))/7
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3341
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[1/(m + p + 1)   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim 
p[a*c*(m + p + 1) + b*d*m + (a*d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] 
 &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S 
implerQ[c + d*x, a + b*x])
 
Maple [A] (verified)

Time = 88.87 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81

method result size
derivativedivides \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )}{d}\) \(105\)
default \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{5}}{5}+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )}{d}\) \(105\)
parallelrisch \(\frac {\left (-420 a^{2}-105 b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (-84 a^{2}+21 b^{2}\right ) \cos \left (5 d x +5 c \right )+15 b^{2} \cos \left (7 d x +7 c \right )+210 a b \sin \left (2 d x +2 c \right )-210 a b \sin \left (4 d x +4 c \right )-70 a b \sin \left (6 d x +6 c \right )+\left (-840 a^{2}-315 b^{2}\right ) \cos \left (d x +c \right )+840 a b x d -1344 a^{2}-384 b^{2}}{6720 d}\) \(136\)
risch \(\frac {a b x}{8}-\frac {a^{2} \cos \left (d x +c \right )}{8 d}-\frac {3 b^{2} \cos \left (d x +c \right )}{64 d}+\frac {b^{2} \cos \left (7 d x +7 c \right )}{448 d}-\frac {a b \sin \left (6 d x +6 c \right )}{96 d}-\frac {\cos \left (5 d x +5 c \right ) a^{2}}{80 d}+\frac {\cos \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{16 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{64 d}+\frac {a b \sin \left (2 d x +2 c \right )}{32 d}\) \(168\)
norman \(\frac {-\frac {14 a^{2}+4 b^{2}}{35 d}+\frac {a b x}{8}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}-\frac {\left (4 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {\left (4 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {\left (6 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {\left (8 a^{2}+8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {\left (22 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d}-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {31 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {31 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {11 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {7 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {21 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {35 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}+\frac {35 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {21 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {7 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) \(411\)
orering \(\text {Expression too large to display}\) \(3490\)

Input:

int(cos(d*x+c)^4*sin(d*x+c)*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/5*a^2*cos(d*x+c)^5+2*a*b*(-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)+b^2*(-1/7*sin(d*x+c)^2 
*cos(d*x+c)^5-2/35*cos(d*x+c)^5))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, a b d x - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:

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

Output:

1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + b^2)*cos(d*x + c)^5 + 105*a*b*d 
*x - 35*(8*a*b*cos(d*x + c)^5 - 2*a*b*cos(d*x + c)^3 - 3*a*b*cos(d*x + c)) 
*sin(d*x + c))/d
 

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin {\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2*cos(c + d*x)**5/(5*d) + a*b*x*sin(c + d*x)**6/8 + 3*a*b*x 
*sin(c + d*x)**4*cos(c + d*x)**2/8 + 3*a*b*x*sin(c + d*x)**2*cos(c + d*x)* 
*4/8 + a*b*x*cos(c + d*x)**6/8 + a*b*sin(c + d*x)**5*cos(c + d*x)/(8*d) + 
a*b*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - a*b*sin(c + d*x)*cos(c + d*x)* 
*5/(8*d) - b**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*b**2*cos(c + d*x 
)**7/(35*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c)**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.63 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {672 \, a^{2} \cos \left (d x + c\right )^{5} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{2}}{3360 \, d} \] Input:

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

Output:

-1/3360*(672*a^2*cos(d*x + c)^5 - 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c 
 - 3*sin(4*d*x + 4*c))*a*b - 96*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*b^2) 
/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{8} \, a b x + \frac {b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {{\left (4 \, a^{2} - b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (8 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \] Input:

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

Output:

1/8*a*b*x + 1/448*b^2*cos(7*d*x + 7*c)/d - 1/96*a*b*sin(6*d*x + 6*c)/d - 1 
/32*a*b*sin(4*d*x + 4*c)/d + 1/32*a*b*sin(2*d*x + 2*c)/d - 1/320*(4*a^2 - 
b^2)*cos(5*d*x + 5*c)/d - 1/64*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/64*(8* 
a^2 + 3*b^2)*cos(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 38.19 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.98 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a\,b\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (6\,a^2-4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{5}+\frac {4\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,a^2+8\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {22\,a^2}{5}-\frac {8\,b^2}{5}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {2\,a^2}{5}+\frac {4\,b^2}{35}-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {31\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {31\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:

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

Output:

(a*b*x)/8 - (tan(c/2 + (d*x)/2)^8*(6*a^2 - 4*b^2) + tan(c/2 + (d*x)/2)^10* 
(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^2*((4*a^2)/5 + (4*b^2)/5) + tan(c/2 + 
 (d*x)/2)^6*(8*a^2 + 8*b^2) + tan(c/2 + (d*x)/2)^4*((22*a^2)/5 - (8*b^2)/5 
) + 2*a^2*tan(c/2 + (d*x)/2)^12 + (2*a^2)/5 + (4*b^2)/35 - (11*a*b*tan(c/2 
 + (d*x)/2)^3)/3 + (31*a*b*tan(c/2 + (d*x)/2)^5)/12 - (31*a*b*tan(c/2 + (d 
*x)/2)^9)/12 + (11*a*b*tan(c/2 + (d*x)/2)^11)/3 - (a*b*tan(c/2 + (d*x)/2)^ 
13)/4 + (a*b*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.48 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b^{2}-280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b -168 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}+192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}+490 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +336 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -168 \cos \left (d x +c \right ) a^{2}-48 \cos \left (d x +c \right ) b^{2}+168 a^{2}+105 a b d x +48 b^{2}}{840 d} \] Input:

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

Output:

( - 120*cos(c + d*x)*sin(c + d*x)**6*b**2 - 280*cos(c + d*x)*sin(c + d*x)* 
*5*a*b - 168*cos(c + d*x)*sin(c + d*x)**4*a**2 + 192*cos(c + d*x)*sin(c + 
d*x)**4*b**2 + 490*cos(c + d*x)*sin(c + d*x)**3*a*b + 336*cos(c + d*x)*sin 
(c + d*x)**2*a**2 - 24*cos(c + d*x)*sin(c + d*x)**2*b**2 - 105*cos(c + d*x 
)*sin(c + d*x)*a*b - 168*cos(c + d*x)*a**2 - 48*cos(c + d*x)*b**2 + 168*a* 
*2 + 105*a*b*d*x + 48*b**2)/(840*d)