\(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1056]

Optimal result

Integrand size = 29, antiderivative size = 163 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{16} \left (2 a^2+b^2\right ) x-\frac {2 a b \cos (c+d x)}{5 d}+\frac {2 a b \cos ^3(c+d x)}{15 d}-\frac {\left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d} \] Output:

1/16*(2*a^2+b^2)*x-2/5*a*b*cos(d*x+c)/d+2/15*a*b*cos(d*x+c)^3/d-1/16*(2*a^ 
2+b^2)*cos(d*x+c)*sin(d*x+c)/d+1/24*(2*a^2-b^2)*cos(d*x+c)*sin(d*x+c)^3/d+ 
1/15*a*b*cos(d*x+c)*sin(d*x+c)^4/d+1/6*cos(d*x+c)*sin(d*x+c)^3*(a+b*sin(d* 
x+c))^2/d
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {120 a^2 c+60 b^2 c+120 a^2 d x+60 b^2 d x-240 a b \cos (c+d x)-40 a b \cos (3 (c+d x))+24 a b \cos (5 (c+d x))-15 b^2 \sin (2 (c+d x))-30 a^2 \sin (4 (c+d x))-15 b^2 \sin (4 (c+d x))+5 b^2 \sin (6 (c+d x))}{960 d} \] Input:

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

Output:

(120*a^2*c + 60*b^2*c + 120*a^2*d*x + 60*b^2*d*x - 240*a*b*Cos[c + d*x] - 
40*a*b*Cos[3*(c + d*x)] + 24*a*b*Cos[5*(c + d*x)] - 15*b^2*Sin[2*(c + d*x) 
] - 30*a^2*Sin[4*(c + d*x)] - 15*b^2*Sin[4*(c + d*x)] + 5*b^2*Sin[6*(c + d 
*x)])/(960*d)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3368, 3042, 3529, 3042, 3512, 3042, 3502, 27, 3042, 3227, 3042, 3113, 2009, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

\(\displaystyle \int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right ) (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^2 \left (1-\sin (c+d x)^2\right ) (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int \sin ^2(c+d x) \left (15 a^2+12 b \sin (c+d x) a-5 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right )dx+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \int \sin (c+d x)^2 \left (15 a^2+12 b \sin (c+d x) a-5 \left (2 a^2-b^2\right ) \sin (c+d x)^2\right )dx+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {1}{4} \int 3 \sin ^2(c+d x) \left (5 \left (2 a^2+b^2\right )+16 a b \sin (c+d x)\right )dx+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \int \sin ^2(c+d x) \left (5 \left (2 a^2+b^2\right )+16 a b \sin (c+d x)\right )dx+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \int \sin (c+d x)^2 \left (5 \left (2 a^2+b^2\right )+16 a b \sin (c+d x)\right )dx+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \int \sin ^2(c+d x)dx+16 a b \int \sin ^3(c+d x)dx\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \int \sin (c+d x)^2dx+16 a b \int \sin (c+d x)^3dx\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \int \sin (c+d x)^2dx-\frac {16 a b \int \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \int \sin (c+d x)^2dx-\frac {16 a b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {16 a b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {3}{4} \left (5 \left (2 a^2+b^2\right ) \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {16 a b \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {5 \left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 d}\right )+\frac {2 a b \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d}\)

Input:

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

Output:

(Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(6*d) + ((2*a*b*Cos[c 
 + d*x]*Sin[c + d*x]^4)/(5*d) + ((5*(2*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x 
]^3)/(4*d) + (3*((-16*a*b*(Cos[c + d*x] - Cos[c + d*x]^3/3))/d + 5*(2*a^2 
+ b^2)*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d))))/4)/5)/6
 

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 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 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3368

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a 
 + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n 
}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3529

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : 
> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 
1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 

Maple [A] (verified)

Time = 31.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {120 a^{2} d x +60 b^{2} d x -240 a b \cos \left (d x +c \right )+5 b^{2} \sin \left (6 d x +6 c \right )+24 a b \cos \left (5 d x +5 c \right )-30 \sin \left (4 d x +4 c \right ) a^{2}-15 \sin \left (4 d x +4 c \right ) b^{2}-40 a b \cos \left (3 d x +3 c \right )-15 b^{2} \sin \left (2 d x +2 c \right )-256 a b}{960 d}\)	\(117\)
risch	\(\frac {b^{2} x}{16}+\frac {a^{2} x}{8}-\frac {a b \cos \left (d x +c \right )}{4 d}+\frac {b^{2} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a b \cos \left (5 d x +5 c \right )}{40 d}-\frac {\sin \left (4 d x +4 c \right ) a^{2}}{32 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{64 d}-\frac {a b \cos \left (3 d x +3 c \right )}{24 d}-\frac {b^{2} \sin \left (2 d x +2 c \right )}{64 d}\)	\(127\)
derivativedivides	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\)	\(141\)
default	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\)	\(141\)
norman	\(\frac {\left (\frac {b^{2}}{16}+\frac {a^{2}}{8}\right ) x +\left (\frac {b^{2}}{16}+\frac {a^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3 b^{2}}{8}+\frac {3 a^{2}}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3 b^{2}}{8}+\frac {3 a^{2}}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {5 b^{2}}{4}+\frac {5 a^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {15 b^{2}}{16}+\frac {15 a^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {15 b^{2}}{16}+\frac {15 a^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {8 a b}{15 d}-\frac {\left (2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {\left (6 a^{2}+19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {\left (6 a^{2}+19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (30 a^{2}-17 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {\left (30 a^{2}-17 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {16 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {16 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\)	\(392\)
orering	\(\text {Expression too large to display}\)	\(2654\)

Input:

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

Output:

1/960*(120*a^2*d*x+60*b^2*d*x-240*a*b*cos(d*x+c)+5*b^2*sin(6*d*x+6*c)+24*a 
*b*cos(5*d*x+5*c)-30*sin(4*d*x+4*c)*a^2-15*sin(4*d*x+4*c)*b^2-40*a*b*cos(3 
*d*x+3*c)-15*b^2*sin(2*d*x+2*c)-256*a*b)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.63 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {96 \, a b \cos \left (d x + c\right )^{5} - 160 \, a b \cos \left (d x + c\right )^{3} + 15 \, {\left (2 \, a^{2} + b^{2}\right )} d x + 5 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:

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

Output:

1/240*(96*a*b*cos(d*x + c)^5 - 160*a*b*cos(d*x + c)^3 + 15*(2*a^2 + b^2)*d 
*x + 5*(8*b^2*cos(d*x + c)^5 - 2*(6*a^2 + 7*b^2)*cos(d*x + c)^3 + 3*(2*a^2 
 + b^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. 309 vs. \(2 (148) = 296\).

Time = 0.36 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.90 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a b \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a b \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a**2*x*sin(c + d*x)**4/8 + a**2*x*sin(c + d*x)**2*cos(c + d*x)* 
*2/4 + a**2*x*cos(c + d*x)**4/8 + a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) 
- a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a*b*sin(c + d*x)**2*cos(c + 
d*x)**3/(3*d) - 4*a*b*cos(c + d*x)**5/(15*d) + b**2*x*sin(c + d*x)**6/16 + 
 3*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*b**2*x*sin(c + d*x)**2*co 
s(c + d*x)**4/16 + b**2*x*cos(c + d*x)**6/16 + b**2*sin(c + d*x)**5*cos(c 
+ d*x)/(16*d) - b**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - b**2*sin(c + 
d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**2*cos 
(c)**2, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.56 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {30 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 128 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{960 \, d} \] Input:

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

Output:

1/960*(30*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2 + 128*(3*cos(d*x + c)^5 - 5 
*cos(d*x + c)^3)*a*b - 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d 
*x + 4*c))*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{16} \, {\left (2 \, a^{2} + b^{2}\right )} x + \frac {a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac {a b \cos \left (d x + c\right )}{4 \, d} + \frac {b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \] Input:

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

Output:

1/16*(2*a^2 + b^2)*x + 1/40*a*b*cos(5*d*x + 5*c)/d - 1/24*a*b*cos(3*d*x + 
3*c)/d - 1/4*a*b*cos(d*x + c)/d + 1/192*b^2*sin(6*d*x + 6*c)/d - 1/64*b^2* 
sin(2*d*x + 2*c)/d - 1/64*(2*a^2 + b^2)*sin(4*d*x + 4*c)/d
 

Mupad [B] (verification not implemented)

Time = 33.96 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.69 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\frac {15\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {15\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {15\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}-\frac {5\,b^2\,\sin \left (6\,c+6\,d\,x\right )}{4}+60\,a\,b\,\cos \left (c+d\,x\right )+10\,a\,b\,\cos \left (3\,c+3\,d\,x\right )-6\,a\,b\,\cos \left (5\,c+5\,d\,x\right )-30\,a^2\,d\,x-15\,b^2\,d\,x}{240\,d} \] Input:

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

Output:

-((15*a^2*sin(4*c + 4*d*x))/2 + (15*b^2*sin(2*c + 2*d*x))/4 + (15*b^2*sin( 
4*c + 4*d*x))/4 - (5*b^2*sin(6*c + 6*d*x))/4 + 60*a*b*cos(c + d*x) + 10*a* 
b*cos(3*c + 3*d*x) - 6*a*b*cos(5*c + 5*d*x) - 30*a^2*d*x - 15*b^2*d*x)/(24 
0*d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.99 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{2}+96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a b +60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2}-10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b -30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-64 \cos \left (d x +c \right ) a b +30 a^{2} d x +64 a b +15 b^{2} d x}{240 d} \] Input:

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

Output:

(40*cos(c + d*x)*sin(c + d*x)**5*b**2 + 96*cos(c + d*x)*sin(c + d*x)**4*a* 
b + 60*cos(c + d*x)*sin(c + d*x)**3*a**2 - 10*cos(c + d*x)*sin(c + d*x)**3 
*b**2 - 32*cos(c + d*x)*sin(c + d*x)**2*a*b - 30*cos(c + d*x)*sin(c + d*x) 
*a**2 - 15*cos(c + d*x)*sin(c + d*x)*b**2 - 64*cos(c + d*x)*a*b + 30*a**2* 
d*x + 64*a*b + 15*b**2*d*x)/(240*d)