\(\int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1055]

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 = 29, antiderivative size = 190 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b x}{8}-\frac {\left (7 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {\left (7 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{12 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a b \cos (c+d x) \sin ^5(c+d x)}{21 d}+\frac {\cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{7 d} \] Output:

1/8*a*b*x-1/35*(7*a^2+4*b^2)*cos(d*x+c)/d+1/105*(7*a^2+4*b^2)*cos(d*x+c)^3 
/d-1/8*a*b*cos(d*x+c)*sin(d*x+c)/d-1/12*a*b*cos(d*x+c)*sin(d*x+c)^3/d+1/35 
*(2*a^2-b^2)*cos(d*x+c)*sin(d*x+c)^4/d+1/21*a*b*cos(d*x+c)*sin(d*x+c)^5/d+ 
1/7*cos(d*x+c)*sin(d*x+c)^4*(a+b*sin(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.69 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {840 a b c+840 a b d x-105 \left (8 a^2+5 b^2\right ) \cos (c+d x)-35 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+84 a^2 \cos (5 (c+d x))+63 b^2 \cos (5 (c+d x))-15 b^2 \cos (7 (c+d x))-210 a b \sin (2 (c+d x))-210 a b \sin (4 (c+d x))+70 a b \sin (6 (c+d x))}{6720 d} \] Input:

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

Output:

(840*a*b*c + 840*a*b*d*x - 105*(8*a^2 + 5*b^2)*Cos[c + d*x] - 35*(4*a^2 + 
b^2)*Cos[3*(c + d*x)] + 84*a^2*Cos[5*(c + d*x)] + 63*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 = 1.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.621, Rules used = {3042, 3368, 3042, 3529, 3042, 3512, 27, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \int 2 \sin ^3(c+d x) \left (9 a^2+7 b \sin (c+d x) a-3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right )dx+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \sin ^3(c+d x) \left (9 a^2+7 b \sin (c+d x) a-3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right )dx+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \sin (c+d x)^3 \left (9 a^2+7 b \sin (c+d x) a-3 \left (2 a^2-b^2\right ) \sin (c+d x)^2\right )dx+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \sin ^3(c+d x) \left (3 \left (7 a^2+4 b^2\right )+35 a b \sin (c+d x)\right )dx+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \sin (c+d x)^3 \left (3 \left (7 a^2+4 b^2\right )+35 a b \sin (c+d x)\right )dx+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (3 \left (7 a^2+4 b^2\right ) \int \sin ^3(c+d x)dx+35 a b \int \sin ^4(c+d x)dx\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (3 \left (7 a^2+4 b^2\right ) \int \sin (c+d x)^3dx+35 a b \int \sin (c+d x)^4dx\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \int \sin (c+d x)^4dx-\frac {3 \left (7 a^2+4 b^2\right ) \int \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \int \sin (c+d x)^4dx-\frac {3 \left (7 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {3 \left (7 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {3 \left (7 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {3 \left (7 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (35 a b \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {3 \left (7 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}\right )+\frac {\sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}\)

Input:

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

Output:

(Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2)/(7*d) + ((a*b*Cos[c + 
 d*x]*Sin[c + d*x]^5)/(3*d) + ((3*(2*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^ 
4)/(5*d) + ((-3*(7*a^2 + 4*b^2)*(Cos[c + d*x] - Cos[c + d*x]^3/3))/d + 35* 
a*b*(-1/4*(Cos[c + d*x]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c 
+ d*x])/(2*d)))/4))/5)/3)/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 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 = 85.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.72

method result size
parallelrisch \(\frac {\left (-140 a^{2}-35 b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (84 a^{2}+63 b^{2}\right ) \cos \left (5 d x +5 c \right )-15 b^{2} \cos \left (7 d x +7 c \right )-210 a b \sin \left (2 d x +2 c \right )-210 a b \sin \left (4 d x +4 c \right )+70 a b \sin \left (6 d x +6 c \right )+\left (-840 a^{2}-525 b^{2}\right ) \cos \left (d x +c \right )+840 a b x d -896 a^{2}-512 b^{2}}{6720 d}\) \(136\)
derivativedivides \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{4}}{7}-\frac {4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{35}-\frac {8 \cos \left (d x +c \right )^{3}}{105}\right )}{d}\) \(150\)
default \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{6}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{8}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{4}}{7}-\frac {4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{35}-\frac {8 \cos \left (d x +c \right )^{3}}{105}\right )}{d}\) \(150\)
risch \(\frac {a b x}{8}-\frac {a^{2} \cos \left (d x +c \right )}{8 d}-\frac {5 b^{2} \cos \left (d x +c \right )}{64 d}-\frac {b^{2} \cos \left (7 d x +7 c \right )}{448 d}+\frac {a b \sin \left (6 d x +6 c \right )}{96 d}+\frac {\cos \left (5 d x +5 c \right ) a^{2}}{80 d}+\frac {3 \cos \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{48 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{192 d}-\frac {a b \sin \left (2 d x +2 c \right )}{32 d}\) \(168\)
norman \(\frac {-\frac {28 a^{2}+16 b^{2}}{105 d}+\frac {a b x}{8}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {\left (8 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {\left (8 a^{2}+16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d}-\frac {\left (20 a^{2}+32 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}-\frac {\left (28 a^{2}+16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d}-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {5 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {97 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {97 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {5 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {7 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {21 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {35 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}+\frac {35 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {21 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {7 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) \(384\)
orering \(\text {Expression too large to display}\) \(3660\)

Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/6720*((-140*a^2-35*b^2)*cos(3*d*x+3*c)+(84*a^2+63*b^2)*cos(5*d*x+5*c)-15 
*b^2*cos(7*d*x+7*c)-210*a*b*sin(2*d*x+2*c)-210*a*b*sin(4*d*x+4*c)+70*a*b*s 
in(6*d*x+6*c)+(-840*a^2-525*b^2)*cos(d*x+c)+840*a*b*x*d-896*a^2-512*b^2)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.55 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 105 \, a b d x + 280 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} - 14 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \] Input:

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

Output:

-1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + 2*b^2)*cos(d*x + c)^5 - 105*a* 
b*d*x + 280*(a^2 + b^2)*cos(d*x + c)^3 - 35*(8*a*b*cos(d*x + c)^5 - 14*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 = 0.47 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.45 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a b x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {8 b^{2} \cos ^{7}{\left (c + d x \right )}}{105 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*a**2*cos(c + d* 
x)**5/(15*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)**4*cos(c + d*x)**3/(3*d) - 4*b**2*sin(c + d*x)**2*cos(c + d*x)**5/(15*d 
) - 8*b**2*cos(c + d*x)**7/(105*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c) 
**3*cos(c)**2, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.55 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {224 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 32 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{3360 \, d} \] Input:

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

Output:

1/3360*(224*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2 - 35*(4*sin(2*d*x + 
2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*a*b - 32*(15*cos(d*x + c)^7 - 
 42*cos(d*x + c)^5 + 35*cos(d*x + c)^3)*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {1}{8} \, a b x - \frac {b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac {{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (8 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(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 + 
3*b^2)*cos(5*d*x + 5*c)/d - 1/192*(4*a^2 + b^2)*cos(3*d*x + 3*c)/d - 1/64* 
(8*a^2 + 5*b^2)*cos(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 38.04 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.23 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a\,b\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {8\,a^2}{3}-\frac {16\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {8\,a^2}{5}+\frac {16\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {20\,a^2}{3}+\frac {32\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {28\,a^2}{15}+\frac {16\,b^2}{15}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {4\,a^2}{15}+\frac {16\,b^2}{105}+\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {97\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {97\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:

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

Output:

(a*b*x)/8 - (tan(c/2 + (d*x)/2)^6*((8*a^2)/3 - (16*b^2)/3) + tan(c/2 + (d* 
x)/2)^4*((8*a^2)/5 + (16*b^2)/5) + tan(c/2 + (d*x)/2)^8*((20*a^2)/3 + (32* 
b^2)/3) + tan(c/2 + (d*x)/2)^2*((28*a^2)/15 + (16*b^2)/15) + 4*a^2*tan(c/2 
 + (d*x)/2)^10 + (4*a^2)/15 + (16*b^2)/105 + (5*a*b*tan(c/2 + (d*x)/2)^3)/ 
3 - (97*a*b*tan(c/2 + (d*x)/2)^5)/12 + (97*a*b*tan(c/2 + (d*x)/2)^9)/12 - 
(5*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 = 0.17 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.01 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b^{2}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b +168 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}-70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b -56 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -112 \cos \left (d x +c \right ) a^{2}-64 \cos \left (d x +c \right ) b^{2}+112 a^{2}+105 a b d x +64 b^{2}}{840 d} \] Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(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 - 24*cos(c + d*x)*sin(c + d*x) 
**4*b**2 - 70*cos(c + d*x)*sin(c + d*x)**3*a*b - 56*cos(c + d*x)*sin(c + d 
*x)**2*a**2 - 32*cos(c + d*x)*sin(c + d*x)**2*b**2 - 105*cos(c + d*x)*sin( 
c + d*x)*a*b - 112*cos(c + d*x)*a**2 - 64*cos(c + d*x)*b**2 + 112*a**2 + 1 
05*a*b*d*x + 64*b**2)/(840*d)