\(\int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [322]

Optimal result

Integrand size = 31, antiderivative size = 193 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {76 a \cos (c+d x)}{495 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d} \] Output:

-76/495*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-38/693*a*cos(d*x+c)*sin(d*x+ 
c)^3/d/(a+a*sin(d*x+c))^(1/2)+2/99*a*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d* 
x+c))^(1/2)+152/3465*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d+2/11*cos(d*x+c)*s 
in(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/d-76/1155*cos(d*x+c)*(a+a*sin(d*x+c))^( 
3/2)/a/d
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.56 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (5657-3540 \cos (2 (c+d x))+315 \cos (4 (c+d x))+7638 \sin (c+d x)-1330 \sin (3 (c+d x)))}{13860 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

-1/13860*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x] 
)]*(5657 - 3540*Cos[2*(c + d*x)] + 315*Cos[4*(c + d*x)] + 7638*Sin[c + d*x 
] - 1330*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.20, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 3358, 3042, 3455, 27, 3042, 3460, 3042, 3249, 3042, 3238, 27, 3042, 3230, 3042, 3125}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3358

\(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin (c+d x)^3 (a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{a^2}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {2}{11} \int \frac {1}{2} \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )dx+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )dx+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a} \left (3 a^2-a^2 \sin (c+d x)\right )dx+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a}dx+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3238

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{11} \left (\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}}{a^2}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}+\frac {1}{11} \left (\frac {2 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}+\frac {19}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )\right )}{a^2}\)

Input:

Int[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

((2*a^2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) + ((2 
*a^3*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (19*a^2 
*((-2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) + (6*( 
(-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d) + ((-14*a^2*Cos[c + d 
*x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + 
 d*x]])/(3*d))/(5*a)))/7))/9)/11)/a^2
 

Defintions of rubi rules used

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 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

rule 3230

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

rule 3238

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), 
x_Symbol] :> Simp[(-Cos[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*(b*(m + 1) - a*Si 
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  ! 
LtQ[m, -2^(-1)]
 

rule 3249

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

rule 3358

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

rule 3455

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

rule 3460

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.44

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{2} \left (315 \sin \left (d x +c \right )^{4}+665 \sin \left (d x +c \right )^{3}+570 \sin \left (d x +c \right )^{2}+456 \sin \left (d x +c \right )+304\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(85\)

Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBO 
SE)
 

Output:

-2/3465*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+665*sin(d*x+c) 
^3+570*sin(d*x+c)^2+456*sin(d*x+c)+304)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/ 
d
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} + 350 \, \cos \left (d x + c\right )^{5} - 500 \, \cos \left (d x + c\right )^{4} - 586 \, \cos \left (d x + c\right )^{3} + 17 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{4} - 535 \, \cos \left (d x + c\right )^{3} + 51 \, \cos \left (d x + c\right )^{2} + 68 \, \cos \left (d x + c\right ) + 136\right )} \sin \left (d x + c\right ) - 68 \, \cos \left (d x + c\right ) - 136\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="f 
ricas")
 

Output:

2/3465*(315*cos(d*x + c)^6 + 350*cos(d*x + c)^5 - 500*cos(d*x + c)^4 - 586 
*cos(d*x + c)^3 + 17*cos(d*x + c)^2 + (315*cos(d*x + c)^5 - 35*cos(d*x + c 
)^4 - 535*cos(d*x + c)^3 + 51*cos(d*x + c)^2 + 68*cos(d*x + c) + 136)*sin( 
d*x + c) - 68*cos(d*x + c) - 136)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) 
 + d*sin(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)**3*(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {8 \, \sqrt {2} {\left (2520 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 7700 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 8910 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4851 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1155 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{3465 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="g 
iac")
 

Output:

8/3465*sqrt(2)*(2520*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2 
*d*x + 1/2*c)^11 - 7700*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 
1/2*d*x + 1/2*c)^9 + 8910*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi 
+ 1/2*d*x + 1/2*c)^7 - 4851*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p 
i + 1/2*d*x + 1/2*c)^5 + 1155*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4 
*pi + 1/2*d*x + 1/2*c)^3)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:

int(cos(c + d*x)^2*sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2),x)
 

Output:

int(cos(c + d*x)^2*sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3}d x \right ) \] Input:

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**3,x)