\(\int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [32]

Optimal result

Integrand size = 23, antiderivative size = 158 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {32 a \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {64 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d} \] Output:

-32/45*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-16/63*a*cos(d*x+c)*sin(d*x+c) 
^3/d/(a+a*sin(d*x+c))^(1/2)-2/9*a*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c 
))^(1/2)+64/315*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-32/105*cos(d*x+c)*(a+a 
*sin(d*x+c))^(3/2)/a/d
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left (-1890 \cos \left (\frac {1}{2} (c+d x)\right )-420 \cos \left (\frac {3}{2} (c+d x)\right )+252 \cos \left (\frac {5}{2} (c+d x)\right )+45 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )+1890 \sin \left (\frac {1}{2} (c+d x)\right )-420 \sin \left (\frac {3}{2} (c+d x)\right )-252 \sin \left (\frac {5}{2} (c+d x)\right )+45 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(Sqrt[a*(1 + Sin[c + d*x])]*(-1890*Cos[(c + d*x)/2] - 420*Cos[(3*(c + d*x) 
)/2] + 252*Cos[(5*(c + d*x))/2] + 45*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + 
 d*x))/2] + 1890*Sin[(c + d*x)/2] - 420*Sin[(3*(c + d*x))/2] - 252*Sin[(5* 
(c + d*x))/2] + 45*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520* 
d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3249, 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.

\(\displaystyle \int \sin ^4(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 \sqrt {a \sin (c+d x)+a}dx\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {8}{9} \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{9} \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3249

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3238

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {8}{9} \left (\frac {6}{7} \left (\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\)

Input:

Int[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(-2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (8*((- 
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
 

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]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right ) \left (35 \sin \left (d x +c \right )^{4}+40 \sin \left (d x +c \right )^{3}+48 \sin \left (d x +c \right )^{2}+64 \sin \left (d x +c \right )+128\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(83\)

Input:

int(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/315*(1+sin(d*x+c))*a*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+40*sin(d*x+c)^3+48* 
sin(d*x+c)^2+64*sin(d*x+c)+128)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} - 118 \, \cos \left (d x + c\right )^{3} + 26 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} - 78 \, \cos \left (d x + c\right )^{2} - 104 \, \cos \left (d x + c\right ) + 107\right )} \sin \left (d x + c\right ) + 211 \, \cos \left (d x + c\right ) + 107\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-2/315*(35*cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 118*cos(d*x + c)^3 + 26*cos 
(d*x + c)^2 - (35*cos(d*x + c)^4 + 40*cos(d*x + c)^3 - 78*cos(d*x + c)^2 - 
 104*cos(d*x + c) + 107)*sin(d*x + c) + 211*cos(d*x + c) + 107)*sqrt(a*sin 
(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
 

Sympy [F]

\[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate(sin(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**4, x)
 

Maxima [F]

\[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{4} \,d x } \] Input:

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(a*sin(d*x + c) + a)*sin(d*x + c)^4, x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.93 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (1890 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 420 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 252 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 45 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \] Input:

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

1/2520*sqrt(2)*(1890*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2 
*d*x + 1/2*c) + 420*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2* 
d*x + 3/2*c) + 252*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d 
*x + 5/2*c) + 45*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x 
 + 7/2*c) + 35*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 
 9/2*c))*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\sin \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2),x)
 

Output:

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{4}d x \right ) \] Input:

int(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**4,x)