\(\int (1-\sin (\frac {2 x}{3}))^{5/2} \, dx\) [395]

Optimal result

Integrand size = 14, antiderivative size = 73 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\frac {32 \cos \left (\frac {2 x}{3}\right )}{5 \sqrt {1-\sin \left (\frac {2 x}{3}\right )}}+\frac {8}{5} \cos \left (\frac {2 x}{3}\right ) \sqrt {1-\sin \left (\frac {2 x}{3}\right )}+\frac {3}{5} \cos \left (\frac {2 x}{3}\right ) \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \] Output:

3/5*cos(2/3*x)*(1-sin(2/3*x))^(3/2)+32/5*cos(2/3*x)/(1-sin(2/3*x))^(1/2)+8 
/5*cos(2/3*x)*(1-sin(2/3*x))^(1/2)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\frac {\left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \left (150 \cos \left (\frac {x}{3}\right )+25 \cos (x)-3 \cos \left (\frac {5 x}{3}\right )+150 \sin \left (\frac {x}{3}\right )-25 \sin (x)-3 \sin \left (\frac {5 x}{3}\right )\right )}{20 \left (\cos \left (\frac {x}{3}\right )-\sin \left (\frac {x}{3}\right )\right )^5} \] Input:

Integrate[(1 - Sin[(2*x)/3])^(5/2),x]
 

Output:

((1 - Sin[(2*x)/3])^(5/2)*(150*Cos[x/3] + 25*Cos[x] - 3*Cos[(5*x)/3] + 150 
*Sin[x/3] - 25*Sin[x] - 3*Sin[(5*x)/3]))/(20*(Cos[x/3] - Sin[x/3])^5)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3126, 3042, 3126, 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.

\(\displaystyle \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2}dx\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {8}{5} \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2}dx+\frac {3}{5} \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \cos \left (\frac {2 x}{3}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{5} \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2}dx+\frac {3}{5} \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \cos \left (\frac {2 x}{3}\right )\)

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

\(\displaystyle \frac {3}{5} \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \cos \left (\frac {2 x}{3}\right )+\frac {8}{5} \left (\sqrt {1-\sin \left (\frac {2 x}{3}\right )} \cos \left (\frac {2 x}{3}\right )+\frac {4 \cos \left (\frac {2 x}{3}\right )}{\sqrt {1-\sin \left (\frac {2 x}{3}\right )}}\right )\)

Input:

Int[(1 - Sin[(2*x)/3])^(5/2),x]
 

Output:

(8*((4*Cos[(2*x)/3])/Sqrt[1 - Sin[(2*x)/3]] + Cos[(2*x)/3]*Sqrt[1 - Sin[(2 
*x)/3]]))/5 + (3*Cos[(2*x)/3]*(1 - Sin[(2*x)/3])^(3/2))/5
 

Defintions of rubi rules used

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 3126

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64

method	result	size
default	\(-\frac {\left (-1+\sin \left (\frac {2 x}{3}\right )\right ) \left (\sin \left (\frac {2 x}{3}\right )+1\right ) \left (3 \sin \left (\frac {2 x}{3}\right )^{2}-14 \sin \left (\frac {2 x}{3}\right )+43\right )}{5 \cos \left (\frac {2 x}{3}\right ) \sqrt {1-\sin \left (\frac {2 x}{3}\right )}}\)	\(47\)

Input:

int((1-sin(2/3*x))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/5*(-1+sin(2/3*x))*(sin(2/3*x)+1)*(3*sin(2/3*x)^2-14*sin(2/3*x)+43)/cos( 
2/3*x)/(1-sin(2/3*x))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=-\frac {{\left (3 \, \cos \left (\frac {2}{3} \, x\right )^{3} - 11 \, \cos \left (\frac {2}{3} \, x\right )^{2} + {\left (3 \, \cos \left (\frac {2}{3} \, x\right )^{2} + 14 \, \cos \left (\frac {2}{3} \, x\right ) - 32\right )} \sin \left (\frac {2}{3} \, x\right ) - 46 \, \cos \left (\frac {2}{3} \, x\right ) - 32\right )} \sqrt {-\sin \left (\frac {2}{3} \, x\right ) + 1}}{5 \, {\left (\cos \left (\frac {2}{3} \, x\right ) - \sin \left (\frac {2}{3} \, x\right ) + 1\right )}} \] Input:

integrate((1-sin(2/3*x))^(5/2),x, algorithm="fricas")
 

Output:

-1/5*(3*cos(2/3*x)^3 - 11*cos(2/3*x)^2 + (3*cos(2/3*x)^2 + 14*cos(2/3*x) - 
 32)*sin(2/3*x) - 46*cos(2/3*x) - 32)*sqrt(-sin(2/3*x) + 1)/(cos(2/3*x) - 
sin(2/3*x) + 1)
 

Sympy [F]

\[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int \left (1 - \sin {\left (\frac {2 x}{3} \right )}\right )^{\frac {5}{2}}\, dx \] Input:

integrate((1-sin(2/3*x))**(5/2),x)
 

Output:

Integral((1 - sin(2*x/3))**(5/2), x)
 

Maxima [F]

\[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int { {\left (-\sin \left (\frac {2}{3} \, x\right ) + 1\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((1-sin(2/3*x))^(5/2),x, algorithm="maxima")
 

Output:

integrate((-sin(2/3*x) + 1)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=-\frac {1}{20} \, \sqrt {2} {\left (150 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) - 25 \, \cos \left (-\frac {3}{4} \, \pi + x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) + 3 \, \cos \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) - 128 \, \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right )\right )} \] Input:

integrate((1-sin(2/3*x))^(5/2),x, algorithm="giac")
 

Output:

-1/20*sqrt(2)*(150*cos(-1/4*pi + 1/3*x)*sgn(sin(-1/4*pi + 1/3*x)) - 25*cos 
(-3/4*pi + x)*sgn(sin(-1/4*pi + 1/3*x)) + 3*cos(-5/4*pi + 5/3*x)*sgn(sin(- 
1/4*pi + 1/3*x)) - 128*sgn(sin(-1/4*pi + 1/3*x)))
 

Mupad [F(-1)]

Timed out. \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int {\left (1-\sin \left (\frac {2\,x}{3}\right )\right )}^{5/2} \,d x \] Input:

int((1 - sin((2*x)/3))^(5/2),x)
 

Output:

int((1 - sin((2*x)/3))^(5/2), x)
 

Reduce [F]

\[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int \sqrt {-\sin \left (\frac {2 x}{3}\right )+1}d x +\int \sqrt {-\sin \left (\frac {2 x}{3}\right )+1}\, \sin \left (\frac {2 x}{3}\right )^{2}d x -2 \left (\int \sqrt {-\sin \left (\frac {2 x}{3}\right )+1}\, \sin \left (\frac {2 x}{3}\right )d x \right ) \] Input:

int((1-sin(2/3*x))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int(sqrt( - sin((2*x)/3) + 1),x) + int(sqrt( - sin((2*x)/3) + 1)*sin((2*x) 
/3)**2,x) - 2*int(sqrt( - sin((2*x)/3) + 1)*sin((2*x)/3),x)