Integrand size = 14, antiderivative size = 73 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\frac {32 \cos \left (\frac {2 x}{3}\right )}{5 \sqrt {1-\sin \left (\frac {2 x}{3}\right )}}+\frac {8}{5} \cos \left (\frac {2 x}{3}\right ) \sqrt {1-\sin \left (\frac {2 x}{3}\right )}+\frac {3}{5} \cos \left (\frac {2 x}{3}\right ) \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \]
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)
Time = 1.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\frac {\left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \left (150 \cos \left (\frac {x}{3}\right )+25 \cos (x)-3 \cos \left (\frac {5 x}{3}\right )+150 \sin \left (\frac {x}{3}\right )-25 \sin (x)-3 \sin \left (\frac {5 x}{3}\right )\right )}{20 \left (\cos \left (\frac {x}{3}\right )-\sin \left (\frac {x}{3}\right )\right )^5} \]
((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)
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 |
\(\displaystyle \frac {3}{5} \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{3/2} \cos \left (\frac {2 x}{3}\right )+\frac {8}{5} \left (\sqrt {1-\sin \left (\frac {2 x}{3}\right )} \cos \left (\frac {2 x}{3}\right )+\frac {4 \cos \left (\frac {2 x}{3}\right )}{\sqrt {1-\sin \left (\frac {2 x}{3}\right )}}\right )\) |
(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
3.4.95.3.1 Defintions of rubi rules used
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]
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]
Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64
method | result | size |
default | \(-\frac {\left (-1+\sin \left (\frac {2 x}{3}\right )\right ) \left (\sin \left (\frac {2 x}{3}\right )+1\right ) \left (3 \left (\sin ^{2}\left (\frac {2 x}{3}\right )\right )-14 \sin \left (\frac {2 x}{3}\right )+43\right )}{5 \cos \left (\frac {2 x}{3}\right ) \sqrt {1-\sin \left (\frac {2 x}{3}\right )}}\) | \(47\) |
-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)
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=-\frac {{\left (3 \, \cos \left (\frac {2}{3} \, x\right )^{3} - 11 \, \cos \left (\frac {2}{3} \, x\right )^{2} + {\left (3 \, \cos \left (\frac {2}{3} \, x\right )^{2} + 14 \, \cos \left (\frac {2}{3} \, x\right ) - 32\right )} \sin \left (\frac {2}{3} \, x\right ) - 46 \, \cos \left (\frac {2}{3} \, x\right ) - 32\right )} \sqrt {-\sin \left (\frac {2}{3} \, x\right ) + 1}}{5 \, {\left (\cos \left (\frac {2}{3} \, x\right ) - \sin \left (\frac {2}{3} \, x\right ) + 1\right )}} \]
-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)
\[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int \left (1 - \sin {\left (\frac {2 x}{3} \right )}\right )^{\frac {5}{2}}\, dx \]
\[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int { {\left (-\sin \left (\frac {2}{3} \, x\right ) + 1\right )}^{\frac {5}{2}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=-\frac {1}{20} \, \sqrt {2} {\left (150 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) - 25 \, \cos \left (-\frac {3}{4} \, \pi + x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) + 3 \, \cos \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, x\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right ) - 128 \, \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, x\right )\right )\right )} \]
-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)))
Timed out. \[ \int \left (1-\sin \left (\frac {2 x}{3}\right )\right )^{5/2} \, dx=\int {\left (1-\sin \left (\frac {2\,x}{3}\right )\right )}^{5/2} \,d x \]