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

Optimal. Leaf size=73 \[ \frac{3}{5} \left (1-\sin \left (\frac{2 x}{3}\right )\right )^{3/2} \cos \left (\frac{2 x}{3}\right )+\frac{8}{5} \sqrt{1-\sin \left (\frac{2 x}{3}\right )} \cos \left (\frac{2 x}{3}\right )+\frac{32 \cos \left (\frac{2 x}{3}\right )}{5 \sqrt{1-\sin \left (\frac{2 x}{3}\right )}} \]

[Out]

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

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Rubi [A]  time = 0.0325273, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2647, 2646} \[ \frac{3}{5} \left (1-\sin \left (\frac{2 x}{3}\right )\right )^{3/2} \cos \left (\frac{2 x}{3}\right )+\frac{8}{5} \sqrt{1-\sin \left (\frac{2 x}{3}\right )} \cos \left (\frac{2 x}{3}\right )+\frac{32 \cos \left (\frac{2 x}{3}\right )}{5 \sqrt{1-\sin \left (\frac{2 x}{3}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sin[(2*x)/3])^(5/2),x]

[Out]

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

Rule 2647

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] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \left (1-\sin \left (\frac{2 x}{3}\right )\right )^{5/2} \, dx &=\frac{3}{5} \cos \left (\frac{2 x}{3}\right ) \left (1-\sin \left (\frac{2 x}{3}\right )\right )^{3/2}+\frac{8}{5} \int \left (1-\sin \left (\frac{2 x}{3}\right )\right )^{3/2} \, dx\\ &=\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}+\frac{32}{15} \int \sqrt{1-\sin \left (\frac{2 x}{3}\right )} \, 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}\\ \end{align*}

Mathematica [A]  time = 0.166133, size = 76, normalized size = 1.04 \[ \frac{\left (1-\sin \left (\frac{2 x}{3}\right )\right )^{5/2} \left (150 \sin \left (\frac{x}{3}\right )-25 \sin (x)-3 \sin \left (\frac{5 x}{3}\right )+150 \cos \left (\frac{x}{3}\right )+25 \cos (x)-3 \cos \left (\frac{5 x}{3}\right )\right )}{20 \left (\cos \left (\frac{x}{3}\right )-\sin \left (\frac{x}{3}\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sin[(2*x)/3])^(5/2),x]

[Out]

((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)

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Maple [A]  time = 0.037, size = 47, normalized size = 0.6 \begin{align*} -{\frac{1}{5} \left ( -1+\sin \left ({\frac{2\,x}{3}} \right ) \right ) \left ( 1+\sin \left ({\frac{2\,x}{3}} \right ) \right ) \left ( 3\, \left ( \sin \left ( 2/3\,x \right ) \right ) ^{2}-14\,\sin \left ( 2/3\,x \right ) +43 \right ) \left ( \cos \left ({\frac{2\,x}{3}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{1-\sin \left ({\frac{2\,x}{3}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-\sin \left (\frac{2}{3} \, x\right ) + 1\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-sin(2/3*x) + 1)^(5/2), x)

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Fricas [A]  time = 1.88725, size = 223, normalized size = 3.05 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out