3.10 \(\int \sin ^{\frac{5}{2}}(b x) \, dx\)

Optimal. Leaf size=41 \[ -\frac{6 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{5 b}-\frac{2 \sin ^{\frac{3}{2}}(b x) \cos (b x)}{5 b} \]

[Out]

(-6*EllipticE[Pi/4 - (b*x)/2, 2])/(5*b) - (2*Cos[b*x]*Sin[b*x]^(3/2))/(5*b)

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Rubi [A]  time = 0.015399, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 2639} \[ -\frac{6 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{5 b}-\frac{2 \sin ^{\frac{3}{2}}(b x) \cos (b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[b*x]^(5/2),x]

[Out]

(-6*EllipticE[Pi/4 - (b*x)/2, 2])/(5*b) - (2*Cos[b*x]*Sin[b*x]^(3/2))/(5*b)

Rule 2635

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \sin ^{\frac{5}{2}}(b x) \, dx &=-\frac{2 \cos (b x) \sin ^{\frac{3}{2}}(b x)}{5 b}+\frac{3}{5} \int \sqrt{\sin (b x)} \, dx\\ &=-\frac{6 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{5 b}-\frac{2 \cos (b x) \sin ^{\frac{3}{2}}(b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0384482, size = 35, normalized size = 0.85 \[ -\frac{2 \left (3 E\left (\left .\frac{1}{4} (\pi -2 b x)\right |2\right )+\sin ^{\frac{3}{2}}(b x) \cos (b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[b*x]^(5/2),x]

[Out]

(-2*(3*EllipticE[(Pi - 2*b*x)/4, 2] + Cos[b*x]*Sin[b*x]^(3/2)))/(5*b)

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Maple [B]  time = 0.039, size = 118, normalized size = 2.9 \begin{align*}{\frac{1}{b\cos \left ( bx \right ) } \left ({\frac{2\, \left ( \sin \left ( bx \right ) \right ) ^{4}}{5}}-{\frac{2\, \left ( \sin \left ( bx \right ) \right ) ^{2}}{5}}-{\frac{6}{5}\sqrt{\sin \left ( bx \right ) +1}\sqrt{-2\,\sin \left ( bx \right ) +2}\sqrt{-\sin \left ( bx \right ) }{\it EllipticE} \left ( \sqrt{\sin \left ( bx \right ) +1},{\frac{\sqrt{2}}{2}} \right ) }+{\frac{3}{5}\sqrt{\sin \left ( bx \right ) +1}\sqrt{-2\,\sin \left ( bx \right ) +2}\sqrt{-\sin \left ( bx \right ) }{\it EllipticF} \left ( \sqrt{\sin \left ( bx \right ) +1},{\frac{\sqrt{2}}{2}} \right ) } \right ){\frac{1}{\sqrt{\sin \left ( bx \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x)^(5/2),x)

[Out]

(2/5*sin(b*x)^4-2/5*sin(b*x)^2-6/5*(sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticE((sin(b
*x)+1)^(1/2),1/2*2^(1/2))+3/5*(sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticF((sin(b*x)+1
)^(1/2),1/2*2^(1/2)))/cos(b*x)/sin(b*x)^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (b x\right )^{2} - 1\right )} \sqrt{\sin \left (b x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x)^(5/2),x, algorithm="fricas")

[Out]

integral(-(cos(b*x)^2 - 1)*sqrt(sin(b*x)), x)

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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(sin(b*x)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x)^(5/2), x)