3.14 \(\int \frac{1}{\sin ^{\frac{3}{2}}(b x)} \, dx\)

Optimal. Leaf size=37 \[ \frac{2 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{b}-\frac{2 \cos (b x)}{b \sqrt{\sin (b x)}} \]

[Out]

(2*EllipticE[Pi/4 - (b*x)/2, 2])/b - (2*Cos[b*x])/(b*Sqrt[Sin[b*x]])

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Rubi [A]  time = 0.0131666, antiderivative size = 37, 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 = {2636, 2639} \[ \frac{2 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{b}-\frac{2 \cos (b x)}{b \sqrt{\sin (b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*EllipticE[Pi/4 - (b*x)/2, 2])/b - (2*Cos[b*x])/(b*Sqrt[Sin[b*x]])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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 \frac{1}{\sin ^{\frac{3}{2}}(b x)} \, dx &=-\frac{2 \cos (b x)}{b \sqrt{\sin (b x)}}-\int \sqrt{\sin (b x)} \, dx\\ &=\frac{2 E\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{b}-\frac{2 \cos (b x)}{b \sqrt{\sin (b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0533583, size = 32, normalized size = 0.86 \[ \frac{2 \left (E\left (\left .\frac{1}{4} (\pi -2 b x)\right |2\right )-\frac{\cos (b x)}{\sqrt{\sin (b x)}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(EllipticE[(Pi - 2*b*x)/4, 2] - Cos[b*x]/Sqrt[Sin[b*x]]))/b

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Maple [A]  time = 0.039, size = 110, normalized size = 3. \begin{align*}{\frac{1}{b\cos \left ( bx \right ) } \left ( 2\,\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},1/2\,\sqrt{2} \right ) -\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 ) -2\, \left ( \cos \left ( bx \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{\sin \left ( bx \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(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))-(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))-2*cos(b*x)^2)/c
os(b*x)/sin(b*x)^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sin ^{\frac{3}{2}}{\left (b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(b*x)**(3/2),x)

[Out]

Integral(sin(b*x)**(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x)^(-3/2), x)