3.15 \(\int \frac{1}{\sin ^{\frac{5}{2}}(b x)} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0149834, 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 = {2636, 2641} \[ -\frac{2 F\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{3 b}-\frac{2 \cos (b x)}{3 b \sin ^{\frac{3}{2}}(b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2641

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

Rubi steps

\begin{align*} \int \frac{1}{\sin ^{\frac{5}{2}}(b x)} \, dx &=-\frac{2 \cos (b x)}{3 b \sin ^{\frac{3}{2}}(b x)}+\frac{1}{3} \int \frac{1}{\sqrt{\sin (b x)}} \, dx\\ &=-\frac{2 F\left (\left .\frac{\pi }{4}-\frac{b x}{2}\right |2\right )}{3 b}-\frac{2 \cos (b x)}{3 b \sin ^{\frac{3}{2}}(b x)}\\ \end{align*}

Mathematica [A]  time = 0.0493103, size = 33, normalized size = 0.8 \[ -\frac{2 \left (F\left (\left .\frac{1}{4} (\pi -2 b x)\right |2\right )+\frac{\cos (b x)}{\sin ^{\frac{3}{2}}(b x)}\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 72, normalized size = 1.8 \begin{align*}{\frac{1}{3\,b\cos \left ( bx \right ) } \left ( \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 ) \sin \left ( bx \right ) -2\, \left ( \cos \left ( bx \right ) \right ) ^{2} \right ) \left ( \sin \left ( bx \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/sin(b*x)^(3/2)*((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))*sin(b*x)-2*cos(b*x)^2)/cos(b*x)/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x)^(-5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{1}{{\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(1/sin(b*x)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sin ^{\frac{5}{2}}{\left (b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(b*x)**(5/2),x)

[Out]

Integral(sin(b*x)**(-5/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x)^(-5/2), x)