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

Optimal. Leaf size=47 \[ \frac{2 F\left (\left .\frac{1}{2} \left (a+b x-\frac{\pi }{2}\right )\right |2\right )}{3 b}-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0163912, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2636, 2641} \[ \frac{2 F\left (\left .\frac{1}{2} \left (a+b x-\frac{\pi }{2}\right )\right |2\right )}{3 b}-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^(-5/2),x]

[Out]

(2*EllipticF[(a - Pi/2 + b*x)/2, 2])/(3*b) - (2*Cos[a + b*x])/(3*b*Sin[a + 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}}(a+b x)} \, dx &=-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(a+b x)}+\frac{1}{3} \int \frac{1}{\sqrt{\sin (a+b x)}} \, dx\\ &=\frac{2 F\left (\left .\frac{1}{2} \left (a-\frac{\pi }{2}+b x\right )\right |2\right )}{3 b}-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.103562, size = 43, normalized size = 0.91 \[ \frac{2 \left (F\left (\left .\frac{1}{4} (2 a+2 b x-\pi )\right |2\right )-\frac{\cos (a+b x)}{\sin ^{\frac{3}{2}}(a+b x)}\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^(-5/2),x]

[Out]

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

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Maple [A]  time = 0.026, size = 88, normalized size = 1.9 \begin{align*}{\frac{1}{3\,b\cos \left ( bx+a \right ) } \left ( \sqrt{\sin \left ( bx+a \right ) +1}\sqrt{-2\,\sin \left ( bx+a \right ) +2}\sqrt{-\sin \left ( bx+a \right ) }{\it EllipticF} \left ( \sqrt{\sin \left ( bx+a \right ) +1},{\frac{\sqrt{2}}{2}} \right ) \sin \left ( bx+a \right ) -2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \left ( \sin \left ( bx+a \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/sin(b*x+a)^(3/2)*((sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticF((sin(b*x+a)+1
)^(1/2),1/2*2^(1/2))*sin(b*x+a)-2*cos(b*x+a)^2)/cos(b*x+a)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(-5/2), x)