3.53 \(\int \frac{1}{(a \csc ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ -\frac{8 \cot (x)}{15 a^2 \sqrt{a \csc ^2(x)}}-\frac{4 \cot (x)}{15 a \left (a \csc ^2(x)\right )^{3/2}}-\frac{\cot (x)}{5 \left (a \csc ^2(x)\right )^{5/2}} \]

[Out]

-Cot[x]/(5*(a*Csc[x]^2)^(5/2)) - (4*Cot[x])/(15*a*(a*Csc[x]^2)^(3/2)) - (8*Cot[x])/(15*a^2*Sqrt[a*Csc[x]^2])

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Rubi [A]  time = 0.026984, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4122, 192, 191} \[ -\frac{8 \cot (x)}{15 a^2 \sqrt{a \csc ^2(x)}}-\frac{4 \cot (x)}{15 a \left (a \csc ^2(x)\right )^{3/2}}-\frac{\cot (x)}{5 \left (a \csc ^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Csc[x]^2)^(-5/2),x]

[Out]

-Cot[x]/(5*(a*Csc[x]^2)^(5/2)) - (4*Cot[x])/(15*a*(a*Csc[x]^2)^(3/2)) - (8*Cot[x])/(15*a^2*Sqrt[a*Csc[x]^2])

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a \csc ^2(x)\right )^{5/2}} \, dx &=-\left (a \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\cot (x)\right )\right )\\ &=-\frac{\cot (x)}{5 \left (a \csc ^2(x)\right )^{5/2}}-\frac{4}{5} \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\cot (x)\right )\\ &=-\frac{\cot (x)}{5 \left (a \csc ^2(x)\right )^{5/2}}-\frac{4 \cot (x)}{15 a \left (a \csc ^2(x)\right )^{3/2}}-\frac{8 \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\cot (x)\right )}{15 a}\\ &=-\frac{\cot (x)}{5 \left (a \csc ^2(x)\right )^{5/2}}-\frac{4 \cot (x)}{15 a \left (a \csc ^2(x)\right )^{3/2}}-\frac{8 \cot (x)}{15 a^2 \sqrt{a \csc ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0342015, size = 36, normalized size = 0.65 \[ -\frac{\sin (x) (150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \sqrt{a \csc ^2(x)}}{240 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Csc[x]^2)^(-5/2),x]

[Out]

-((150*Cos[x] - 25*Cos[3*x] + 3*Cos[5*x])*Sqrt[a*Csc[x]^2]*Sin[x])/(240*a^3)

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Maple [A]  time = 0.081, size = 39, normalized size = 0.7 \begin{align*}{\frac{\sqrt{4}\sin \left ( x \right ) \left ( 3\, \left ( \cos \left ( x \right ) \right ) ^{2}-9\,\cos \left ( x \right ) +8 \right ) }{30\, \left ( -1+\cos \left ( x \right ) \right ) ^{3}} \left ( -{\frac{a}{ \left ( \cos \left ( x \right ) \right ) ^{2}-1}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*csc(x)^2)^(5/2),x)

[Out]

1/30*4^(1/2)*sin(x)*(3*cos(x)^2-9*cos(x)+8)/(-1+cos(x))^3/(-a/(cos(x)^2-1))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*csc(x)^2)^(-5/2), x)

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Fricas [A]  time = 0.480975, size = 109, normalized size = 1.98 \begin{align*} -\frac{{\left (3 \, \cos \left (x\right )^{5} - 10 \, \cos \left (x\right )^{3} + 15 \, \cos \left (x\right )\right )} \sqrt{-\frac{a}{\cos \left (x\right )^{2} - 1}} \sin \left (x\right )}{15 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*cos(x)^5 - 10*cos(x)^3 + 15*cos(x))*sqrt(-a/(cos(x)^2 - 1))*sin(x)/a^3

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Sympy [A]  time = 32.7597, size = 61, normalized size = 1.11 \begin{align*} - \frac{8 \cot ^{5}{\left (x \right )}}{15 a^{\frac{5}{2}} \left (\csc ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} - \frac{4 \cot ^{3}{\left (x \right )}}{3 a^{\frac{5}{2}} \left (\csc ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} - \frac{\cot{\left (x \right )}}{a^{\frac{5}{2}} \left (\csc ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)**2)**(5/2),x)

[Out]

-8*cot(x)**5/(15*a**(5/2)*(csc(x)**2)**(5/2)) - 4*cot(x)**3/(3*a**(5/2)*(csc(x)**2)**(5/2)) - cot(x)/(a**(5/2)
*(csc(x)**2)**(5/2))

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Giac [A]  time = 1.44929, size = 76, normalized size = 1.38 \begin{align*} -\frac{16 \,{\left (\frac{10 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )\right ) \tan \left (\frac{1}{2} \, x\right )^{4} + 5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )\right ) \tan \left (\frac{1}{2} \, x\right )^{2} + \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )\right )}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{5}} - \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )\right )\right )}}{15 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*((10*sgn(tan(1/2*x))*tan(1/2*x)^4 + 5*sgn(tan(1/2*x))*tan(1/2*x)^2 + sgn(tan(1/2*x)))/(tan(1/2*x)^2 + 1
)^5 - sgn(tan(1/2*x)))/a^(5/2)