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

Optimal. Leaf size=43 \[ -\frac {8 \cot (x)}{15 \sqrt {\csc ^2(x)}}-\frac {4 \cot (x)}{15 \csc ^2(x)^{3/2}}-\frac {\cot (x)}{5 \csc ^2(x)^{5/2}} \]

[Out]

-1/5*cot(x)/(csc(x)^2)^(5/2)-4/15*cot(x)/(csc(x)^2)^(3/2)-8/15*cot(x)/(csc(x)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4122, 192, 191} \[ -\frac {8 \cot (x)}{15 \sqrt {\csc ^2(x)}}-\frac {4 \cot (x)}{15 \csc ^2(x)^{3/2}}-\frac {\cot (x)}{5 \csc ^2(x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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
]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.72 \[ -\frac {(150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \csc (x)}{240 \sqrt {\csc ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*((150*Cos[x] - 25*Cos[3*x] + 3*Cos[5*x])*Csc[x])/Sqrt[Csc[x]^2]

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fricas [A]  time = 0.58, size = 17, normalized size = 0.40 \[ -\frac {1}{5} \, \cos \relax (x)^{5} + \frac {2}{3} \, \cos \relax (x)^{3} - \cos \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*cos(x)^5 + 2/3*cos(x)^3 - cos(x)

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giac [A]  time = 0.33, size = 61, normalized size = 1.42 \[ -\frac {16 \, {\left (\frac {5 \, {\left (\cos \relax (x) - 1\right )} \mathrm {sgn}\left (\sin \relax (x)\right )}{\cos \relax (x) + 1} - \frac {10 \, {\left (\cos \relax (x) - 1\right )}^{2} \mathrm {sgn}\left (\sin \relax (x)\right )}{{\left (\cos \relax (x) + 1\right )}^{2}} - \mathrm {sgn}\left (\sin \relax (x)\right )\right )}}{15 \, {\left (\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1} - 1\right )}^{5}} + \frac {16}{15} \, \mathrm {sgn}\left (\sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*(5*(cos(x) - 1)*sgn(sin(x))/(cos(x) + 1) - 10*(cos(x) - 1)^2*sgn(sin(x))/(cos(x) + 1)^2 - sgn(sin(x)))/
((cos(x) - 1)/(cos(x) + 1) - 1)^5 + 16/15*sgn(sin(x))

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maple [A]  time = 0.43, size = 38, normalized size = 0.88 \[ \frac {\sin \relax (x ) \left (3 \left (\cos ^{2}\relax (x )\right )-9 \cos \relax (x )+8\right ) \sqrt {4}}{30 \left (-1+\cos \relax (x )\right )^{3} \left (-\frac {1}{-1+\cos ^{2}\relax (x )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.68, size = 17, normalized size = 0.40 \[ -\frac {1}{80} \, \cos \left (5 \, x\right ) + \frac {5}{48} \, \cos \left (3 \, x\right ) - \frac {5}{8} \, \cos \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*cos(5*x) + 5/48*cos(3*x) - 5/8*cos(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\frac {1}{{\sin \relax (x)}^2}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(1/sin(x)^2)^(5/2), x)

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sympy [A]  time = 9.32, size = 46, normalized size = 1.07 \[ - \frac {8 \cot ^{5}{\relax (x )}}{15 \left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} - \frac {4 \cot ^{3}{\relax (x )}}{3 \left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} - \frac {\cot {\relax (x )}}{\left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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