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]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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}{\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*}

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Mathematica [A]  time = 0.03, 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]

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

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fricas [A]  time = 0.60, size = 37, normalized size = 0.67 \[ -\frac {{\left (3 \, \cos \relax (x)^{5} - 10 \, \cos \relax (x)^{3} + 15 \, \cos \relax (x)\right )} \sqrt {-\frac {a}{\cos \relax (x)^{2} - 1}} \sin \relax (x)}{15 \, a^{3}} \]

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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giac [A]  time = 0.50, size = 56, normalized size = 1.02 \[ -\frac {16 \, {\left (\frac {10 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + 5 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2} + \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right )}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{5}} - \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right )\right )}}{15 \, a^{\frac {5}{2}}} \]

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)

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maple [A]  time = 0.42, size = 39, normalized size = 0.71 \[ \frac {\sin \relax (x ) \left (3 \left (\cos ^{2}\relax (x )\right )-9 \cos \relax (x )+8\right ) \sqrt {4}}{30 \left (-1+\cos \relax (x )\right )^{3} \left (-\frac {a}{-1+\cos ^{2}\relax (x )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*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)*a)^(5/2)*4^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \csc \relax (x)^{2}\right )^{\frac {5}{2}}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/sin(x)^2)^(5/2), x)

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sympy [A]  time = 9.41, size = 61, normalized size = 1.11 \[ - \frac {8 \cot ^{5}{\relax (x )}}{15 a^{\frac {5}{2}} \left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} - \frac {4 \cot ^{3}{\relax (x )}}{3 a^{\frac {5}{2}} \left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} - \frac {\cot {\relax (x )}}{a^{\frac {5}{2}} \left (\csc ^{2}{\relax (x )}\right )^{\frac {5}{2}}} \]

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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