3.410 \(\int \frac {\csc (x)}{\sin ^{\frac {3}{2}}(2 x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {4 \sin (x)}{3 \sqrt {\sin (2 x)}}-\frac {2 \cos (x)}{3 \sin ^{\frac {3}{2}}(2 x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4308, 4303, 4292} \[ \frac {4 \sin (x)}{3 \sqrt {\sin (2 x)}}-\frac {2 \cos (x)}{3 \sin ^{\frac {3}{2}}(2 x)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/Sin[2*x]^(3/2),x]

[Out]

(-2*Cos[x])/(3*Sin[2*x]^(3/2)) + (4*Sin[x])/(3*Sqrt[Sin[2*x]])

Rule 4292

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[((e*Sin[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[a + b*x]*(g*Sin[c + d
*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4308

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[a + b*x]*(g*S
in[c + d*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ
[p] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\csc (x)}{\sin ^{\frac {3}{2}}(2 x)} \, dx &=2 \int \frac {\cos (x)}{\sin ^{\frac {5}{2}}(2 x)} \, dx\\ &=-\frac {2 \cos (x)}{3 \sin ^{\frac {3}{2}}(2 x)}+\frac {4}{3} \int \frac {\sin (x)}{\sin ^{\frac {3}{2}}(2 x)} \, dx\\ &=-\frac {2 \cos (x)}{3 \sin ^{\frac {3}{2}}(2 x)}+\frac {4 \sin (x)}{3 \sqrt {\sin (2 x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.83 \[ \sqrt {\sin (2 x)} \left (\frac {\sec (x)}{2}-\frac {1}{6} \cot (x) \csc (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/Sin[2*x]^(3/2),x]

[Out]

(-1/6*(Cot[x]*Csc[x]) + Sec[x]/2)*Sqrt[Sin[2*x]]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc (x)}{\sin ^{\frac {3}{2}}(2 x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Csc[x]/Sin[2*x]^(3/2),x]

[Out]

Could not integrate

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fricas [B]  time = 0.61, size = 43, normalized size = 1.48 \[ \frac {4 \, \cos \relax (x)^{3} + \sqrt {2} {\left (4 \, \cos \relax (x)^{2} - 3\right )} \sqrt {\cos \relax (x) \sin \relax (x)} - 4 \, \cos \relax (x)}{6 \, {\left (\cos \relax (x)^{3} - \cos \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)/sin(2*x)^(3/2),x, algorithm="fricas")

[Out]

1/6*(4*cos(x)^3 + sqrt(2)*(4*cos(x)^2 - 3)*sqrt(cos(x)*sin(x)) - 4*cos(x))/(cos(x)^3 - cos(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin \left (2 \, x\right )^{\frac {3}{2}} \sin \relax (x)}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)/sin(2*x)^(3/2),x, algorithm="giac")

[Out]

integrate(1/(sin(2*x)^(3/2)*sin(x)), x)

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maple [C]  time = 0.16, size = 121, normalized size = 4.17

method	result	size
default	\(-\frac {\sqrt {-\frac {\tan \left (\frac {x}{2}\right )}{\tan ^{2}\left (\frac {x}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \left (2 \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {x}{2}\right )-\left (\tan ^{4}\left (\frac {x}{2}\right )\right )+1\right )}{12 \tan \left (\frac {x}{2}\right ) \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(x)/sin(2*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^2-1)/tan(1/2*x)*(2*(tan(1/2*x)+1)^(1/2)*(-2*tan(1/2*x)+
2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*x)-tan(1/2*x)^4+1)/(tan(1/2*x
)*(tan(1/2*x)^2-1))^(1/2)/(tan(1/2*x)^3-tan(1/2*x))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sin \left (2 \, x\right )^{\frac {3}{2}} \sin \relax (x)}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)/sin(2*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(sin(2*x)^(3/2)*sin(x)), x)

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mupad [B]  time = 0.43, size = 29, normalized size = 1.00 \[ -\frac {\sqrt {\sin \left (2\,x\right )}\,\left (2\,\cos \left (2\,x\right )-1\right )}{6\,\left (\cos \relax (x)-{\cos \relax (x)}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(2*x)^(3/2)*sin(x)),x)

[Out]

-(sin(2*x)^(1/2)*(2*cos(2*x) - 1))/(6*(cos(x) - cos(x)^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)/sin(2*x)**(3/2),x)

[Out]

Timed out

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