3.59 \(\int \frac {1}{(a \csc ^3(x))^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac {14 \cos (x)}{45 a \sqrt {a \csc ^3(x)}}-\frac {2 \sin ^2(x) \cos (x)}{9 a \sqrt {a \csc ^3(x)}}-\frac {14 E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right )}{15 a \sin ^{\frac {3}{2}}(x) \sqrt {a \csc ^3(x)}} \]

[Out]

-14/45*cos(x)/a/(a*csc(x)^3)^(1/2)-14/15*(sin(1/4*Pi+1/2*x)^2)^(1/2)/sin(1/4*Pi+1/2*x)*EllipticE(cos(1/4*Pi+1/
2*x),2^(1/2))/a/sin(x)^(3/2)/(a*csc(x)^3)^(1/2)-2/9*cos(x)*sin(x)^2/a/(a*csc(x)^3)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4123, 3769, 3771, 2639} \[ -\frac {14 \cos (x)}{45 a \sqrt {a \csc ^3(x)}}-\frac {2 \sin ^2(x) \cos (x)}{9 a \sqrt {a \csc ^3(x)}}-\frac {14 E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right )}{15 a \sin ^{\frac {3}{2}}(x) \sqrt {a \csc ^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*Cos[x])/(45*a*Sqrt[a*Csc[x]^3]) - (14*EllipticE[Pi/4 - x/2, 2])/(15*a*Sqrt[a*Csc[x]^3]*Sin[x]^(3/2)) - (2
*Cos[x]*Sin[x]^2)/(9*a*Sqrt[a*Csc[x]^3])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \csc ^3(x)\right )^{3/2}} \, dx &=-\frac {(-\csc (x))^{3/2} \int \frac {1}{(-\csc (x))^{9/2}} \, dx}{a \sqrt {a \csc ^3(x)}}\\ &=-\frac {2 \cos (x) \sin ^2(x)}{9 a \sqrt {a \csc ^3(x)}}-\frac {\left (7 (-\csc (x))^{3/2}\right ) \int \frac {1}{(-\csc (x))^{5/2}} \, dx}{9 a \sqrt {a \csc ^3(x)}}\\ &=-\frac {14 \cos (x)}{45 a \sqrt {a \csc ^3(x)}}-\frac {2 \cos (x) \sin ^2(x)}{9 a \sqrt {a \csc ^3(x)}}-\frac {\left (7 (-\csc (x))^{3/2}\right ) \int \frac {1}{\sqrt {-\csc (x)}} \, dx}{15 a \sqrt {a \csc ^3(x)}}\\ &=-\frac {14 \cos (x)}{45 a \sqrt {a \csc ^3(x)}}-\frac {2 \cos (x) \sin ^2(x)}{9 a \sqrt {a \csc ^3(x)}}+\frac {7 \int \sqrt {\sin (x)} \, dx}{15 a \sqrt {a \csc ^3(x)} \sin ^{\frac {3}{2}}(x)}\\ &=-\frac {14 \cos (x)}{45 a \sqrt {a \csc ^3(x)}}-\frac {14 E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right )}{15 a \sqrt {a \csc ^3(x)} \sin ^{\frac {3}{2}}(x)}-\frac {2 \cos (x) \sin ^2(x)}{9 a \sqrt {a \csc ^3(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 52, normalized size = 0.66 \[ \frac {\sin ^{\frac {3}{2}}(x) (5 \cos (3 x)-33 \cos (x))-84 E\left (\left .\frac {1}{4} (\pi -2 x)\right |2\right )}{90 \sin ^{\frac {9}{2}}(x) \left (a \csc ^3(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-84*EllipticE[(Pi - 2*x)/4, 2] + (-33*Cos[x] + 5*Cos[3*x])*Sin[x]^(3/2))/(90*(a*Csc[x]^3)^(3/2)*Sin[x]^(9/2))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \csc \relax (x)^{3}}}{a^{2} \csc \relax (x)^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*csc(x)^3)/(a^2*csc(x)^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*csc(x)^3)^(-3/2), x)

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maple [C]  time = 0.79, size = 361, normalized size = 4.57 \[ -\frac {\left (42 \cos \relax (x ) \sqrt {2}\, \sqrt {-\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}}\, \sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {i \cos \relax (x )-i-\sin \relax (x )}{\sin \relax (x )}}\, \EllipticE \left (\sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}, \frac {\sqrt {2}}{2}\right )-21 \cos \relax (x ) \sqrt {2}\, \sqrt {-\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}}\, \sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {i \cos \relax (x )-i-\sin \relax (x )}{\sin \relax (x )}}\, \EllipticF \left (\sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}, \frac {\sqrt {2}}{2}\right )+10 \left (\cos ^{5}\relax (x )\right )+42 \sqrt {2}\, \sqrt {-\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}}\, \sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {i \cos \relax (x )-i-\sin \relax (x )}{\sin \relax (x )}}\, \EllipticE \left (\sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {2}\, \sqrt {-\frac {i \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}}\, \sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}\, \sqrt {-\frac {i \cos \relax (x )-i-\sin \relax (x )}{\sin \relax (x )}}\, \EllipticF \left (\sqrt {\frac {i \cos \relax (x )-i+\sin \relax (x )}{\sin \relax (x )}}, \frac {\sqrt {2}}{2}\right )-34 \left (\cos ^{3}\relax (x )\right )+66 \cos \relax (x )-42\right ) \sqrt {8}}{45 \left (-\frac {2 a}{\sin \relax (x ) \left (-1+\cos ^{2}\relax (x )\right )}\right )^{\frac {3}{2}} \sin \relax (x )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*csc(x)^3)^(3/2),x)

[Out]

-1/45*(42*cos(x)*2^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*((I*cos(x)-I+sin(x))/sin(x))^(1/2)*(-(I*cos(x)-I-sin(x)
)/sin(x))^(1/2)*EllipticE(((I*cos(x)-I+sin(x))/sin(x))^(1/2),1/2*2^(1/2))-21*cos(x)*2^(1/2)*(-I*(-1+cos(x))/si
n(x))^(1/2)*((I*cos(x)-I+sin(x))/sin(x))^(1/2)*(-(I*cos(x)-I-sin(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)-I+sin(
x))/sin(x))^(1/2),1/2*2^(1/2))+10*cos(x)^5+42*2^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*((I*cos(x)-I+sin(x))/sin(x
))^(1/2)*(-(I*cos(x)-I-sin(x))/sin(x))^(1/2)*EllipticE(((I*cos(x)-I+sin(x))/sin(x))^(1/2),1/2*2^(1/2))-21*2^(1
/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*((I*cos(x)-I+sin(x))/sin(x))^(1/2)*(-(I*cos(x)-I-sin(x))/sin(x))^(1/2)*Ellip
ticF(((I*cos(x)-I+sin(x))/sin(x))^(1/2),1/2*2^(1/2))-34*cos(x)^3+66*cos(x)-42)/(-2/sin(x)/(-1+cos(x)^2)*a)^(3/
2)/sin(x)^5*8^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*csc(x)^3)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {a}{{\sin \relax (x)}^3}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/sin(x)^3)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \csc ^{3}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)**3)**(3/2),x)

[Out]

Integral((a*csc(x)**3)**(-3/2), x)

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