3.63 \(\int (a \csc ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=62 \[ -\frac {1}{5} a \cos ^2(x) \cot ^3(x) \sqrt {a \csc ^4(x)}-\frac {2}{3} a \cos ^2(x) \cot (x) \sqrt {a \csc ^4(x)}-a \sin (x) \cos (x) \sqrt {a \csc ^4(x)} \]

[Out]

-2/3*a*cos(x)^2*cot(x)*(a*csc(x)^4)^(1/2)-1/5*a*cos(x)^2*cot(x)^3*(a*csc(x)^4)^(1/2)-a*cos(x)*sin(x)*(a*csc(x)
^4)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4123, 3767} \[ -\frac {1}{5} a \cos ^2(x) \cot ^3(x) \sqrt {a \csc ^4(x)}-\frac {2}{3} a \cos ^2(x) \cot (x) \sqrt {a \csc ^4(x)}-a \sin (x) \cos (x) \sqrt {a \csc ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Csc[x]^4)^(3/2),x]

[Out]

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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 \left (a \csc ^4(x)\right )^{3/2} \, dx &=\left (a \sqrt {a \csc ^4(x)} \sin ^2(x)\right ) \int \csc ^6(x) \, dx\\ &=-\left (\left (a \sqrt {a \csc ^4(x)} \sin ^2(x)\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (x)\right )\right )\\ &=-\frac {2}{3} a \cos ^2(x) \cot (x) \sqrt {a \csc ^4(x)}-\frac {1}{5} a \cos ^2(x) \cot ^3(x) \sqrt {a \csc ^4(x)}-a \cos (x) \sqrt {a \csc ^4(x)} \sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.53 \[ -\frac {1}{15} a \sin (x) \cos (x) \left (3 \csc ^4(x)+4 \csc ^2(x)+8\right ) \sqrt {a \csc ^4(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Csc[x]^4)^(3/2),x]

[Out]

-1/15*(a*Cos[x]*Sqrt[a*Csc[x]^4]*(8 + 4*Csc[x]^2 + 3*Csc[x]^4)*Sin[x])

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fricas [A]  time = 0.57, size = 52, normalized size = 0.84 \[ \frac {{\left (8 \, a \cos \relax (x)^{5} - 20 \, a \cos \relax (x)^{3} + 15 \, a \cos \relax (x)\right )} \sqrt {\frac {a}{\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1}}}{15 \, {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csc(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/15*(8*a*cos(x)^5 - 20*a*cos(x)^3 + 15*a*cos(x))*sqrt(a/(cos(x)^4 - 2*cos(x)^2 + 1))/((cos(x)^2 - 1)*sin(x))

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giac [A]  time = 0.29, size = 23, normalized size = 0.37 \[ -\frac {{\left (15 \, \tan \relax (x)^{4} + 10 \, \tan \relax (x)^{2} + 3\right )} a^{\frac {3}{2}}}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csc(x)^4)^(3/2),x, algorithm="giac")

[Out]

-1/15*(15*tan(x)^4 + 10*tan(x)^2 + 3)*a^(3/2)/tan(x)^5

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maple [A]  time = 0.42, size = 29, normalized size = 0.47 \[ -\frac {\left (8 \left (\cos ^{4}\relax (x )\right )-20 \left (\cos ^{2}\relax (x )\right )+15\right ) \cos \relax (x ) \sin \relax (x ) \left (\frac {a}{\sin \relax (x )^{4}}\right )^{\frac {3}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*csc(x)^4)^(3/2),x)

[Out]

-1/15*(8*cos(x)^4-20*cos(x)^2+15)*cos(x)*sin(x)*(a/sin(x)^4)^(3/2)

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maxima [A]  time = 0.48, size = 30, normalized size = 0.48 \[ -\frac {15 \, a^{\frac {3}{2}} \tan \relax (x)^{4} + 10 \, a^{\frac {3}{2}} \tan \relax (x)^{2} + 3 \, a^{\frac {3}{2}}}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csc(x)^4)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*a^(3/2)*tan(x)^4 + 10*a^(3/2)*tan(x)^2 + 3*a^(3/2))/tan(x)^5

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mupad [B]  time = 0.67, size = 44, normalized size = 0.71 \[ \frac {\frac {a^{3/2}\,8{}\mathrm {i}}{15}-\frac {4\,a^{3/2}\,\left (2\,{\sin \left (2\,x\right )}^3-9\,\sin \left (2\,x\right )+3\,\sin \left (4\,x\right )+2{}\mathrm {i}\right )}{15}}{{\left (\cos \left (2\,x\right )-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/sin(x)^4)^(3/2),x)

[Out]

((a^(3/2)*8i)/15 - (4*a^(3/2)*(3*sin(4*x) - 9*sin(2*x) + 2*sin(2*x)^3 + 2i))/15)/(cos(2*x) - 1)^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \csc ^{4}{\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csc(x)**4)**(3/2),x)

[Out]

Integral((a*csc(x)**4)**(3/2), x)

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