3.33 \(\int (a \cot ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=70 \[ \frac {1}{3} a \cot (x) \sqrt {a \cot ^4(x)}-\frac {1}{5} a \cot ^3(x) \sqrt {a \cot ^4(x)}-a x \tan ^2(x) \sqrt {a \cot ^4(x)}-a \tan (x) \sqrt {a \cot ^4(x)} \]

[Out]

1/3*a*cot(x)*(a*cot(x)^4)^(1/2)-1/5*a*cot(x)^3*(a*cot(x)^4)^(1/2)-a*(a*cot(x)^4)^(1/2)*tan(x)-a*x*(a*cot(x)^4)
^(1/2)*tan(x)^2

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Rubi [A]  time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3658, 3473, 8} \[ -\frac {1}{5} a \cot ^3(x) \sqrt {a \cot ^4(x)}+\frac {1}{3} a \cot (x) \sqrt {a \cot ^4(x)}-a x \tan ^2(x) \sqrt {a \cot ^4(x)}-a \tan (x) \sqrt {a \cot ^4(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 39, normalized size = 0.56 \[ -\frac {1}{15} \tan ^6(x) \left (a \cot ^4(x)\right )^{3/2} \left (15 x+\cot (x) \left (3 \csc ^4(x)-11 \csc ^2(x)+23\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((a*Cot[x]^4)^(3/2)*(15*x + Cot[x]*(23 - 11*Csc[x]^2 + 3*Csc[x]^4))*Tan[x]^6)

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fricas [A]  time = 1.39, size = 110, normalized size = 1.57 \[ \frac {{\left (23 \, a \cos \left (2 \, x\right )^{3} - a \cos \left (2 \, x\right )^{2} - 11 \, a \cos \left (2 \, x\right ) + 15 \, {\left (a x \cos \left (2 \, x\right )^{2} - 2 \, a x \cos \left (2 \, x\right ) + a x\right )} \sin \left (2 \, x\right ) + 13 \, a\right )} \sqrt {\frac {a \cos \left (2 \, x\right )^{2} + 2 \, a \cos \left (2 \, x\right ) + a}{\cos \left (2 \, x\right )^{2} - 2 \, \cos \left (2 \, x\right ) + 1}}}{15 \, {\left (\cos \left (2 \, x\right )^{2} - 1\right )} \sin \left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(23*a*cos(2*x)^3 - a*cos(2*x)^2 - 11*a*cos(2*x) + 15*(a*x*cos(2*x)^2 - 2*a*x*cos(2*x) + a*x)*sin(2*x) + 1
3*a)*sqrt((a*cos(2*x)^2 + 2*a*cos(2*x) + a)/(cos(2*x)^2 - 2*cos(2*x) + 1))/((cos(2*x)^2 - 1)*sin(2*x))

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giac [A]  time = 2.41, size = 57, normalized size = 0.81 \[ \frac {1}{480} \, {\left (3 \, \tan \left (\frac {1}{2} \, x\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, x\right )^{3} - 480 \, x - \frac {330 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 35 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 3}{\tan \left (\frac {1}{2} \, x\right )^{5}} + 330 \, \tan \left (\frac {1}{2} \, x\right )\right )} a^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*tan(1/2*x)^5 - 35*tan(1/2*x)^3 - 480*x - (330*tan(1/2*x)^4 - 35*tan(1/2*x)^2 + 3)/tan(1/2*x)^5 + 330*
tan(1/2*x))*a^(3/2)

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maple [A]  time = 0.17, size = 40, normalized size = 0.57 \[ \frac {\left (a \left (\cot ^{4}\relax (x )\right )\right )^{\frac {3}{2}} \left (-3 \left (\cot ^{5}\relax (x )\right )+5 \left (\cot ^{3}\relax (x )\right )+\frac {15 \pi }{2}-15 \,\mathrm {arccot}\left (\cot \relax (x )\right )-15 \cot \relax (x )\right )}{15 \cot \relax (x )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(a*cot(x)^4)^(3/2)*(-3*cot(x)^5+5*cot(x)^3+15/2*Pi-15*arccot(cot(x))-15*cot(x))/cot(x)^6

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maxima [A]  time = 0.48, size = 37, normalized size = 0.53 \[ -a^{\frac {3}{2}} x - \frac {15 \, a^{\frac {3}{2}} \tan \relax (x)^{4} - 5 \, a^{\frac {3}{2}} \tan \relax (x)^{2} + 3 \, a^{\frac {3}{2}}}{15 \, \tan \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {cot}\relax (x)}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*cot(x)^4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*cot(x)**4)**(3/2), x)

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