3.36 \(\int \frac {1}{(a \cot ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac {\cot (x)}{a \sqrt {a \cot ^4(x)}}-\frac {x \cot ^2(x)}{a \sqrt {a \cot ^4(x)}}+\frac {\tan ^3(x)}{5 a \sqrt {a \cot ^4(x)}}-\frac {\tan (x)}{3 a \sqrt {a \cot ^4(x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 77, 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} \[ -\frac {x \cot ^2(x)}{a \sqrt {a \cot ^4(x)}}+\frac {\cot (x)}{a \sqrt {a \cot ^4(x)}}+\frac {\tan ^3(x)}{5 a \sqrt {a \cot ^4(x)}}-\frac {\tan (x)}{3 a \sqrt {a \cot ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 42, normalized size = 0.55 \[ \frac {-15 x \cot ^2(x)+23 \cot (x)+\csc (x) \sec (x) \left (3 \sec ^2(x)-11\right )}{15 a \sqrt {a \cot ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*Cot[x] - 15*x*Cot[x]^2 + Csc[x]*Sec[x]*(-11 + 3*Sec[x]^2))/(15*a*Sqrt[a*Cot[x]^4])

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fricas [B]  time = 1.37, size = 142, normalized size = 1.84 \[ \frac {{\left (15 \, x \cos \left (2 \, x\right )^{4} + 30 \, x \cos \left (2 \, x\right )^{3} - 30 \, x \cos \left (2 \, x\right ) - {\left (23 \, \cos \left (2 \, x\right )^{3} + \cos \left (2 \, x\right )^{2} - 11 \, \cos \left (2 \, x\right ) - 13\right )} \sin \left (2 \, x\right ) - 15 \, x\right )} \sqrt {\frac {a \cos \left (2 \, x\right )^{2} + 2 \, a \cos \left (2 \, x\right ) + a}{\cos \left (2 \, x\right )^{2} - 2 \, \cos \left (2 \, x\right ) + 1}}}{15 \, {\left (a^{2} \cos \left (2 \, x\right )^{4} + 4 \, a^{2} \cos \left (2 \, x\right )^{3} + 6 \, a^{2} \cos \left (2 \, x\right )^{2} + 4 \, a^{2} \cos \left (2 \, x\right ) + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/15*(15*x*cos(2*x)^4 + 30*x*cos(2*x)^3 - 30*x*cos(2*x) - (23*cos(2*x)^3 + cos(2*x)^2 - 11*cos(2*x) - 13)*sin(
2*x) - 15*x)*sqrt((a*cos(2*x)^2 + 2*a*cos(2*x) + a)/(cos(2*x)^2 - 2*cos(2*x) + 1))/(a^2*cos(2*x)^4 + 4*a^2*cos
(2*x)^3 + 6*a^2*cos(2*x)^2 + 4*a^2*cos(2*x) + a^2)

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giac [A]  time = 0.42, size = 43, normalized size = 0.56 \[ -\frac {\frac {15 \, x}{\sqrt {a}} - \frac {3 \, a^{2} \tan \relax (x)^{5} - 5 \, a^{2} \tan \relax (x)^{3} + 15 \, a^{2} \tan \relax (x)}{a^{\frac {5}{2}}}}{15 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)^4)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 42, normalized size = 0.55 \[ \frac {\cot \relax (x ) \left (15 \left (\frac {\pi }{2}-\mathrm {arccot}\left (\cot \relax (x )\right )\right ) \left (\cot ^{5}\relax (x )\right )+15 \left (\cot ^{4}\relax (x )\right )-5 \left (\cot ^{2}\relax (x )\right )+3\right )}{15 \left (a \left (\cot ^{4}\relax (x )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cot(x)^4)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.45, size = 29, normalized size = 0.38 \[ \frac {3 \, \tan \relax (x)^{5} - 5 \, \tan \relax (x)^{3} + 15 \, \tan \relax (x)}{15 \, a^{\frac {3}{2}}} - \frac {x}{a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)^4)^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cot(x)^4)^(3/2),x)

[Out]

int(1/(a*cot(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cot ^{4}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)**4)**(3/2),x)

[Out]

Integral((a*cot(x)**4)**(-3/2), x)

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