3.831 \(\int \cos ^4(x) \cot ^2(x) \, dx\)

Optimal. Leaf size=32 \[ -\frac {15 x}{8}-\frac {15 \cot (x)}{8}+\frac {1}{4} \cos ^4(x) \cot (x)+\frac {5}{8} \cos ^2(x) \cot (x) \]

[Out]

-15/8*x-15/8*cot(x)+5/8*cos(x)^2*cot(x)+1/4*cos(x)^4*cot(x)

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2591, 288, 321, 203} \[ -\frac {15 x}{8}-\frac {15 \cot (x)}{8}+\frac {1}{4} \cos ^4(x) \cot (x)+\frac {5}{8} \cos ^2(x) \cot (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4*Cot[x]^2,x]

[Out]

(-15*x)/8 - (15*Cot[x])/8 + (5*Cos[x]^2*Cot[x])/8 + (Cos[x]^4*Cot[x])/4

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \cos ^4(x) \cot ^2(x) \, dx &=-\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\cot (x)\right )\\ &=\frac {1}{4} \cos ^4(x) \cot (x)-\frac {5}{4} \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (x)\right )\\ &=\frac {5}{8} \cos ^2(x) \cot (x)+\frac {1}{4} \cos ^4(x) \cot (x)-\frac {15}{8} \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\cot (x)\right )\\ &=-\frac {15 \cot (x)}{8}+\frac {5}{8} \cos ^2(x) \cot (x)+\frac {1}{4} \cos ^4(x) \cot (x)+\frac {15}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (x)\right )\\ &=-\frac {15 x}{8}-\frac {15 \cot (x)}{8}+\frac {5}{8} \cos ^2(x) \cot (x)+\frac {1}{4} \cos ^4(x) \cot (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.81 \[ -\frac {15 x}{8}-\frac {1}{2} \sin (2 x)-\frac {1}{32} \sin (4 x)-\cot (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4*Cot[x]^2,x]

[Out]

(-15*x)/8 - Cot[x] - Sin[2*x]/2 - Sin[4*x]/32

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fricas [A]  time = 1.63, size = 28, normalized size = 0.88 \[ \frac {2 \, \cos \relax (x)^{5} + 5 \, \cos \relax (x)^{3} - 15 \, x \sin \relax (x) - 15 \, \cos \relax (x)}{8 \, \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cot(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 31, normalized size = 0.97 \[ -\frac {15}{8} \, x - \frac {7 \, \tan \relax (x)^{3} + 9 \, \tan \relax (x)}{8 \, {\left (\tan \relax (x)^{2} + 1\right )}^{2}} - \frac {1}{\tan \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cot(x)^2,x, algorithm="giac")

[Out]

-15/8*x - 1/8*(7*tan(x)^3 + 9*tan(x))/(tan(x)^2 + 1)^2 - 1/tan(x)

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maple [A]  time = 0.04, size = 34, normalized size = 1.06 \[ -\frac {\cos ^{7}\relax (x )}{\sin \relax (x )}-\left (\cos ^{5}\relax (x )+\frac {5 \left (\cos ^{3}\relax (x )\right )}{4}+\frac {15 \cos \relax (x )}{8}\right ) \sin \relax (x )-\frac {15 x}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4*cot(x)^2,x)

[Out]

-1/sin(x)*cos(x)^7-(cos(x)^5+5/4*cos(x)^3+15/8*cos(x))*sin(x)-15/8*x

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maxima [A]  time = 0.43, size = 35, normalized size = 1.09 \[ -\frac {15}{8} \, x - \frac {15 \, \tan \relax (x)^{4} + 25 \, \tan \relax (x)^{2} + 8}{8 \, {\left (\tan \relax (x)^{5} + 2 \, \tan \relax (x)^{3} + \tan \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4*cot(x)^2,x, algorithm="maxima")

[Out]

-15/8*x - 1/8*(15*tan(x)^4 + 25*tan(x)^2 + 8)/(tan(x)^5 + 2*tan(x)^3 + tan(x))

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mupad [B]  time = 3.00, size = 26, normalized size = 0.81 \[ \frac {\frac {{\cos \relax (x)}^5}{4}+\frac {5\,{\cos \relax (x)}^3}{8}-\frac {15\,\cos \relax (x)}{8}}{\sin \relax (x)}-\frac {15\,x}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4*cot(x)^2,x)

[Out]

((5*cos(x)^3)/8 - (15*cos(x))/8 + cos(x)^5/4)/sin(x) - (15*x)/8

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sympy [A]  time = 0.06, size = 36, normalized size = 1.12 \[ - \frac {15 x}{8} - \frac {5 \sin {\relax (x )} \cos ^{3}{\relax (x )}}{4} - \frac {15 \sin {\relax (x )} \cos {\relax (x )}}{8} - \frac {\cos ^{5}{\relax (x )}}{\sin {\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**4*cot(x)**2,x)

[Out]

-15*x/8 - 5*sin(x)*cos(x)**3/4 - 15*sin(x)*cos(x)/8 - cos(x)**5/sin(x)

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