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

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Rubi [A]  time = 0.0336789, antiderivative size = 32, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 0.0219386, 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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Maple [A]  time = 0.01, size = 34, normalized size = 1.1 \begin{align*} -{\frac{ \left ( \cos \left ( x \right ) \right ) ^{7}}{\sin \left ( x \right ) }}- \left ( \left ( \cos \left ( x \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( x \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( x \right ) }{8}} \right ) \sin \left ( x \right ) -{\frac{15\,x}{8}} \end{align*}

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 = 1.44403, size = 47, normalized size = 1.47 \begin{align*} -\frac{15}{8} \, x - \frac{15 \, \tan \left (x\right )^{4} + 25 \, \tan \left (x\right )^{2} + 8}{8 \,{\left (\tan \left (x\right )^{5} + 2 \, \tan \left (x\right )^{3} + \tan \left (x\right )\right )}} \end{align*}

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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Fricas [A]  time = 2.46964, size = 86, normalized size = 2.69 \begin{align*} \frac{2 \, \cos \left (x\right )^{5} + 5 \, \cos \left (x\right )^{3} - 15 \, x \sin \left (x\right ) - 15 \, \cos \left (x\right )}{8 \, \sin \left (x\right )} \end{align*}

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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Sympy [A]  time = 0.066273, size = 36, normalized size = 1.12 \begin{align*} - \frac{15 x}{8} - \frac{5 \sin{\left (x \right )} \cos ^{3}{\left (x \right )}}{4} - \frac{15 \sin{\left (x \right )} \cos{\left (x \right )}}{8} - \frac{\cos ^{5}{\left (x \right )}}{\sin{\left (x \right )}} \end{align*}

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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Giac [A]  time = 1.1096, size = 42, normalized size = 1.31 \begin{align*} -\frac{15}{8} \, x - \frac{7 \, \tan \left (x\right )^{3} + 9 \, \tan \left (x\right )}{8 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{2}} - \frac{1}{\tan \left (x\right )} \end{align*}

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)