3.111 \(\int \cos (x) \cot (3 x) \, dx\)

Optimal. Leaf size=45 \[ \cos (x)+\frac {1}{6} \log (1-2 \cos (x))+\frac {1}{6} \log (1-\cos (x))-\frac {1}{6} \log (\cos (x)+1)-\frac {1}{6} \log (2 \cos (x)+1) \]

[Out]

cos(x)+1/6*ln(1-2*cos(x))+1/6*ln(1-cos(x))-1/6*ln(1+cos(x))-1/6*ln(1+2*cos(x))

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Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1279, 1161, 616, 31} \[ \cos (x)+\frac {1}{6} \log (1-2 \cos (x))+\frac {1}{6} \log (1-\cos (x))-\frac {1}{6} \log (\cos (x)+1)-\frac {1}{6} \log (2 \cos (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Cot[3*x],x]

[Out]

Cos[x] + Log[1 - 2*Cos[x]]/6 + Log[1 - Cos[x]]/6 - Log[1 + Cos[x]]/6 - Log[1 + 2*Cos[x]]/6

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.04 \[ \cos (x)+\frac {1}{3} \log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {1}{3} \log \left (\cos \left (\frac {x}{2}\right )\right )+\frac {1}{6} \log (1-2 \cos (x))-\frac {1}{6} \log (2 \cos (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Cot[3*x],x]

[Out]

Cos[x] - Log[Cos[x/2]]/3 + Log[1 - 2*Cos[x]]/6 - Log[1 + 2*Cos[x]]/6 + Log[Sin[x/2]]/3

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fricas [A]  time = 3.04, size = 39, normalized size = 0.87 \[ \cos \relax (x) - \frac {1}{6} \, \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + \frac {1}{6} \, \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + \frac {1}{6} \, \log \left (-2 \, \cos \relax (x) + 1\right ) - \frac {1}{6} \, \log \left (-2 \, \cos \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cot(3*x),x, algorithm="fricas")

[Out]

cos(x) - 1/6*log(1/2*cos(x) + 1/2) + 1/6*log(-1/2*cos(x) + 1/2) + 1/6*log(-2*cos(x) + 1) - 1/6*log(-2*cos(x) -
1)

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giac [A]  time = 0.13, size = 39, normalized size = 0.87 \[ \cos \relax (x) - \frac {1}{6} \, \log \left (\cos \relax (x) + 1\right ) + \frac {1}{6} \, \log \left (-\cos \relax (x) + 1\right ) - \frac {1}{6} \, \log \left ({\left | 2 \, \cos \relax (x) + 1 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | 2 \, \cos \relax (x) - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cot(3*x),x, algorithm="giac")

[Out]

cos(x) - 1/6*log(cos(x) + 1) + 1/6*log(-cos(x) + 1) - 1/6*log(abs(2*cos(x) + 1)) + 1/6*log(abs(2*cos(x) - 1))

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maple [A]  time = 0.32, size = 36, normalized size = 0.80 \[ \frac {\ln \left (2 \cos \relax (x )-1\right )}{6}-\frac {\ln \left (1+2 \cos \relax (x )\right )}{6}+\frac {\ln \left (-1+\cos \relax (x )\right )}{6}-\frac {\ln \left (1+\cos \relax (x )\right )}{6}+\cos \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*cot(3*x),x)

[Out]

1/6*ln(2*cos(x)-1)-1/6*ln(1+2*cos(x))+1/6*ln(-1+cos(x))-1/6*ln(1+cos(x))+cos(x)

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maxima [B]  time = 0.42, size = 131, normalized size = 2.91 \[ \cos \relax (x) - \frac {1}{12} \, \log \left (2 \, {\left (\cos \relax (x) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \cos \relax (x)^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (2 \, x\right ) \sin \relax (x) + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + \frac {1}{12} \, \log \left (-2 \, {\left (\cos \relax (x) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \cos \relax (x)^{2} + \sin \left (2 \, x\right )^{2} - 2 \, \sin \left (2 \, x\right ) \sin \relax (x) + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - \frac {1}{6} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + \frac {1}{6} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cot(3*x),x, algorithm="maxima")

[Out]

cos(x) - 1/12*log(2*(cos(x) + 1)*cos(2*x) + cos(2*x)^2 + cos(x)^2 + sin(2*x)^2 + 2*sin(2*x)*sin(x) + sin(x)^2
+ 2*cos(x) + 1) + 1/12*log(-2*(cos(x) - 1)*cos(2*x) + cos(2*x)^2 + cos(x)^2 + sin(2*x)^2 - 2*sin(2*x)*sin(x) +
sin(x)^2 - 2*cos(x) + 1) - 1/6*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + 1/6*log(cos(x)^2 + sin(x)^2 - 2*cos(
x) + 1)

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mupad [B]  time = 2.37, size = 39, normalized size = 0.87 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{3}+\frac {\mathrm {atanh}\left (\frac {8}{183\,\left (\frac {488\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{243}-\frac {56}{81}\right )}+\frac {121}{122}\right )}{3}+\frac {2}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(3*x)*cos(x),x)

[Out]

log(tan(x/2))/3 + atanh(8/(183*((488*tan(x/2)^2)/243 - 56/81)) + 121/122)/3 + 2/(tan(x/2)^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cot(3*x),x)

[Out]

Integral(cos(x)*cot(3*x), x)

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