3.112 \(\int \cos (x) \cot (4 x) \, dx\)

Optimal. Leaf size=28 \[ \cos (x)-\frac {1}{4} \tanh ^{-1}(\cos (x))-\frac {\tanh ^{-1}\left (\sqrt {2} \cos (x)\right )}{2 \sqrt {2}} \]

[Out]

-1/4*arctanh(cos(x))+cos(x)-1/4*arctanh(cos(x)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1676, 1166, 207} \[ \cos (x)-\frac {1}{4} \tanh ^{-1}(\cos (x))-\frac {\tanh ^{-1}\left (\sqrt {2} \cos (x)\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[x]]/4 - ArcTanh[Sqrt[2]*Cos[x]]/(2*Sqrt[2]) + Cos[x]

Rule 207

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

Rule 1166

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

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rubi steps

\begin {align*} \int \cos (x) \cot (4 x) \, dx &=-\operatorname {Subst}\left (\int \frac {-1+8 x^2-8 x^4}{4-12 x^2+8 x^4} \, dx,x,\cos (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (-1+\frac {3-4 x^2}{4-12 x^2+8 x^4}\right ) \, dx,x,\cos (x)\right )\\ &=\cos (x)-\operatorname {Subst}\left (\int \frac {3-4 x^2}{4-12 x^2+8 x^4} \, dx,x,\cos (x)\right )\\ &=\cos (x)+2 \operatorname {Subst}\left (\int \frac {1}{-8+8 x^2} \, dx,x,\cos (x)\right )+2 \operatorname {Subst}\left (\int \frac {1}{-4+8 x^2} \, dx,x,\cos (x)\right )\\ &=-\frac {1}{4} \tanh ^{-1}(\cos (x))-\frac {\tanh ^{-1}\left (\sqrt {2} \cos (x)\right )}{2 \sqrt {2}}+\cos (x)\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 73, normalized size = 2.61 \[ \frac {1}{4} \left (4 \cos (x)+\log \left (\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )+(-1-i) (-1)^{3/4} \tanh ^{-1}\left (\frac {\tan \left (\frac {x}{2}\right )-1}{\sqrt {2}}\right )-(1-i) \sqrt [4]{-1} \tanh ^{-1}\left (\frac {\tan \left (\frac {x}{2}\right )+1}{\sqrt {2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 - I)*(-1)^(3/4)*ArcTanh[(-1 + Tan[x/2])/Sqrt[2]] - (1 - I)*(-1)^(1/4)*ArcTanh[(1 + Tan[x/2])/Sqrt[2]] + 4
*Cos[x] - Log[Cos[x/2]] + Log[Sin[x/2]])/4

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fricas [B]  time = 1.01, size = 53, normalized size = 1.89 \[ \frac {1}{8} \, \sqrt {2} \log \left (\frac {2 \, \cos \relax (x)^{2} - 2 \, \sqrt {2} \cos \relax (x) + 1}{2 \, \cos \relax (x)^{2} - 1}\right ) + \cos \relax (x) - \frac {1}{8} \, \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + \frac {1}{8} \, \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log((2*cos(x)^2 - 2*sqrt(2)*cos(x) + 1)/(2*cos(x)^2 - 1)) + cos(x) - 1/8*log(1/2*cos(x) + 1/2) + 1
/8*log(-1/2*cos(x) + 1/2)

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giac [B]  time = 0.14, size = 50, normalized size = 1.79 \[ \frac {1}{8} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \cos \relax (x) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \cos \relax (x) \right |}}\right ) + \cos \relax (x) - \frac {1}{8} \, \log \left (\cos \relax (x) + 1\right ) + \frac {1}{8} \, \log \left (-\cos \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(abs(-2*sqrt(2) + 4*cos(x))/abs(2*sqrt(2) + 4*cos(x))) + cos(x) - 1/8*log(cos(x) + 1) + 1/8*log
(-cos(x) + 1)

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maple [A]  time = 0.41, size = 30, normalized size = 1.07 \[ \frac {\ln \left (-1+\cos \relax (x )\right )}{8}-\frac {\arctanh \left (\cos \relax (x ) \sqrt {2}\right ) \sqrt {2}}{4}-\frac {\ln \left (1+\cos \relax (x )\right )}{8}+\cos \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*ln(-1+cos(x))-1/4*arctanh(cos(x)*2^(1/2))*2^(1/2)-1/8*ln(1+cos(x))+cos(x)

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maxima [B]  time = 0.42, size = 165, normalized size = 5.89 \[ -\frac {1}{16} \, \sqrt {2} \log \left (2 \, \sqrt {2} \sin \left (2 \, x\right ) \sin \relax (x) + 2 \, {\left (\sqrt {2} \cos \relax (x) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \relax (x)^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \relax (x)^{2} + 2 \, \sqrt {2} \cos \relax (x) + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (-2 \, \sqrt {2} \sin \left (2 \, x\right ) \sin \relax (x) - 2 \, {\left (\sqrt {2} \cos \relax (x) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \relax (x)^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \relax (x)^{2} - 2 \, \sqrt {2} \cos \relax (x) + 1\right ) + \cos \relax (x) - \frac {1}{8} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + \frac {1}{8} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*log(2*sqrt(2)*sin(2*x)*sin(x) + 2*(sqrt(2)*cos(x) + 1)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(
2*x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 1) + 1/16*sqrt(2)*log(-2*sqrt(2)*sin(2*x)*sin(x) - 2*(sqrt(2)*cos(x)
- 1)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*x)^2 + 2*sin(x)^2 - 2*sqrt(2)*cos(x) + 1) + cos(x) - 1/8*log(c
os(x)^2 + sin(x)^2 + 2*cos(x) + 1) + 1/8*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)

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mupad [B]  time = 2.35, size = 67, normalized size = 2.39 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{4}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {7\,\sqrt {2}}{8\,\left (\frac {29\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4}-\frac {5}{4}\right )}-\frac {41\,\sqrt {2}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,\left (\frac {29\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4}-\frac {5}{4}\right )}\right )}{4}+\frac {2}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(x/2))/4 - (2^(1/2)*atanh((7*2^(1/2))/(8*((29*tan(x/2)^2)/4 - 5/4)) - (41*2^(1/2)*tan(x/2)^2)/(8*((29*t
an(x/2)^2)/4 - 5/4))))/4 + 2/(tan(x/2)^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)*cot(4*x), x)

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