3.125 \(\int \cos (x) \csc (4 x) \, dx\)

Optimal. Leaf size=26 \[ \frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{2 \sqrt{2}}-\frac{1}{4} \tanh ^{-1}(\cos (x)) \]

[Out]

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

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Rubi [A]  time = 0.0257339, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1093, 206} \[ \frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{2 \sqrt{2}}-\frac{1}{4} \tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1093

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

Rule 206

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

Rubi steps

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

Mathematica [C]  time = 0.0578497, size = 66, normalized size = 2.54 \[ \frac{1}{4} \left (\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )+(1+i) (-1)^{3/4} \tanh ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )-1}{\sqrt{2}}\right )+\sqrt{2} \tanh ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )+1}{\sqrt{2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 28, normalized size = 1.1 \begin{align*}{\frac{{\it Artanh} \left ( \cos \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{4}}-{\frac{\ln \left ( 1+\cos \left ( x \right ) \right ) }{8}}+{\frac{\ln \left ( -1+\cos \left ( x \right ) \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.5529, size = 220, normalized size = 8.46 \begin{align*} \frac{1}{16} \, \sqrt{2} \log \left (2 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{2} \cos \left (x\right ) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) - \frac{1}{16} \, \sqrt{2} \log \left (-2 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) - 2 \,{\left (\sqrt{2} \cos \left (x\right ) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) - \frac{1}{8} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + \frac{1}{8} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*csc(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) - 1/8*log(cos(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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Fricas [B]  time = 2.47266, size = 174, normalized size = 6.69 \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (-\frac{2 \, \cos \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 1}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \frac{1}{8} \, \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) + \frac{1}{8} \, \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*csc(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)) - 1/8*log(1/2*cos(x) + 1/2) + 1/8*log(-
1/2*cos(x) + 1/2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.17051, size = 88, normalized size = 3.38 \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} - \frac{2 \,{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}{{\left | 4 \, \sqrt{2} - \frac{2 \,{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}\right ) + \frac{1}{8} \, \log \left (-\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(abs(-4*sqrt(2) - 2*(cos(x) - 1)/(cos(x) + 1) - 6)/abs(4*sqrt(2) - 2*(cos(x) - 1)/(cos(x) + 1)
- 6)) + 1/8*log(-(cos(x) - 1)/(cos(x) + 1))