3.94 \(\int \csc (6 x) \sin (x) \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{6} \tanh ^{-1}(\sin (x))+\frac{1}{6} \tanh ^{-1}(2 \sin (x))-\frac{\tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

ArcTanh[Sin[x]]/6 + ArcTanh[2*Sin[x]]/6 - ArcTanh[(2*Sin[x])/Sqrt[3]]/(2*Sqrt[3])

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Rubi [A]  time = 0.0463048, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {12, 2057, 207} \[ \frac{1}{6} \tanh ^{-1}(\sin (x))+\frac{1}{6} \tanh ^{-1}(2 \sin (x))-\frac{\tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[6*x]*Sin[x],x]

[Out]

ArcTanh[Sin[x]]/6 + ArcTanh[2*Sin[x]]/6 - ArcTanh[(2*Sin[x])/Sqrt[3]]/(2*Sqrt[3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2057

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(u /. x -> x^2)^p, x], x
] /;  !SumQ[NonfreeFactors[u, x]]] /; PolyQ[P, x^2] && ILtQ[p, 0]

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])

Rubi steps

\begin{align*} \int \csc (6 x) \sin (x) \, dx &=\operatorname{Subst}\left (\int \frac{1}{2 \left (3-19 x^2+32 x^4-16 x^6\right )} \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{3-19 x^2+32 x^4-16 x^6} \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{1}{3 \left (-1+x^2\right )}+\frac{2}{-3+4 x^2}-\frac{2}{3 \left (-1+4 x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\left (\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sin (x)\right )\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-1+4 x^2} \, dx,x,\sin (x)\right )+\operatorname{Subst}\left (\int \frac{1}{-3+4 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{1}{6} \tanh ^{-1}(\sin (x))+\frac{1}{6} \tanh ^{-1}(2 \sin (x))-\frac{\tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{3}}\right )}{2 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0374529, size = 30, normalized size = 0.83 \[ \frac{1}{6} \left (\tanh ^{-1}(\sin (x))+\tanh ^{-1}(2 \sin (x))-\sqrt{3} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[6*x]*Sin[x],x]

[Out]

(ArcTanh[Sin[x]] + ArcTanh[2*Sin[x]] - Sqrt[3]*ArcTanh[(2*Sin[x])/Sqrt[3]])/6

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Maple [A]  time = 0.093, size = 47, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{12}}-{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) }{12}}+{\frac{\ln \left ( 1+2\,\sin \left ( x \right ) \right ) }{12}}-{\frac{\ln \left ( -1+2\,\sin \left ( x \right ) \right ) }{12}}-{\frac{\sqrt{3}}{6}{\it Artanh} \left ({\frac{2\,\sin \left ( x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(6*x)*sin(x),x)

[Out]

1/12*ln(1+sin(x))-1/12*ln(sin(x)-1)+1/12*ln(1+2*sin(x))-1/12*ln(-1+2*sin(x))-1/6*arctanh(2/3*sin(x)*3^(1/2))*3
^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{24} \, \sqrt{3} \log \left (\frac{4}{3} \, \cos \left (x\right )^{2} + \frac{4}{3} \, \sin \left (x\right )^{2} + \frac{4}{3} \, \sqrt{3} \sin \left (x\right ) + \frac{4}{3} \, \cos \left (x\right ) + \frac{4}{3}\right ) - \frac{1}{24} \, \sqrt{3} \log \left (\frac{4}{3} \, \cos \left (x\right )^{2} + \frac{4}{3} \, \sin \left (x\right )^{2} + \frac{4}{3} \, \sqrt{3} \sin \left (x\right ) - \frac{4}{3} \, \cos \left (x\right ) + \frac{4}{3}\right ) + \frac{1}{24} \, \sqrt{3} \log \left (\frac{4}{3} \, \cos \left (x\right )^{2} + \frac{4}{3} \, \sin \left (x\right )^{2} - \frac{4}{3} \, \sqrt{3} \sin \left (x\right ) + \frac{4}{3} \, \cos \left (x\right ) + \frac{4}{3}\right ) + \frac{1}{24} \, \sqrt{3} \log \left (\frac{4}{3} \, \cos \left (x\right )^{2} + \frac{4}{3} \, \sin \left (x\right )^{2} - \frac{4}{3} \, \sqrt{3} \sin \left (x\right ) - \frac{4}{3} \, \cos \left (x\right ) + \frac{4}{3}\right ) + \int -\frac{{\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \cos \left (4 \, x\right ) -{\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - \cos \left (2 \, x\right ) \cos \left (x\right ) +{\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - \sin \left (2 \, x\right ) \sin \left (x\right ) + \cos \left (x\right )}{6 \,{\left (2 \,{\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 2 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) - 1\right )}}\,{d x} + \frac{1}{12} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - \frac{1}{12} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(6*x)*sin(x),x, algorithm="maxima")

[Out]

-1/24*sqrt(3)*log(4/3*cos(x)^2 + 4/3*sin(x)^2 + 4/3*sqrt(3)*sin(x) + 4/3*cos(x) + 4/3) - 1/24*sqrt(3)*log(4/3*
cos(x)^2 + 4/3*sin(x)^2 + 4/3*sqrt(3)*sin(x) - 4/3*cos(x) + 4/3) + 1/24*sqrt(3)*log(4/3*cos(x)^2 + 4/3*sin(x)^
2 - 4/3*sqrt(3)*sin(x) + 4/3*cos(x) + 4/3) + 1/24*sqrt(3)*log(4/3*cos(x)^2 + 4/3*sin(x)^2 - 4/3*sqrt(3)*sin(x)
- 4/3*cos(x) + 4/3) + integrate(-1/6*((cos(3*x) + cos(x))*cos(4*x) - (cos(2*x) - 1)*cos(3*x) - cos(2*x)*cos(x
) + (sin(3*x) + sin(x))*sin(4*x) - sin(3*x)*sin(2*x) - sin(2*x)*sin(x) + cos(x))/(2*(cos(2*x) - 1)*cos(4*x) -
cos(4*x)^2 - cos(2*x)^2 - sin(4*x)^2 + 2*sin(4*x)*sin(2*x) - sin(2*x)^2 + 2*cos(2*x) - 1), x) + 1/12*log(cos(x
)^2 + sin(x)^2 + 2*sin(x) + 1) - 1/12*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1)

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Fricas [B]  time = 2.73993, size = 231, normalized size = 6.42 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (-\frac{4 \, \cos \left (x\right )^{2} + 4 \, \sqrt{3} \sin \left (x\right ) - 7}{4 \, \cos \left (x\right )^{2} - 1}\right ) + \frac{1}{12} \, \log \left (2 \, \sin \left (x\right ) + 1\right ) + \frac{1}{12} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{12} \, \log \left (-\sin \left (x\right ) + 1\right ) - \frac{1}{12} \, \log \left (-2 \, \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(6*x)*sin(x),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*log(-(4*cos(x)^2 + 4*sqrt(3)*sin(x) - 7)/(4*cos(x)^2 - 1)) + 1/12*log(2*sin(x) + 1) + 1/12*log(si
n(x) + 1) - 1/12*log(-sin(x) + 1) - 1/12*log(-2*sin(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (x \right )} \csc{\left (6 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(6*x)*sin(x),x)

[Out]

Integral(sin(x)*csc(6*x), x)

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Giac [B]  time = 1.19834, size = 92, normalized size = 2.56 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (\frac{{\left | -4 \, \sqrt{3} + 8 \, \sin \left (x\right ) \right |}}{{\left | 4 \, \sqrt{3} + 8 \, \sin \left (x\right ) \right |}}\right ) + \frac{1}{12} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{12} \, \log \left (-\sin \left (x\right ) + 1\right ) + \frac{1}{12} \, \log \left ({\left | 2 \, \sin \left (x\right ) + 1 \right |}\right ) - \frac{1}{12} \, \log \left ({\left | 2 \, \sin \left (x\right ) - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(6*x)*sin(x),x, algorithm="giac")

[Out]

1/12*sqrt(3)*log(abs(-4*sqrt(3) + 8*sin(x))/abs(4*sqrt(3) + 8*sin(x))) + 1/12*log(sin(x) + 1) - 1/12*log(-sin(
x) + 1) + 1/12*log(abs(2*sin(x) + 1)) - 1/12*log(abs(2*sin(x) - 1))