### 3.210 $$\int \coth (6 x) \sinh (x) \, dx$$

Optimal. Leaf size=38 $\sinh (x)-\frac{1}{6} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tan ^{-1}(2 \sinh (x))-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{3}}\right )}{2 \sqrt{3}}$

[Out]

-ArcTan[Sinh[x]]/6 - ArcTan[2*Sinh[x]]/6 - ArcTan[(2*Sinh[x])/Sqrt[3]]/(2*Sqrt[3]) + Sinh[x]

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Rubi [A]  time = 0.0773604, antiderivative size = 38, 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, 2073, 203} $\sinh (x)-\frac{1}{6} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tan ^{-1}(2 \sinh (x))-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{3}}\right )}{2 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[6*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/6 - ArcTan[2*Sinh[x]]/6 - ArcTan[(2*Sinh[x])/Sqrt[3]]/(2*Sqrt[3]) + Sinh[x]

Rule 12

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

Rule 2073

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

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 \coth (6 x) \sinh (x) \, dx &=\operatorname{Subst}\left (\int \frac{1+18 x^2+48 x^4+32 x^6}{2 \left (3+19 x^2+32 x^4+16 x^6\right )} \, dx,x,\sinh (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1+18 x^2+48 x^4+32 x^6}{3+19 x^2+32 x^4+16 x^6} \, dx,x,\sinh (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (2-\frac{1}{3 \left (1+x^2\right )}-\frac{2}{3 \left (1+4 x^2\right )}-\frac{2}{3+4 x^2}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1+4 x^2} \, dx,x,\sinh (x)\right )-\operatorname{Subst}\left (\int \frac{1}{3+4 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{6} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tan ^{-1}(2 \sinh (x))-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{3}}\right )}{2 \sqrt{3}}+\sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.0578494, size = 38, normalized size = 1. $\sinh (x)-\frac{1}{6} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tan ^{-1}(2 \sinh (x))-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{3}}\right )}{2 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[6*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/6 - ArcTan[2*Sinh[x]]/6 - ArcTan[(2*Sinh[x])/Sqrt[3]]/(2*Sqrt[3]) + Sinh[x]

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Maple [B]  time = 0.112, size = 172, normalized size = 4.5 \begin{align*}{\frac{\sqrt{3}}{12-6\,\sqrt{3}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{4-2\,\sqrt{3}}} \right ) }-{\frac{2}{12-6\,\sqrt{3}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{4-2\,\sqrt{3}}} \right ) }-{\frac{\sqrt{3}}{12+6\,\sqrt{3}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{4+2\,\sqrt{3}}} \right ) }-{\frac{2}{12+6\,\sqrt{3}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{4+2\,\sqrt{3}}} \right ) }-{\frac{\sqrt{3}}{6}\arctan \left ({\frac{\sqrt{3}}{3}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{1}{3}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }- \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}-{\frac{\sqrt{3}}{6}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \sqrt{3} \right ) }- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(6*x)*sinh(x),x)

[Out]

1/3*3^(1/2)/(4-2*3^(1/2))*arctan(2*tanh(1/2*x)/(4-2*3^(1/2)))-2/3/(4-2*3^(1/2))*arctan(2*tanh(1/2*x)/(4-2*3^(1
/2)))-1/3*3^(1/2)/(4+2*3^(1/2))*arctan(2*tanh(1/2*x)/(4+2*3^(1/2)))-2/3/(4+2*3^(1/2))*arctan(2*tanh(1/2*x)/(4+
2*3^(1/2)))-1/6*3^(1/2)*arctan(1/3*tanh(1/2*x)*3^(1/2))-1/3*arctan(tanh(1/2*x))-1/(tanh(1/2*x)-1)-1/6*3^(1/2)*
arctan(tanh(1/2*x)*3^(1/2))-1/(tanh(1/2*x)+1)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(6*x)*sinh(x),x, algorithm="maxima")

[Out]

1/2*(e^(2*x) - 1)*e^(-x) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^x + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^x
- 1)) - 1/3*arctan(e^x) - 1/2*integrate(1/3*(e^(3*x) + e^x)/(e^(4*x) - e^(2*x) + 1), x)

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Fricas [B]  time = 2.23425, size = 624, normalized size = 16.42 \begin{align*} -\frac{{\left (\sqrt{3} \cosh \left (x\right ) + \sqrt{3} \sinh \left (x\right )\right )} \arctan \left (\frac{1}{3} \, \sqrt{3} \cosh \left (x\right ) + \frac{1}{3} \, \sqrt{3} \sinh \left (x\right )\right ) -{\left (\sqrt{3} \cosh \left (x\right ) + \sqrt{3} \sinh \left (x\right )\right )} \arctan \left (-\frac{\sqrt{3} \cosh \left (x\right )^{2} + 2 \, \sqrt{3} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{3} \sinh \left (x\right )^{2} + 2 \, \sqrt{3}}{3 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) -{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (-\frac{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 3 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 3 \, \cosh \left (x\right )^{2} - 6 \, \cosh \left (x\right ) \sinh \left (x\right ) - 3 \, \sinh \left (x\right )^{2} + 3}{6 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(6*x)*sinh(x),x, algorithm="fricas")

[Out]

-1/6*((sqrt(3)*cosh(x) + sqrt(3)*sinh(x))*arctan(1/3*sqrt(3)*cosh(x) + 1/3*sqrt(3)*sinh(x)) - (sqrt(3)*cosh(x)
+ sqrt(3)*sinh(x))*arctan(-1/3*(sqrt(3)*cosh(x)^2 + 2*sqrt(3)*cosh(x)*sinh(x) + sqrt(3)*sinh(x)^2 + 2*sqrt(3)
)/(cosh(x) - sinh(x))) - (cosh(x) + sinh(x))*arctan(-(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)/(cosh(x) - si
nh(x))) + 3*(cosh(x) + sinh(x))*arctan(cosh(x) + sinh(x)) - 3*cosh(x)^2 - 6*cosh(x)*sinh(x) - 3*sinh(x)^2 + 3)
/(cosh(x) + sinh(x))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(6*x)*sinh(x),x)

[Out]

Integral(sinh(x)*coth(6*x), x)

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Giac [B]  time = 1.2043, size = 92, normalized size = 2.42 \begin{align*} -\frac{1}{6} \, \pi - \frac{1}{12} \, \sqrt{3}{\left (\pi + 2 \, \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} - \frac{1}{6} \, \arctan \left ({\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac{1}{6} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(6*x)*sinh(x),x, algorithm="giac")

[Out]

-1/6*pi - 1/12*sqrt(3)*(pi + 2*arctan(1/3*sqrt(3)*(e^(2*x) - 1)*e^(-x))) - 1/6*arctan((e^(2*x) - 1)*e^(-x)) -
1/6*arctan(1/2*(e^(2*x) - 1)*e^(-x)) - 1/2*e^(-x) + 1/2*e^x