3.208 \(\int \coth (4 x) \sinh (x) \, dx\)

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

[Out]

-ArcTan[Sinh[x]]/4 - ArcTan[Sqrt[2]*Sinh[x]]/(2*Sqrt[2]) + Sinh[x]

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Rubi [A]  time = 0.0531672, antiderivative size = 28, normalized size of antiderivative = 1., 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, 203} \[ \sinh (x)-\frac{1}{4} \tan ^{-1}(\sinh (x))-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[4*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/4 - ArcTan[Sqrt[2]*Sinh[x]]/(2*Sqrt[2]) + Sinh[x]

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

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

Mathematica [A]  time = 0.029861, size = 28, normalized size = 1. \[ \sinh (x)-\frac{1}{4} \tan ^{-1}(\sinh (x))-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[4*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/4 - ArcTan[Sqrt[2]*Sinh[x]]/(2*Sqrt[2]) + Sinh[x]

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Maple [B]  time = 0.092, size = 143, normalized size = 5.1 \begin{align*} -{\frac{\sqrt{2}}{4+4\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{2+2\,\sqrt{2}}} \right ) }-{\frac{1}{2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{2+2\,\sqrt{2}}} \right ) }+{\frac{\sqrt{2}}{-4+4\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{-2+2\,\sqrt{2}}} \right ) }-{\frac{1}{-2+2\,\sqrt{2}}\arctan \left ( 2\,{\frac{\tanh \left ( x/2 \right ) }{-2+2\,\sqrt{2}}} \right ) }-{\frac{1}{2}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }- \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}- \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(4*x)*sinh(x),x)

[Out]

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

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Maxima [B]  time = 1.56177, size = 81, normalized size = 2.89 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{\left (-x\right )}\right )}\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{\left (-x\right )}\right )}\right ) + \frac{1}{2} \, \arctan \left (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(4*x)*sinh(x),x, algorithm="maxima")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^(-x))) + 1/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^(-x))) +
 1/2*arctan(e^(-x)) - 1/2*e^(-x) + 1/2*e^x

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Fricas [B]  time = 2.31449, size = 493, normalized size = 17.61 \begin{align*} -\frac{{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \cosh \left (x\right ) + \frac{1}{2} \, \sqrt{2} \sinh \left (x\right )\right ) -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \arctan \left (-\frac{\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2} + \sqrt{2}}{2 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) + 2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} + 2}{4 \,{\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(4*x)*sinh(x),x, algorithm="fricas")

[Out]

-1/4*((sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*arctan(1/2*sqrt(2)*cosh(x) + 1/2*sqrt(2)*sinh(x)) - (sqrt(2)*cosh(x)
 + sqrt(2)*sinh(x))*arctan(-1/2*(sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 + sqrt(2))/
(cosh(x) - sinh(x))) + 2*(cosh(x) + sinh(x))*arctan(cosh(x) + sinh(x)) - 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*s
inh(x)^2 + 2)/(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 (4 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(4*x)*sinh(x),x)

[Out]

Integral(sinh(x)*coth(4*x), x)

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Giac [B]  time = 1.21276, size = 73, normalized size = 2.61 \begin{align*} -\frac{1}{8} \, \pi - \frac{1}{8} \, \sqrt{2}{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} - \frac{1}{4} \, \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(4*x)*sinh(x),x, algorithm="giac")

[Out]

-1/8*pi - 1/8*sqrt(2)*(pi + 2*arctan(1/2*sqrt(2)*(e^(2*x) - 1)*e^(-x))) - 1/4*arctan(1/2*(e^(2*x) - 1)*e^(-x))
 - 1/2*e^(-x) + 1/2*e^x