### 3.206 $$\int \coth (2 x) \sinh (x) \, dx$$

Optimal. Leaf size=10 $\sinh (x)-\frac{1}{2} \tan ^{-1}(\sinh (x))$

[Out]

-ArcTan[Sinh[x]]/2 + Sinh[x]

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Rubi [A]  time = 0.0230417, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {388, 203} $\sinh (x)-\frac{1}{2} \tan ^{-1}(\sinh (x))$

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[2*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/2 + Sinh[x]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 (2 x) \sinh (x) \, dx &=\operatorname{Subst}\left (\int \frac{1+2 x^2}{2+2 x^2} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname{Subst}\left (\int \frac{1}{2+2 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{2} \tan ^{-1}(\sinh (x))+\sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.0084669, size = 10, normalized size = 1. $\sinh (x)-\frac{1}{2} \tan ^{-1}(\sinh (x))$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[2*x]*Sinh[x],x]

[Out]

-ArcTan[Sinh[x]]/2 + Sinh[x]

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Maple [A]  time = 0.019, size = 9, normalized size = 0.9 \begin{align*} \sinh \left ( x \right ) -\arctan \left ({{\rm e}^{x}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(2*x)*sinh(x),x)

[Out]

sinh(x)-arctan(exp(x))

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Maxima [A]  time = 1.51668, size = 22, normalized size = 2.2 \begin{align*} \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(2*x)*sinh(x),x, algorithm="maxima")

[Out]

arctan(e^(-x)) - 1/2*e^(-x) + 1/2*e^x

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Fricas [B]  time = 2.05331, size = 167, normalized size = 16.7 \begin{align*} -\frac{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \cosh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) \sinh \left (x\right ) - \sinh \left (x\right )^{2} + 1}{2 \,{\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(2*x)*sinh(x),x, algorithm="fricas")

[Out]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(2*x)*sinh(x),x)

[Out]

Integral(sinh(x)*coth(2*x), x)

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Giac [A]  time = 1.15817, size = 22, normalized size = 2.2 \begin{align*} -\arctan \left (e^{x}\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(2*x)*sinh(x),x, algorithm="giac")

[Out]

-arctan(e^x) - 1/2*e^(-x) + 1/2*e^x