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

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

[Out]

-1/2*arctan(sinh(x))+sinh(x)

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ \sinh (x)-\frac {1}{2} \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[Sinh[x]] + Sinh[x]

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fricas [B]  time = 0.44, size = 42, normalized size = 4.20 \[ -\frac {2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - \cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) - \sinh \relax (x)^{2} + 1}{2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

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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giac [A]  time = 0.13, size = 16, normalized size = 1.60 \[ -\arctan \left (e^{x}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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

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maple [A]  time = 0.14, size = 9, normalized size = 0.90 \[ \sinh \relax (x )-\arctan \left ({\mathrm e}^{x}\right ) \]

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 = 0.44, size = 16, normalized size = 1.60 \[ \arctan \left (e^{\left (-x\right )}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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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mupad [B]  time = 0.05, size = 16, normalized size = 1.60 \[ \frac {{\mathrm {e}}^x}{2}-\mathrm {atan}\left ({\mathrm {e}}^x\right )-\frac {{\mathrm {e}}^{-x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/2 - atan(exp(x)) - exp(-x)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \coth {\left (2 x \right )}\, dx \]

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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