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]

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

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Rubi [A]  time = 0.05, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 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 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 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

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*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 1.00 \[ \sinh (x)-\frac {1}{4} \tan ^{-1}(\sinh (x))-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*ArcTan[Sinh[x]] - ArcTan[Sqrt[2]*Sinh[x]]/(2*Sqrt[2]) + Sinh[x]

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fricas [B]  time = 0.50, size = 128, normalized size = 4.57 \[ -\frac {{\left (\sqrt {2} \cosh \relax (x) + \sqrt {2} \sinh \relax (x)\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} \cosh \relax (x) + \frac {1}{2} \, \sqrt {2} \sinh \relax (x)\right ) - {\left (\sqrt {2} \cosh \relax (x) + \sqrt {2} \sinh \relax (x)\right )} \arctan \left (-\frac {\sqrt {2} \cosh \relax (x)^{2} + 2 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x) + \sqrt {2} \sinh \relax (x)^{2} + \sqrt {2}}{2 \, {\left (\cosh \relax (x) - \sinh \relax (x)\right )}}\right ) + 2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} + 2}{4 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

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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giac [B]  time = 0.14, size = 54, normalized size = 1.93 \[ -\frac {1}{8} \, \pi - \frac {1}{8} \, \sqrt {2} {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} - \frac {1}{4} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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

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maple [B]  time = 0.31, size = 143, normalized size = 5.11 \[ -\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {\sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{-2+2 \sqrt {2}}\right )}{-4+4 \sqrt {2}}-\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{-2+2 \sqrt {2}}\right )}{-2+2 \sqrt {2}}-\frac {\sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{2 \left (2+2 \sqrt {2}\right )}-\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{2+2 \sqrt {2}}-\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{2}-\frac {1}{\tanh \left (\frac {x}{2}\right )-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(tanh(1/2*x)+1)+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*t
anh(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))*arcta
n(2*tanh(1/2*x)/(2+2*2^(1/2)))-1/2*arctan(tanh(1/2*x))-1/(tanh(1/2*x)-1)

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maxima [B]  time = 0.46, size = 60, normalized size = 2.14 \[ \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (-x\right )}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (-x\right )}\right )}\right ) + \frac {1}{2} \, \arctan \left (e^{\left (-x\right )}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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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mupad [B]  time = 1.43, size = 52, normalized size = 1.86 \[ \frac {{\mathrm {e}}^x}{2}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}-\frac {{\mathrm {e}}^{-x}}{2}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,{\mathrm {e}}^x}{2}+\frac {\sqrt {2}\,{\mathrm {e}}^{3\,x}}{2}\right )}{4}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,{\mathrm {e}}^x}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/2 - atan(exp(x))/2 - exp(-x)/2 - (2^(1/2)*atan((2^(1/2)*exp(x))/2 + (2^(1/2)*exp(3*x))/2))/4 - (2^(1/2)
*atan((2^(1/2)*exp(x))/2))/4

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \coth {\left (4 x \right )}\, dx \]

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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