3.235 \(\int \cosh (x) \coth (3 x) \, dx\)

Optimal. Leaf size=45 \[ \cosh (x)+\frac {1}{6} \log (1-2 \cosh (x))+\frac {1}{6} \log (1-\cosh (x))-\frac {1}{6} \log (\cosh (x)+1)-\frac {1}{6} \log (2 \cosh (x)+1) \]

[Out]

cosh(x)+1/6*ln(1-2*cosh(x))+1/6*ln(1-cosh(x))-1/6*ln(1+cosh(x))-1/6*ln(1+2*cosh(x))

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1279, 1161, 616, 31} \[ \cosh (x)+\frac {1}{6} \log (1-2 \cosh (x))+\frac {1}{6} \log (1-\cosh (x))-\frac {1}{6} \log (\cosh (x)+1)-\frac {1}{6} \log (2 \cosh (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Coth[3*x],x]

[Out]

Cosh[x] + Log[1 - 2*Cosh[x]]/6 + Log[1 - Cosh[x]]/6 - Log[1 + Cosh[x]]/6 - Log[1 + 2*Cosh[x]]/6

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])

Rubi steps

\begin {align*} \int \cosh (x) \coth (3 x) \, dx &=-\operatorname {Subst}\left (\int \frac {x^2 \left (3-4 x^2\right )}{1-5 x^2+4 x^4} \, dx,x,\cosh (x)\right )\\ &=\cosh (x)+\frac {1}{4} \operatorname {Subst}\left (\int \frac {-4+8 x^2}{1-5 x^2+4 x^4} \, dx,x,\cosh (x)\right )\\ &=\cosh (x)+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2}-\frac {x}{2}+x^2} \, dx,x,\cosh (x)\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2}+\frac {x}{2}+x^2} \, dx,x,\cosh (x)\right )\\ &=\cosh (x)+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,\cosh (x)\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2}+x} \, dx,x,\cosh (x)\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{2}+x} \, dx,x,\cosh (x)\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\cosh (x)\right )\\ &=\cosh (x)+\frac {1}{6} \log (1-2 \cosh (x))+\frac {1}{6} \log (1-\cosh (x))-\frac {1}{6} \log (1+\cosh (x))-\frac {1}{6} \log (1+2 \cosh (x))\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.04 \[ \cosh (x)+\frac {1}{3} \log \left (\sinh \left (\frac {x}{2}\right )\right )-\frac {1}{3} \log \left (\cosh \left (\frac {x}{2}\right )\right )+\frac {1}{6} \log (1-2 \cosh (x))-\frac {1}{6} \log (2 \cosh (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Coth[3*x],x]

[Out]

Cosh[x] - Log[Cosh[x/2]]/3 + Log[1 - 2*Cosh[x]]/6 - Log[1 + 2*Cosh[x]]/6 + Log[Sinh[x/2]]/3

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fricas [B]  time = 0.44, size = 104, normalized size = 2.31 \[ \frac {3 \, \cosh \relax (x)^{2} - {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\frac {2 \, \cosh \relax (x) + 1}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\frac {2 \, \cosh \relax (x) - 1}{\cosh \relax (x) - \sinh \relax (x)}\right ) - 2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + 2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) + 6 \, \cosh \relax (x) \sinh \relax (x) + 3 \, \sinh \relax (x)^{2} + 3}{6 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*coth(3*x),x, algorithm="fricas")

[Out]

1/6*(3*cosh(x)^2 - (cosh(x) + sinh(x))*log((2*cosh(x) + 1)/(cosh(x) - sinh(x))) + (cosh(x) + sinh(x))*log((2*c
osh(x) - 1)/(cosh(x) - sinh(x))) - 2*(cosh(x) + sinh(x))*log(cosh(x) + sinh(x) + 1) + 2*(cosh(x) + sinh(x))*lo
g(cosh(x) + sinh(x) - 1) + 6*cosh(x)*sinh(x) + 3*sinh(x)^2 + 3)/(cosh(x) + sinh(x))

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giac [A]  time = 0.12, size = 55, normalized size = 1.22 \[ \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} - \frac {1}{6} \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) - \frac {1}{6} \, \log \left (e^{\left (-x\right )} + e^{x} + 1\right ) + \frac {1}{6} \, \log \left (e^{\left (-x\right )} + e^{x} - 1\right ) + \frac {1}{6} \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*coth(3*x),x, algorithm="giac")

[Out]

1/2*e^(-x) + 1/2*e^x - 1/6*log(e^(-x) + e^x + 2) - 1/6*log(e^(-x) + e^x + 1) + 1/6*log(e^(-x) + e^x - 1) + 1/6
*log(e^(-x) + e^x - 2)

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maple [A]  time = 0.24, size = 50, normalized size = 1.11 \[ \frac {{\mathrm e}^{x}}{2}+\frac {{\mathrm e}^{-x}}{2}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{3}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{3}+\frac {\ln \left (1-{\mathrm e}^{x}+{\mathrm e}^{2 x}\right )}{6}-\frac {\ln \left (1+{\mathrm e}^{x}+{\mathrm e}^{2 x}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*coth(3*x),x)

[Out]

1/2*exp(x)+1/2*exp(-x)+1/3*ln(exp(x)-1)-1/3*ln(exp(x)+1)+1/6*ln(1-exp(x)+exp(2*x))-1/6*ln(1+exp(x)+exp(2*x))

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maxima [A]  time = 0.44, size = 57, normalized size = 1.27 \[ \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} - \frac {1}{6} \, \log \left (e^{\left (-x\right )} + e^{\left (-2 \, x\right )} + 1\right ) - \frac {1}{3} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {1}{3} \, \log \left (e^{\left (-x\right )} - 1\right ) + \frac {1}{6} \, \log \left (-e^{\left (-x\right )} + e^{\left (-2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*coth(3*x),x, algorithm="maxima")

[Out]

1/2*e^(-x) + 1/2*e^x - 1/6*log(e^(-x) + e^(-2*x) + 1) - 1/3*log(e^(-x) + 1) + 1/3*log(e^(-x) - 1) + 1/6*log(-e
^(-x) + e^(-2*x) + 1)

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mupad [B]  time = 0.06, size = 57, normalized size = 1.27 \[ \frac {\ln \left (6-6\,{\mathrm {e}}^x\right )}{3}-\frac {\ln \left (-6\,{\mathrm {e}}^x-6\right )}{3}+\frac {{\mathrm {e}}^{-x}}{2}+\frac {\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{2\,x}-1\right )}{6}-\frac {\ln \left (-{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x-1\right )}{6}+\frac {{\mathrm {e}}^x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(3*x)*cosh(x),x)

[Out]

log(6 - 6*exp(x))/3 - log(- 6*exp(x) - 6)/3 + exp(-x)/2 + log(exp(x) - exp(2*x) - 1)/6 - log(- exp(2*x) - exp(
x) - 1)/6 + exp(x)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh {\relax (x )} \coth {\left (3 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*coth(3*x),x)

[Out]

Integral(cosh(x)*coth(3*x), x)

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