\(\int e^x \coth (3 x) \, dx\) [220]

Optimal result

Integrand size = 8, antiderivative size = 85 \[ \int e^x \coth (3 x) \, dx=e^x+\frac {\arctan \left (\frac {1-2 e^x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1+2 e^x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \text {arctanh}\left (e^x\right )}{3}+\frac {1}{6} \log \left (1-e^x+e^{2 x}\right )-\frac {1}{6} \log \left (1+e^x+e^{2 x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2320, 396, 216, 648, 632, 210, 642, 212} \[ \int e^x \coth (3 x) \, dx=\frac {\arctan \left (\frac {1-2 e^x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 e^x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \text {arctanh}\left (e^x\right )}{3}+e^x+\frac {1}{6} \log \left (-e^x+e^{2 x}+1\right )-\frac {1}{6} \log \left (e^x+e^{2 x}+1\right ) \]

[In]

Int[E^x*Coth[3*x],x]

[Out]

E^x + ArcTan[(1 - 2*E^x)/Sqrt[3]]/Sqrt[3] - ArcTan[(1 + 2*E^x)/Sqrt[3]]/Sqrt[3] - (2*ArcTanh[E^x])/3 + Log[1 -
E^x + E^(2*x)]/6 - Log[1 + E^x + E^(2*x)]/6

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a/b, n]], s = Denominator[Rt[-a/b, n
]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 396

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 632

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

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-1-x^6}{1-x^6} \, dx,x,e^x\right ) \\ & = e^x-2 \text {Subst}\left (\int \frac {1}{1-x^6} \, dx,x,e^x\right ) \\ & = e^x-\frac {2}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {2}{3} \text {Subst}\left (\int \frac {1-\frac {x}{2}}{1-x+x^2} \, dx,x,e^x\right )-\frac {2}{3} \text {Subst}\left (\int \frac {1+\frac {x}{2}}{1+x+x^2} \, dx,x,e^x\right ) \\ & = e^x-\frac {2 \text {arctanh}\left (e^x\right )}{3}+\frac {1}{6} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,e^x\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,e^x\right ) \\ & = e^x-\frac {2 \text {arctanh}\left (e^x\right )}{3}+\frac {1}{6} \log \left (1-e^x+e^{2 x}\right )-\frac {1}{6} \log \left (1+e^x+e^{2 x}\right )+\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 e^x\right )+\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 e^x\right ) \\ & = e^x-\frac {\arctan \left (\frac {-1+2 e^x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1+2 e^x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \text {arctanh}\left (e^x\right )}{3}+\frac {1}{6} \log \left (1-e^x+e^{2 x}\right )-\frac {1}{6} \log \left (1+e^x+e^{2 x}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.26 \[ \int e^x \coth (3 x) \, dx=e^x-2 e^x \operatorname {Hypergeometric2F1}\left (\frac {1}{6},1,\frac {7}{6},e^{6 x}\right ) \]

[In]

Integrate[E^x*Coth[3*x],x]

[Out]

E^x - 2*E^x*Hypergeometric2F1[1/6, 1, 7/6, E^(6*x)]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.62

method	result	size
risch	\({\mathrm e}^{x}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{3}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{6}+\frac {i \ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{6}-\frac {i \ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{6}+\frac {i \ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{6}-\frac {i \ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{3}\)	\(138\)

[In]

int(exp(x)*coth(3*x),x,method=_RETURNVERBOSE)

[Out]

exp(x)+1/3*ln(exp(x)-1)+1/6*ln(exp(x)-1/2-1/2*I*3^(1/2))+1/6*I*ln(exp(x)-1/2-1/2*I*3^(1/2))*3^(1/2)+1/6*ln(exp
(x)-1/2+1/2*I*3^(1/2))-1/6*I*ln(exp(x)-1/2+1/2*I*3^(1/2))*3^(1/2)-1/6*ln(exp(x)+1/2-1/2*I*3^(1/2))+1/6*I*ln(ex
p(x)+1/2-1/2*I*3^(1/2))*3^(1/2)-1/6*ln(exp(x)+1/2+1/2*I*3^(1/2))-1/6*I*ln(exp(x)+1/2+1/2*I*3^(1/2))*3^(1/2)-1/
3*ln(exp(x)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33 \[ \int e^x \coth (3 x) \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \cosh \left (x\right ) + \frac {2}{3} \, \sqrt {3} \sinh \left (x\right ) + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \cosh \left (x\right ) + \frac {2}{3} \, \sqrt {3} \sinh \left (x\right ) - \frac {1}{3} \, \sqrt {3}\right ) + \cosh \left (x\right ) - \frac {1}{6} \, \log \left (\frac {2 \, \cosh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \frac {1}{6} \, \log \left (\frac {2 \, \cosh \left (x\right ) - 1}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - \frac {1}{3} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{3} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + \sinh \left (x\right ) \]

[In]

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

[Out]

-1/3*sqrt(3)*arctan(2/3*sqrt(3)*cosh(x) + 2/3*sqrt(3)*sinh(x) + 1/3*sqrt(3)) - 1/3*sqrt(3)*arctan(2/3*sqrt(3)*
cosh(x) + 2/3*sqrt(3)*sinh(x) - 1/3*sqrt(3)) + cosh(x) - 1/6*log((2*cosh(x) + 1)/(cosh(x) - sinh(x))) + 1/6*lo
g((2*cosh(x) - 1)/(cosh(x) - sinh(x))) - 1/3*log(cosh(x) + sinh(x) + 1) + 1/3*log(cosh(x) + sinh(x) - 1) + sin
h(x)

Sympy [F]

\[ \int e^x \coth (3 x) \, dx=\int e^{x} \coth {\left (3 x \right )}\, dx \]

[In]

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

[Out]

Integral(exp(x)*coth(3*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int e^x \coth (3 x) \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) + e^{x} - \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{3} \, \log \left (e^{x} + 1\right ) + \frac {1}{3} \, \log \left (e^{x} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int e^x \coth (3 x) \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) + e^{x} - \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{3} \, \log \left (e^{x} + 1\right ) + \frac {1}{3} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.84 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int e^x \coth (3 x) \, dx=\frac {\ln \left (2-2\,{\mathrm {e}}^x\right )}{3}-\frac {\ln \left (-2\,{\mathrm {e}}^x-2\right )}{3}+\frac {\ln \left ({\left (2\,{\mathrm {e}}^x-1\right )}^2+3\right )}{6}-\frac {\ln \left ({\left (2\,{\mathrm {e}}^x+1\right )}^2+3\right )}{6}+{\mathrm {e}}^x-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (2\,{\mathrm {e}}^x-1\right )}{3}\right )}{3}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (2\,{\mathrm {e}}^x+1\right )}{3}\right )}{3} \]

[In]

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

[Out]

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