### 3.941 $$\int e^x \coth (2 x) \text{csch}(2 x) \, dx$$

Optimal. Leaf size=34 $\frac{e^{3 x}}{1-e^{4 x}}+\frac{1}{2} \tan ^{-1}\left (e^x\right )-\frac{1}{2} \tanh ^{-1}\left (e^x\right )$

[Out]

E^(3*x)/(1 - E^(4*x)) + ArcTan[E^x]/2 - ArcTanh[E^x]/2

________________________________________________________________________________________

Rubi [A]  time = 0.0283708, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {2282, 12, 457, 298, 203, 206} $\frac{e^{3 x}}{1-e^{4 x}}+\frac{1}{2} \tan ^{-1}\left (e^x\right )-\frac{1}{2} \tanh ^{-1}\left (e^x\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[E^x*Coth[2*x]*Csch[2*x],x]

[Out]

E^(3*x)/(1 - E^(4*x)) + ArcTan[E^x]/2 - ArcTanh[E^x]/2

Rule 2282

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]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 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])

Rule 206

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

Rubi steps

\begin{align*} \int e^x \coth (2 x) \text{csch}(2 x) \, dx &=\operatorname{Subst}\left (\int \frac{2 x^2 \left (1+x^4\right )}{\left (1-x^4\right )^2} \, dx,x,e^x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x^2 \left (1+x^4\right )}{\left (1-x^4\right )^2} \, dx,x,e^x\right )\\ &=\frac{e^{3 x}}{1-e^{4 x}}-\operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,e^x\right )\\ &=\frac{e^{3 x}}{1-e^{4 x}}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^x\right )\\ &=\frac{e^{3 x}}{1-e^{4 x}}+\frac{1}{2} \tan ^{-1}\left (e^x\right )-\frac{1}{2} \tanh ^{-1}\left (e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0558284, size = 31, normalized size = 0.91 $\frac{1}{2} \left (-\frac{2 e^{3 x}}{e^{4 x}-1}+\tan ^{-1}\left (e^x\right )-\tanh ^{-1}\left (e^x\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^x*Coth[2*x]*Csch[2*x],x]

[Out]

((-2*E^(3*x))/(-1 + E^(4*x)) + ArcTan[E^x] - ArcTanh[E^x])/2

________________________________________________________________________________________

Maple [C]  time = 0.061, size = 48, normalized size = 1.4 \begin{align*} -{\frac{{{\rm e}^{3\,x}}}{{{\rm e}^{4\,x}}-1}}-{\frac{\ln \left ({{\rm e}^{x}}+1 \right ) }{4}}+{\frac{\ln \left ({{\rm e}^{x}}-1 \right ) }{4}}+{\frac{i}{4}}\ln \left ({{\rm e}^{x}}+i \right ) -{\frac{i}{4}}\ln \left ({{\rm e}^{x}}-i \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*coth(2*x)*csch(2*x),x)

[Out]

-exp(3*x)/(exp(4*x)-1)-1/4*ln(exp(x)+1)+1/4*ln(exp(x)-1)+1/4*I*ln(exp(x)+I)-1/4*I*ln(exp(x)-I)

________________________________________________________________________________________

Maxima [A]  time = 1.55826, size = 46, normalized size = 1.35 \begin{align*} -\frac{e^{\left (3 \, x\right )}}{e^{\left (4 \, x\right )} - 1} + \frac{1}{2} \, \arctan \left (e^{x}\right ) - \frac{1}{4} \, \log \left (e^{x} + 1\right ) + \frac{1}{4} \, \log \left (e^{x} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(2*x)*csch(2*x),x, algorithm="maxima")

[Out]

-e^(3*x)/(e^(4*x) - 1) + 1/2*arctan(e^x) - 1/4*log(e^x + 1) + 1/4*log(e^x - 1)

________________________________________________________________________________________

Fricas [B]  time = 1.84365, size = 745, normalized size = 21.91 \begin{align*} -\frac{4 \, \cosh \left (x\right )^{3} + 12 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 12 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 4 \, \sinh \left (x\right )^{3} - 2 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) +{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{4 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(2*x)*csch(2*x),x, algorithm="fricas")

[Out]

-1/4*(4*cosh(x)^3 + 12*cosh(x)^2*sinh(x) + 12*cosh(x)*sinh(x)^2 + 4*sinh(x)^3 - 2*(cosh(x)^4 + 4*cosh(x)^3*sin
h(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*arctan(cosh(x) + sinh(x)) + (cosh(x)^4 + 4
*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*log(cosh(x) + sinh(x) + 1) -
(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*log(cosh(x) +
sinh(x) - 1))/(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \coth{\left (2 x \right )} \operatorname{csch}{\left (2 x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(2*x)*csch(2*x),x)

[Out]

Integral(exp(x)*coth(2*x)*csch(2*x), x)

________________________________________________________________________________________

Giac [A]  time = 1.16302, size = 47, normalized size = 1.38 \begin{align*} -\frac{e^{\left (3 \, x\right )}}{e^{\left (4 \, x\right )} - 1} + \frac{1}{2} \, \arctan \left (e^{x}\right ) - \frac{1}{4} \, \log \left (e^{x} + 1\right ) + \frac{1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(2*x)*csch(2*x),x, algorithm="giac")

[Out]

-e^(3*x)/(e^(4*x) - 1) + 1/2*arctan(e^x) - 1/4*log(e^x + 1) + 1/4*log(abs(e^x - 1))