3.312 \(\int e^x \text {csch}(2 x) \, dx\)

Optimal. Leaf size=11 \[ \tan ^{-1}\left (e^x\right )-\tanh ^{-1}\left (e^x\right ) \]

[Out]

arctan(exp(x))-arctanh(exp(x))

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2282, 12, 298, 203, 206} \[ \tan ^{-1}\left (e^x\right )-\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[E^x] - ArcTanh[E^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 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])

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 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]]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \[ \tan ^{-1}\left (e^x\right )-\tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[E^x] - ArcTanh[E^x]

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fricas [B]  time = 0.76, size = 25, normalized size = 2.27 \[ \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - \frac {1}{2} \, \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + \frac {1}{2} \, \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 19, normalized size = 1.73 \[ \arctan \left (e^{x}\right ) - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.08, size = 34, normalized size = 3.09 \[ -\frac {\ln \left ({\mathrm e}^{x}+1\right )}{2}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{2}+\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(exp(x)+1)+1/2*ln(exp(x)-1)+1/2*I*ln(exp(x)+I)-1/2*I*ln(exp(x)-I)

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maxima [A]  time = 0.41, size = 18, normalized size = 1.64 \[ \arctan \left (e^{x}\right ) - \frac {1}{2} \, \log \left (e^{x} + 1\right ) + \frac {1}{2} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 26, normalized size = 2.36 \[ \frac {\ln \left (4\,{\mathrm {e}}^x-4\right )}{2}-\frac {\ln \left (-4\,{\mathrm {e}}^x-4\right )}{2}-\mathrm {atan}\left ({\mathrm {e}}^{-x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/sinh(2*x),x)

[Out]

log(4*exp(x) - 4)/2 - log(- 4*exp(x) - 4)/2 - atan(exp(-x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \operatorname {csch}{\left (2 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*csch(2*x), x)

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