3.875 \(\int e^{2 (a+b x)} \text {csch}(a+b x) \, dx\)

Optimal. Leaf size=26 \[ \frac {2 e^{a+b x}}{b}-\frac {2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

2*exp(b*x+a)/b-2*arctanh(exp(b*x+a))/b

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2282, 12, 321, 207} \[ \frac {2 e^{a+b x}}{b}-\frac {2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Csch[a + b*x],x]

[Out]

(2*E^(a + b*x))/b - (2*ArcTanh[E^(a + b*x)])/b

Rule 12

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

Rule 207

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.88 \[ \frac {2 \left (e^{a+b x}-\tanh ^{-1}\left (e^{a+b x}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*(a + b*x))*Csch[a + b*x],x]

[Out]

(2*(E^(a + b*x) - ArcTanh[E^(a + b*x)]))/b

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fricas [B]  time = 0.51, size = 53, normalized size = 2.04 \[ \frac {2 \, \cosh \left (b x + a\right ) - \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, \sinh \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.12, size = 50, normalized size = 1.92 \[ -\frac {{\left (e^{\left (-2 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) - e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right ) - 2 \, e^{\left (b x - a\right )}\right )} e^{\left (2 \, a\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 40, normalized size = 1.54 \[ \frac {2 \,{\mathrm e}^{b x +a}}{b}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*csch(b*x+a),x)

[Out]

2*exp(b*x+a)/b+1/b*ln(exp(b*x+a)-1)-1/b*ln(1+exp(b*x+a))

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maxima [A]  time = 0.31, size = 45, normalized size = 1.73 \[ \frac {2 \, e^{\left (b x + a\right )}}{b} - \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 39, normalized size = 1.50 \[ \frac {2\,{\mathrm {e}}^{a+b\,x}}{b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*a + 2*b*x)/sinh(a + b*x),x)

[Out]

(2*exp(a + b*x))/b - (2*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*csch(b*x+a),x)

[Out]

exp(2*a)*Integral(exp(2*b*x)*csch(a + b*x), x)

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