\(\int e^{a+b x} \text {csch}^2(a+b x) \, dx\) [306]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 42 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 12, 294, 212} \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b} \]

[In]

Int[E^(a + b*x)*Csch[a + b*x]^2,x]

[Out]

(2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (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 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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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}& = \frac {\text {Subst}\left (\int \frac {4 x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {4 \text {Subst}\left (\int \frac {x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=\frac {-\frac {2 e^{a+b x}}{-1+e^{2 (a+b x)}}-2 \text {arctanh}\left (e^{a+b x}\right )}{b} \]

[In]

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

[Out]

((-2*E^(a + b*x))/(-1 + E^(2*(a + b*x))) - 2*ArcTanh[E^(a + b*x)])/b

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.60

method result size
derivativedivides \(\frac {-2 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )-\frac {1}{\sinh \left (b x +a \right )}}{b}\) \(25\)
default \(\frac {-2 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )-\frac {1}{\sinh \left (b x +a \right )}}{b}\) \(25\)
risch \(-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b}\) \(53\)

[In]

int(exp(b*x+a)*csch(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (37) = 74\).

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 3.74 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=-\frac {{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=e^{a} \int e^{b x} \operatorname {csch}^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=-\frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b} - \frac {2 \, e^{\left (b x + a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=-\frac {\frac {2 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + \log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int e^{a+b x} \text {csch}^2(a+b x) \, dx=-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

[In]

int(exp(a + b*x)/sinh(a + b*x)^2,x)

[Out]

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