3.209 \(\int e^{a+b x} \tanh (a+b x) \, dx\)

Optimal. Leaf size=25 \[ \frac{e^{a+b x}}{b}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0148009, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2282, 388, 203} \[ \frac{e^{a+b x}}{b}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Tanh[a + b*x],x]

[Out]

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

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 388

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0144078, size = 22, normalized size = 0.88 \[ \frac{e^{a+b x}-2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 34, normalized size = 1.4 \begin{align*}{\frac{\sinh \left ( bx+a \right ) }{b}}-2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}}+{\frac{\cosh \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*tanh(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.61474, size = 31, normalized size = 1.24 \begin{align*} -\frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac{e^{\left (b x + a\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.92056, size = 105, normalized size = 4.2 \begin{align*} -\frac{2 \, \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \tanh{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a),x)

[Out]

exp(a)*Integral(exp(b*x)*tanh(a + b*x), x)

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Giac [A]  time = 1.52603, size = 31, normalized size = 1.24 \begin{align*} -\frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right ) - e^{\left (b x + a\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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