\(\int e^{a+b x} \tanh (a+b x) \, dx\) [209]

Optimal result

Integrand size = 14, antiderivative size = 25 \[ \int e^{a+b x} \tanh (a+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 \arctan \left (e^{a+b x}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2320, 396, 209} \[ \int e^{a+b x} \tanh (a+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 \arctan \left (e^{a+b x}\right )}{b} \]

[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 209

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

Rule 396

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int e^{a+b x} \tanh (a+b x) \, dx=\frac {e^{a+b x}-2 \arctan \left (e^{a+b x}\right )}{b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\sinh \left (b x +a \right )-2 \arctan \left ({\mathrm e}^{b x +a}\right )+\cosh \left (b x +a \right )}{b}\)	\(27\)
default	\(\frac {\sinh \left (b x +a \right )-2 \arctan \left ({\mathrm e}^{b x +a}\right )+\cosh \left (b x +a \right )}{b}\)	\(27\)
risch	\(\frac {{\mathrm e}^{b x +a}}{b}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b}\)	\(44\)

[In]

int(exp(b*x+a)*tanh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int e^{a+b x} \tanh (a+b x) \, dx=-\frac {2 \, \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}{b} \]

[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

Sympy [F]

\[ \int e^{a+b x} \tanh (a+b x) \, dx=e^{a} \int e^{b x} \tanh {\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int e^{a+b x} \tanh (a+b x) \, dx=-\frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int e^{a+b x} \tanh (a+b x) \, dx=-\frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right ) - e^{\left (b x + a\right )}}{b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int e^{a+b x} \tanh (a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}} \]

[In]

int(exp(a + b*x)*tanh(a + b*x),x)

[Out]

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