3.208 \(\int e^{a+b x} \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=51 \[ \frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0349531, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 390, 288, 203} \[ \frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Tanh[a + b*x]^2,x]

[Out]

E^(a + b*x)/b + (2*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))) - (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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 288

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

Mathematica [A]  time = 0.0875275, size = 40, normalized size = 0.78 \[ \frac{e^{a+b x} \left (\frac{2}{e^{2 (a+b x)}+1}+1\right )-2 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 56, normalized size = 1.1 \begin{align*} -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{b\cosh \left ( bx+a \right ) }}+2\,{\frac{\cosh \left ( bx+a \right ) }{b}}+{\frac{\sinh \left ( bx+a \right ) }{b}}-2\,{\frac{\arctan \left ({{\rm e}^{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)^2,x)

[Out]

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

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Maxima [A]  time = 1.59225, size = 63, 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} + \frac{2 \, e^{\left (b x + a\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.98645, size = 433, normalized size = 8.49 \begin{align*} \frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} - 2 \,{\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 )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 3 \,{\left (\cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + 3 \, \cosh \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 - 2*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*si
nh(b*x + a) + sinh(b*x + a)^2 + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + 3*(cosh(b*x + a)^2 + 1)*sinh(b*x +
a) + 3*cosh(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \tanh ^{2}{\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)**2,x)

[Out]

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

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Giac [A]  time = 2.20872, size = 55, normalized size = 1.08 \begin{align*} \frac{\frac{2 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1} - 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)^2,x, algorithm="giac")

[Out]

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