3.230 \(\int e^{c (a+b x)} \tanh (d+e x) \, dx\)

Optimal. Leaf size=67 \[ \frac{e^{c (a+b x)}}{b c}-\frac{2 e^{c (a+b x)} \, _2F_1\left (1,\frac{b c}{2 e};\frac{b c}{2 e}+1;-e^{2 (d+e x)}\right )}{b c} \]

[Out]

E^(c*(a + b*x))/(b*c) - (2*E^(c*(a + b*x))*Hypergeometric2F1[1, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*(d + e*x))
])/(b*c)

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Rubi [A]  time = 0.0708198, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {5484, 2194, 2251} \[ \frac{e^{c (a+b x)}}{b c}-\frac{2 e^{c (a+b x)} \, _2F_1\left (1,\frac{b c}{2 e};\frac{b c}{2 e}+1;-e^{2 (d+e x)}\right )}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Tanh[d + e*x],x]

[Out]

E^(c*(a + b*x))/(b*c) - (2*E^(c*(a + b*x))*Hypergeometric2F1[1, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*(d + e*x))
])/(b*c)

Rule 5484

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tanh[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandIntegrand[(F^(c*(
a + b*x))*(-1 + E^(2*(d + e*x)))^n)/(1 + E^(2*(d + e*x)))^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && Integer
Q[n]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int e^{c (a+b x)} \tanh (d+e x) \, dx &=\int \left (e^{c (a+b x)}-\frac{2 e^{c (a+b x)}}{1+e^{2 (d+e x)}}\right ) \, dx\\ &=-\left (2 \int \frac{e^{c (a+b x)}}{1+e^{2 (d+e x)}} \, dx\right )+\int e^{c (a+b x)} \, dx\\ &=\frac{e^{c (a+b x)}}{b c}-\frac{2 e^{c (a+b x)} \, _2F_1\left (1,\frac{b c}{2 e};1+\frac{b c}{2 e};-e^{2 (d+e x)}\right )}{b c}\\ \end{align*}

Mathematica [B]  time = 1.80816, size = 141, normalized size = 2.1 \[ \frac{e^{c (a+b x)} \left (2 b c e^{2 (d+e x)} \, _2F_1\left (1,\frac{b c}{2 e}+1;\frac{b c}{2 e}+2;-e^{2 (d+e x)}\right )-(b c+2 e) \left (2 e^{2 d} \, _2F_1\left (1,\frac{b c}{2 e};\frac{b c}{2 e}+1;-e^{2 (d+e x)}\right )-e^{2 d}+1\right )\right )}{b c \left (e^{2 d}+1\right ) (b c+2 e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c*(a + b*x))*(2*b*c*E^(2*(d + e*x))*Hypergeometric2F1[1, 1 + (b*c)/(2*e), 2 + (b*c)/(2*e), -E^(2*(d + e*x)
)] - (b*c + 2*e)*(1 - E^(2*d) + 2*E^(2*d)*Hypergeometric2F1[1, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*(d + e*x))]
)))/(b*c*(b*c + 2*e)*(1 + E^(2*d)))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c \left ( bx+a \right ) }}\tanh \left ( ex+d \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*tanh(e*x+d),x)

[Out]

int(exp(c*(b*x+a))*tanh(e*x+d),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 4 \, e \int \frac{e^{\left (b c x + a c\right )}}{b c +{\left (b c e^{\left (4 \, d\right )} - 2 \, e e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 2 \,{\left (b c e^{\left (2 \, d\right )} - 2 \, e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} - 2 \, e}\,{d x} - \frac{{\left (b c e^{\left (a c\right )} + 2 \, e e^{\left (a c\right )} -{\left (b c e^{\left (a c + 2 \, d\right )} - 2 \, e e^{\left (a c + 2 \, d\right )}\right )} e^{\left (2 \, e x\right )}\right )} e^{\left (b c x\right )}}{b^{2} c^{2} - 2 \, b c e +{\left (b^{2} c^{2} e^{\left (2 \, d\right )} - 2 \, b c e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e*integrate(e^(b*c*x + a*c)/(b*c + (b*c*e^(4*d) - 2*e*e^(4*d))*e^(4*e*x) + 2*(b*c*e^(2*d) - 2*e*e^(2*d))*e^(
2*e*x) - 2*e), x) - (b*c*e^(a*c) + 2*e*e^(a*c) - (b*c*e^(a*c + 2*d) - 2*e*e^(a*c + 2*d))*e^(2*e*x))*e^(b*c*x)/
(b^2*c^2 - 2*b*c*e + (b^2*c^2*e^(2*d) - 2*b*c*e*e^(2*d))*e^(2*e*x))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (e^{\left (b c x + a c\right )} \tanh \left (e x + d\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(b*c*x + a*c)*tanh(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*tanh(e*x+d),x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*tanh(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^((b*x + a)*c)*tanh(e*x + d), x)