3.3.30 \(\int e^{c (a+b x)} \tanh (d+e x) \, dx\) [230]

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

[Out]

exp(c*(b*x+a))/b/c-2*exp(c*(b*x+a))*hypergeom([1, 1/2*b*c/e],[1+1/2*b*c/e],-exp(2*e*x+2*d))/b/c

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 = {5592, 2225, 2283} \begin {gather*} \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} \end {gather*}

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 2225

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 2283

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

Rule 5592

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]

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] Leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(67)=134\).
time = 1.13, size = 141, normalized size = 2.10 \begin {gather*} \frac {e^{c (a+b x)} \left (2 b c e^{2 (d+e x)} \, _2F_1\left (1,1+\frac {b c}{2 e};2+\frac {b c}{2 e};-e^{2 (d+e x)}\right )-(b c+2 e) \left (1-e^{2 d}+2 e^{2 d} \, _2F_1\left (1,\frac {b c}{2 e};1+\frac {b c}{2 e};-e^{2 (d+e x)}\right )\right )\right )}{b c (b c+2 e) \left (1+e^{2 d}\right )} \end {gather*}

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)))

________________________________________________________________________________________

Maple [F]
time = 1.03, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{c \left (b x +a \right )} \tanh \left (e x +d \right )\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-(b*c*e^(a*c) - (b*c*e^(a*c + 2*d) - 2*e^(a*c + 2*d + 1))*e^(2*x*e) + 2*e^(a*c + 1))*e^(b*c*x)/(b^2*c^2 - 2*b*
c*e + (b^2*c^2*e^(2*d) - 2*b*c*e^(2*d + 1))*e^(2*x*e)) - 4*integrate(-e^(b*c*x + a*c + 1)/(b*c + (b*c*e^(4*d)
- 2*e^(4*d + 1))*e^(4*x*e) + 2*(b*c*e^(2*d) - 2*e^(2*d + 1))*e^(2*x*e) - 2*e), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(x*e + d), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a c} \int e^{b c x} \tanh {\left (d + e x \right )}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,\mathrm {tanh}\left (d+e\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*tanh(d + e*x),x)

[Out]

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

________________________________________________________________________________________