### 3.191 $$\int \cosh (a+b x) \tanh (c+d x) \, dx$$

Optimal. Leaf size=120 $\frac{e^{-a-b x} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 (c+d x)}\right )}{b}-\frac{e^{a+b x} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 (c+d x)}\right )}{b}-\frac{e^{-a-b x}}{2 b}+\frac{e^{a+b x}}{2 b}$

[Out]

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

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Rubi [A]  time = 0.100245, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {5604, 2194, 2251} $\frac{e^{-a-b x} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 (c+d x)}\right )}{b}-\frac{e^{a+b x} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 (c+d x)}\right )}{b}-\frac{e^{-a-b x}}{2 b}+\frac{e^{a+b x}}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]*Tanh[c + d*x],x]

[Out]

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

Rule 5604

Int[Cosh[(a_.) + (b_.)*(x_)]*Tanh[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[1/(E^(a + b*x)*2) + E^(a + b*x)/2 - 1/
(E^(a + b*x)*(1 + E^(2*(c + d*x)))) - E^(a + b*x)/(1 + E^(2*(c + d*x))), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
^2 - d^2, 0]

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

Mathematica [A]  time = 11.3145, size = 103, normalized size = 0.86 $\frac{e^{-a-b x} \left (-2 e^{2 (a+b x)} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 (c+d x)}\right )+e^{2 (a+b x)}+2 \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 (c+d x)}\right )-1\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]*Tanh[c + d*x],x]

[Out]

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

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int \cosh \left ( bx+a \right ) \tanh \left ( dx+c \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}}{2 \, b} - \frac{1}{2} \, \int \frac{2 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{e^{\left (b x + 2 \, d x + a + 2 \, c\right )} + e^{\left (b x + a\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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