### 3.188 $$\int \sinh (a+b x) \tanh (c+d x) \, dx$$

Optimal. Leaf size=121 $-\frac{e^{-a-b x} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 (c+d x)}\right )}{b}-\frac{e^{a+b x} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 (c+d x)}\right )}{b}+\frac{e^{-a-b x}}{2 b}+\frac{e^{a+b x}}{2 b}$

[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 = 0.106408, antiderivative size = 121, 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 = {5601, 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[Sinh[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 5601

Int[Sinh[(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] && Ne
Q[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 \sinh (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\\ &=-\left (\frac{1}{2} \int e^{-a-b x} \, dx\right )+\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 [B]  time = 14.1913, size = 278, normalized size = 2.3 $\frac{e^{-a-b x-c} \left ((b-2 d) \left (2 b \text{sech}(c) e^{2 (a+x (b+d)+c)} \, _2F_1\left (1,\frac{b}{2 d}+1;\frac{b}{2 d}+2;-e^{2 (c+d x)}\right )-(b+2 d) \left (\left (e^{2 a}+2 e^{2 c}+1\right ) \text{sech}(c) \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 (c+d x)}\right )+2 \text{sech}(c) e^{2 (a+b x+c)} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 (c+d x)}\right )-4 \tanh (c) e^{a+b x+c} \cosh (a+b x)-e^{2 a} \text{sech}(c)-\text{sech}(c)\right )\right )-\left (e^{2 a}-1\right ) b (b+2 d) \text{sech}(c) e^{2 (c+d x)} \, _2F_1\left (1,1-\frac{b}{2 d};2-\frac{b}{2 d};-e^{2 (c+d x)}\right )\right )}{4 \left (b^3-4 b d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(E^(-a - c - b*x)*(-(b*(b + 2*d)*E^(2*(c + d*x))*(-1 + E^(2*a))*Hypergeometric2F1[1, 1 - b/(2*d), 2 - b/(2*d),
-E^(2*(c + d*x))]*Sech[c]) + (b - 2*d)*(2*b*E^(2*(a + c + (b + d)*x))*Hypergeometric2F1[1, 1 + b/(2*d), 2 + b
/(2*d), -E^(2*(c + d*x))]*Sech[c] - (b + 2*d)*(-Sech[c] - E^(2*a)*Sech[c] + (1 + E^(2*a) + 2*E^(2*c))*Hypergeo
metric2F1[1, -b/(2*d), 1 - b/(2*d), -E^(2*(c + d*x))]*Sech[c] + 2*E^(2*(a + c + b*x))*Hypergeometric2F1[1, b/(
2*d), 1 + b/(2*d), -E^(2*(c + d*x))]*Sech[c] - 4*E^(a + c + b*x)*Cosh[a + b*x]*Tanh[c]))))/(4*(b^3 - 4*b*d^2))

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int \sinh \left ( bx+a \right ) \tanh \left ( dx+c \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sinh(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(sinh(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 (\sinh \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(sinh(b*x+a)*tanh(d*x+c),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \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(sinh(b*x+a)*tanh(d*x+c),x, algorithm="giac")

[Out]

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