### 3.189 $$\int \coth (c+d x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=117 $-\frac{e^{-a-b x} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};e^{2 (c+d x)}\right )}{b}-\frac{e^{a+b x} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;e^{2 (c+d x)}\right )}{b}+\frac{e^{-a-b x}}{2 b}+\frac{e^{a+b x}}{2 b}$

[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 = 0.108786, antiderivative size = 117, 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 = {5603, 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[Coth[c + d*x]*Sinh[a + b*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 5603

Int[Coth[(c_.) + (d_.)*(x_)]*Sinh[(a_.) + (b_.)*(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 \coth (c+d x) \sinh (a+b 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 = 6.7317, size = 240, normalized size = 2.05 $\frac{e^{-a-b x+2 c} \left (b e^{2 d x} \, _2F_1\left (1,1-\frac{b}{2 d};2-\frac{b}{2 d};e^{2 (c+d x)}\right )-(b-2 d) \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};e^{2 (c+d x)}\right )\right )}{b \left (e^{2 c}-1\right ) (b-2 d)}-\frac{e^{a+2 c} \left (\frac{e^{b x} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;e^{2 (c+d x)}\right )}{b}-\frac{e^{x (b+2 d)} \, _2F_1\left (1,\frac{b}{2 d}+1;\frac{b}{2 d}+2;e^{2 (c+d x)}\right )}{b+2 d}\right )}{e^{2 c}-1}+\frac{\cosh (a) \coth (c) \cosh (b x)}{b}+\frac{\sinh (a) \coth (c) \sinh (b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Cosh[a]*Cosh[b*x]*Coth[c])/b + (E^(-a + 2*c - b*x)*(b*E^(2*d*x)*Hypergeometric2F1[1, 1 - b/(2*d), 2 - b/(2*d)
, E^(2*(c + d*x))] - (b - 2*d)*Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), E^(2*(c + d*x))]))/(b*(b - 2*d)*(-1
+ E^(2*c))) - (E^(a + 2*c)*(-((E^((b + 2*d)*x)*Hypergeometric2F1[1, 1 + b/(2*d), 2 + b/(2*d), E^(2*(c + d*x))
])/(b + 2*d)) + (E^(b*x)*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), E^(2*(c + d*x))])/b))/(-1 + E^(2*c)) + (Co
th[c]*Sinh[a]*Sinh[b*x])/b

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Maple [F]  time = 0.083, size = 0, normalized size = 0. \begin{align*} \int{\rm coth} \left (dx+c\right )\sinh \left ( bx+a \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(coth(d*x+c)*sinh(b*x+a),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{e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (b x + d x + a + c\right )} + e^{\left (b x + a\right )}}\,{d x} + \frac{1}{2} \, \int \frac{e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (b x + d x + a + c\right )} - e^{\left (b x + a\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)*sinh(b*x+a),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)*sinh(b*x+a),x, algorithm="fricas")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)*sinh(b*x+a),x, algorithm="giac")

[Out]

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