### 3.950 $$\int e^{c+d x} \coth (a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=101 $\frac{4 e^{a+x (b+d)+c} \, _2F_1\left (2,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}-\frac{2 e^{a+x (b+d)+c} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}$

[Out]

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

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Rubi [A]  time = 0.249677, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {5511, 2251} $\frac{4 e^{a+x (b+d)+c} \, _2F_1\left (2,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}-\frac{2 e^{a+x (b+d)+c} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(c + d*x)*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

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

Rule 5511

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]

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+d x} \coth (a+b x) \text{csch}(a+b x) \, dx &=\int \left (\frac{4 e^{a+c+(b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^2}+\frac{2 e^{a+c+(b+d) x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=2 \int \frac{e^{a+c+(b+d) x}}{-1+e^{2 (a+b x)}} \, dx+4 \int \frac{e^{a+c+(b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^2} \, dx\\ &=-\frac{2 e^{a+c+(b+d) x} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}+\frac{4 e^{a+c+(b+d) x} \, _2F_1\left (2,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}\\ \end{align*}

Mathematica [A]  time = 0.656673, size = 92, normalized size = 0.91 $\frac{e^c \text{csch}(a) \left (-2 d e^{x (b+d)} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 b x} (\cosh (a)+\sinh (a))^2\right )-(b+d) e^{d x} (\cosh (a)-\sinh (a)) \text{csch}(a+b x)\right )}{b (\coth (a)-1) (b+d)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(c + d*x)*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

(E^c*Csch[a]*(-2*d*E^((b + d)*x)*Hypergeometric2F1[1, (b + d)/(2*b), (3*b + d)/(2*b), E^(2*b*x)*(Cosh[a] + Sin
h[a])^2] - (b + d)*E^(d*x)*Csch[a + b*x]*(Cosh[a] - Sinh[a])))/(b*(b + d)*(-1 + Coth[a]))

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{dx+c}}\cosh \left ( bx+a \right ) \left ({\rm csch} \left (bx+a\right ) \right ) ^{2}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^2,x)

[Out]

int(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 16 \, b d \int -\frac{e^{\left (b x + d x + a + c\right )}}{3 \, b^{2} - 4 \, b d + d^{2} -{\left (3 \, b^{2} - 4 \, b d + d^{2}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \,{\left (3 \, b^{2} - 4 \, b d + d^{2}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \,{\left (3 \, b^{2} - 4 \, b d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} - \frac{2 \,{\left ({\left (3 \, b e^{c} - d e^{c}\right )} e^{\left (3 \, b x + 3 \, a\right )} -{\left (3 \, b e^{c} + d e^{c}\right )} e^{\left (b x + a\right )}\right )} e^{\left (d x\right )}}{3 \, b^{2} - 4 \, b d + d^{2} +{\left (3 \, b^{2} - 4 \, b d + d^{2}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 2 \,{\left (3 \, b^{2} - 4 \, b d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

16*b*d*integrate(-e^(b*x + d*x + a + c)/(3*b^2 - 4*b*d + d^2 - (3*b^2 - 4*b*d + d^2)*e^(6*b*x + 6*a) + 3*(3*b^
2 - 4*b*d + d^2)*e^(4*b*x + 4*a) - 3*(3*b^2 - 4*b*d + d^2)*e^(2*b*x + 2*a)), x) - 2*((3*b*e^c - d*e^c)*e^(3*b*
x + 3*a) - (3*b*e^c + d*e^c)*e^(b*x + a))*e^(d*x)/(3*b^2 - 4*b*d + d^2 + (3*b^2 - 4*b*d + d^2)*e^(4*b*x + 4*a)
- 2*(3*b^2 - 4*b*d + d^2)*e^(2*b*x + 2*a))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{2} e^{\left (d x + c\right )}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(cosh(b*x + a)*csch(b*x + a)^2*e^(d*x + c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{2} e^{\left (d x + c\right )}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)*csch(b*x + a)^2*e^(d*x + c), x)