### 3.958 $$\int e^{c+d x} \coth ^2(a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=151 $-\frac{2 e^{a+x (b+d)+c} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}+\frac{8 e^{a+x (b+d)+c} \, _2F_1\left (2,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}-\frac{8 e^{a+x (b+d)+c} \, _2F_1\left (3,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}$

[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) + (8*
E^(a + c + (b + d)*x)*Hypergeometric2F1[2, (b + d)/(2*b), (3*b + d)/(2*b), E^(2*(a + b*x))])/(b + d) - (8*E^(a
+ c + (b + d)*x)*Hypergeometric2F1[3, (b + d)/(2*b), (3*b + d)/(2*b), E^(2*(a + b*x))])/(b + d)

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Rubi [A]  time = 0.343, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {5511, 2251} $-\frac{2 e^{a+x (b+d)+c} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}+\frac{8 e^{a+x (b+d)+c} \, _2F_1\left (2,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}-\frac{8 e^{a+x (b+d)+c} \, _2F_1\left (3,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(c + d*x)*Coth[a + b*x]^2*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) + (8*
E^(a + c + (b + d)*x)*Hypergeometric2F1[2, (b + d)/(2*b), (3*b + d)/(2*b), E^(2*(a + b*x))])/(b + d) - (8*E^(a
+ c + (b + d)*x)*Hypergeometric2F1[3, (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 ^2(a+b x) \text{csch}(a+b x) \, dx &=\int \left (\frac{8 e^{a+c+(b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^3}+\frac{8 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+8 \int \frac{e^{a+c+(b+d) x}}{\left (-1+e^{2 (a+b x)}\right )^3} \, dx+8 \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{8 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}-\frac{8 e^{a+c+(b+d) x} \, _2F_1\left (3,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )}{b+d}\\ \end{align*}

Mathematica [A]  time = 1.32094, size = 111, normalized size = 0.74 $-\frac{e^{c-\frac{a d}{b}} \left (2 \left (b^2+d^2\right ) e^{\frac{(b+d) (a+b x)}{b}} \, _2F_1\left (1,\frac{b+d}{2 b};\frac{3 b+d}{2 b};e^{2 (a+b x)}\right )+(b+d) e^{d \left (\frac{a}{b}+x\right )} \text{csch}(a+b x) (b \coth (a+b x)+d)\right )}{2 b^2 (b+d)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(c + d*x)*Coth[a + b*x]^2*Csch[a + b*x],x]

[Out]

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

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Maple [F]  time = 0.136, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{dx+c}} \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ({\rm csch} \left (bx+a\right ) \right ) ^{3}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a)^3,x)

[Out]

int(exp(d*x+c)*cosh(b*x+a)^2*csch(b*x+a)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -48 \,{\left (b^{3} e^{c} + b d^{2} e^{c}\right )} \int \frac{e^{\left (b x + d x + a\right )}}{15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3} +{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (8 \, b x + 8 \, a\right )} - 4 \,{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 6 \,{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 4 \,{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac{2 \,{\left ({\left (15 \, b^{2} e^{c} - 8 \, b d e^{c} + d^{2} e^{c}\right )} e^{\left (5 \, b x + 5 \, a\right )} - 2 \,{\left (10 \, b^{2} e^{c} + 3 \, b d e^{c} - d^{2} e^{c}\right )} e^{\left (3 \, b x + 3 \, a\right )} +{\left (9 \, b^{2} e^{c} + 14 \, b d e^{c} + d^{2} e^{c}\right )} e^{\left (b x + a\right )}\right )} e^{\left (d x\right )}}{15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3} -{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \,{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \,{\left (15 \, b^{3} - 23 \, b^{2} d + 9 \, b d^{2} - d^{3}\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)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-48*(b^3*e^c + b*d^2*e^c)*integrate(e^(b*x + d*x + a)/(15*b^3 - 23*b^2*d + 9*b*d^2 - d^3 + (15*b^3 - 23*b^2*d
+ 9*b*d^2 - d^3)*e^(8*b*x + 8*a) - 4*(15*b^3 - 23*b^2*d + 9*b*d^2 - d^3)*e^(6*b*x + 6*a) + 6*(15*b^3 - 23*b^2*
d + 9*b*d^2 - d^3)*e^(4*b*x + 4*a) - 4*(15*b^3 - 23*b^2*d + 9*b*d^2 - d^3)*e^(2*b*x + 2*a)), x) + 2*((15*b^2*e
^c - 8*b*d*e^c + d^2*e^c)*e^(5*b*x + 5*a) - 2*(10*b^2*e^c + 3*b*d*e^c - d^2*e^c)*e^(3*b*x + 3*a) + (9*b^2*e^c
+ 14*b*d*e^c + d^2*e^c)*e^(b*x + a))*e^(d*x)/(15*b^3 - 23*b^2*d + 9*b*d^2 - d^3 - (15*b^3 - 23*b^2*d + 9*b*d^2
- d^3)*e^(6*b*x + 6*a) + 3*(15*b^3 - 23*b^2*d + 9*b*d^2 - d^3)*e^(4*b*x + 4*a) - 3*(15*b^3 - 23*b^2*d + 9*b*d
^2 - d^3)*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 )^{2} \operatorname{csch}\left (b x + a\right )^{3} 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)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(cosh(b*x + a)^2*csch(b*x + a)^3*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)**2*csch(b*x+a)**3,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 )^{2} \operatorname{csch}\left (b x + a\right )^{3} 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)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^2*csch(b*x + a)^3*e^(d*x + c), x)