3.10.50 \(\int e^{c+d x} \coth (a+b x) \text {csch}(a+b x) \, dx\) [950]

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

[Out]

-2*exp(a+c+(b+d)*x)*hypergeom([1, 1/2*(b+d)/b],[1/2*(3*b+d)/b],exp(2*b*x+2*a))/(b+d)+4*exp(a+c+(b+d)*x)*hyperg
eom([2, 1/2*(b+d)/b],[1/2*(3*b+d)/b],exp(2*b*x+2*a))/(b+d)

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Rubi [A]
time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5622, 2283} \begin {gather*} \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} \end {gather*}

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 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
+ 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 5622

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]

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*}

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Mathematica [A]
time = 0.45, size = 92, normalized size = 0.91 \begin {gather*} \frac {e^c \text {csch}(a) \left (-2 d e^{(b+d) x} \, _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} \text {csch}(a+b x) (\cosh (a)-\sinh (a))\right )}{b (b+d) (-1+\coth (a))} \end {gather*}

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 = 0.98, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{d x +c} \cosh \left (b x +a \right ) \mathrm {csch}\left (b x +a \right )^{2}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3433 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {e}}^{c+d\,x}}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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