3.231 \(\int e^{c (a+b x)} \coth (d+e x) \, dx\)

Optimal. Leaf size=65 \[ \frac {e^{c (a+b x)}}{b c}-\frac {2 e^{c (a+b x)} \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c} \]

[Out]

exp(c*(b*x+a))/b/c-2*exp(c*(b*x+a))*hypergeom([1, 1/2*b*c/e],[1+1/2*b*c/e],exp(2*e*x+2*d))/b/c

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5485, 2194, 2251} \[ \frac {e^{c (a+b x)}}{b c}-\frac {2 e^{c (a+b x)} \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Coth[d + e*x],x]

[Out]

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

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])

Rule 5485

Int[Coth[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(F^(c*(
a + b*x))*(1 + E^(2*(d + e*x)))^n)/(-1 + E^(2*(d + e*x)))^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && Integer
Q[n]

Rubi steps

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

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Mathematica [B]  time = 2.54, size = 134, normalized size = 2.06 \[ \frac {e^{c (a+b x)} \left (2 b c e^{2 (d+e x)} \, _2F_1\left (1,\frac {b c}{2 e}+1;\frac {b c}{2 e}+2;e^{2 (d+e x)}\right )+(b c+2 e) \left (-2 e^{2 d} \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )+e^{2 d}+1\right )\right )}{b c \left (e^{2 d}-1\right ) (b c+2 e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\coth \left (e x + d\right ) e^{\left (b c x + a c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \coth \left (e x + d\right ) e^{\left ({\left (b x + a\right )} c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{c \left (b x +a \right )} \coth \left (e x +d \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 4 \, e \int \frac {e^{\left (b c x + a c\right )}}{b c + {\left (b c e^{\left (4 \, d\right )} - 2 \, e e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} - 2 \, {\left (b c e^{\left (2 \, d\right )} - 2 \, e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} - 2 \, e}\,{d x} - \frac {{\left (b c e^{\left (a c\right )} + 2 \, e e^{\left (a c\right )} + {\left (b c e^{\left (a c + 2 \, d\right )} - 2 \, e e^{\left (a c + 2 \, d\right )}\right )} e^{\left (2 \, e x\right )}\right )} e^{\left (b c x\right )}}{b^{2} c^{2} - 2 \, b c e - {\left (b^{2} c^{2} e^{\left (2 \, d\right )} - 2 \, b c e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {coth}\left (d+e\,x\right )\,{\mathrm {e}}^{c\,\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d + e*x)*exp(c*(a + b*x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int e^{b c x} \coth {\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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