3.233 \(\int e^{c (a+b x)} \coth ^3(d+e x) \, dx\)

Optimal. Leaf size=161 \[ -\frac {6 e^{c (a+b x)} \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \, _2F_1\left (2,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \, _2F_1\left (3,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}+\frac {e^{c (a+b x)}}{b c} \]

[Out]

exp(c*(b*x+a))/b/c-6*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+12*exp(c*(b*x+a
))*hypergeom([2, 1/2*b*c/e],[1+1/2*b*c/e],exp(2*e*x+2*d))/b/c-8*exp(c*(b*x+a))*hypergeom([3, 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 = 0.18, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5485, 2194, 2251} \[ -\frac {6 e^{c (a+b x)} \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \, _2F_1\left (2,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \, _2F_1\left (3,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )}{b c}+\frac {e^{c (a+b x)}}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Coth[d + e*x]^3,x]

[Out]

E^(c*(a + b*x))/(b*c) - (6*E^(c*(a + b*x))*Hypergeometric2F1[1, (b*c)/(2*e), 1 + (b*c)/(2*e), E^(2*(d + e*x))]
)/(b*c) + (12*E^(c*(a + b*x))*Hypergeometric2F1[2, (b*c)/(2*e), 1 + (b*c)/(2*e), E^(2*(d + e*x))])/(b*c) - (8*
E^(c*(a + b*x))*Hypergeometric2F1[3, (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 ^3(d+e x) \, dx &=\int \left (e^{c (a+b x)}+\frac {8 e^{c (a+b x)}}{\left (-1+e^{2 (d+e x)}\right )^3}+\frac {12 e^{c (a+b x)}}{\left (-1+e^{2 (d+e x)}\right )^2}+\frac {6 e^{c (a+b x)}}{-1+e^{2 (d+e x)}}\right ) \, dx\\ &=6 \int \frac {e^{c (a+b x)}}{-1+e^{2 (d+e x)}} \, dx+8 \int \frac {e^{c (a+b x)}}{\left (-1+e^{2 (d+e x)}\right )^3} \, dx+12 \int \frac {e^{c (a+b x)}}{\left (-1+e^{2 (d+e x)}\right )^2} \, dx+\int e^{c (a+b x)} \, dx\\ &=\frac {e^{c (a+b x)}}{b c}-\frac {6 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}+\frac {12 e^{c (a+b x)} \, _2F_1\left (2,\frac {b c}{2 e};1+\frac {b c}{2 e};e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \, _2F_1\left (3,\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 [A]  time = 3.61, size = 185, normalized size = 1.15 \[ \frac {1}{2} e^{c (a+b x)} \left (\frac {2 e^{2 d} \left (b^2 c^2+2 e^2\right ) \left (b c e^{2 e x} \, _2F_1\left (1,\frac {b c}{2 e}+1;\frac {b c}{2 e}+2;e^{2 (d+e x)}\right )-(b c+2 e) \, _2F_1\left (1,\frac {b c}{2 e};\frac {b c}{2 e}+1;e^{2 (d+e x)}\right )\right )}{b c \left (e^{2 d}-1\right ) e^2 (b c+2 e)}+\frac {b c \text {csch}(d) \sinh (e x) \text {csch}(d+e x)}{e^2}+\frac {2 \coth (d)}{b c}-\frac {\text {csch}^2(d+e x)}{e}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*Coth[d + e*x]^3,x]

[Out]

(E^(c*(a + b*x))*((2*Coth[d])/(b*c) - Csch[d + e*x]^2/e + (2*(b^2*c^2 + 2*e^2)*E^(2*d)*(b*c*E^(2*e*x)*Hypergeo
metric2F1[1, 1 + (b*c)/(2*e), 2 + (b*c)/(2*e), E^(2*(d + e*x))] - (b*c + 2*e)*Hypergeometric2F1[1, (b*c)/(2*e)
, 1 + (b*c)/(2*e), E^(2*(d + e*x))]))/(b*c*e^2*(b*c + 2*e)*(-1 + E^(2*d))) + (b*c*Csch[d]*Csch[d + e*x]*Sinh[e
*x])/e^2))/2

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\coth \left (e x + d\right )^{3} e^{\left (b c x + a c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*coth(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(coth(e*x + d)^3*e^(b*c*x + a*c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \coth \left (e x + d\right )^{3} e^{\left ({\left (b x + a\right )} c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*coth(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(coth(e*x + d)^3*e^((b*x + a)*c), x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{c \left (b x +a \right )} \left (\coth ^{3}\left (e x +d \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*coth(e*x+d)^3,x)

[Out]

int(exp(c*(b*x+a))*coth(e*x+d)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 48 \, {\left (b^{2} c^{2} e e^{\left (a c\right )} + 2 \, e^{3} e^{\left (a c\right )}\right )} \int \frac {e^{\left (b c x\right )}}{b^{3} c^{3} - 12 \, b^{2} c^{2} e + 44 \, b c e^{2} - 48 \, e^{3} + {\left (b^{3} c^{3} e^{\left (8 \, d\right )} - 12 \, b^{2} c^{2} e e^{\left (8 \, d\right )} + 44 \, b c e^{2} e^{\left (8 \, d\right )} - 48 \, e^{3} e^{\left (8 \, d\right )}\right )} e^{\left (8 \, e x\right )} - 4 \, {\left (b^{3} c^{3} e^{\left (6 \, d\right )} - 12 \, b^{2} c^{2} e e^{\left (6 \, d\right )} + 44 \, b c e^{2} e^{\left (6 \, d\right )} - 48 \, e^{3} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 6 \, {\left (b^{3} c^{3} e^{\left (4 \, d\right )} - 12 \, b^{2} c^{2} e e^{\left (4 \, d\right )} + 44 \, b c e^{2} e^{\left (4 \, d\right )} - 48 \, e^{3} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} - 4 \, {\left (b^{3} c^{3} e^{\left (2 \, d\right )} - 12 \, b^{2} c^{2} e e^{\left (2 \, d\right )} + 44 \, b c e^{2} e^{\left (2 \, d\right )} - 48 \, e^{3} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}}\,{d x} - \frac {{\left (b^{3} c^{3} e^{\left (a c\right )} + 36 \, b^{2} c^{2} e e^{\left (a c\right )} + 44 \, b c e^{2} e^{\left (a c\right )} + 48 \, e^{3} e^{\left (a c\right )} + {\left (b^{3} c^{3} e^{\left (a c + 6 \, d\right )} - 12 \, b^{2} c^{2} e e^{\left (a c + 6 \, d\right )} + 44 \, b c e^{2} e^{\left (a c + 6 \, d\right )} - 48 \, e^{3} e^{\left (a c + 6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 3 \, {\left (b^{3} c^{3} e^{\left (a c + 4 \, d\right )} - 8 \, b^{2} c^{2} e e^{\left (a c + 4 \, d\right )} + 4 \, b c e^{2} e^{\left (a c + 4 \, d\right )} + 48 \, e^{3} e^{\left (a c + 4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 3 \, {\left (b^{3} c^{3} e^{\left (a c + 2 \, d\right )} - 28 \, b c e^{2} e^{\left (a c + 2 \, d\right )} - 48 \, e^{3} e^{\left (a c + 2 \, d\right )}\right )} e^{\left (2 \, e x\right )}\right )} e^{\left (b c x\right )}}{b^{4} c^{4} - 12 \, b^{3} c^{3} e + 44 \, b^{2} c^{2} e^{2} - 48 \, b c e^{3} - {\left (b^{4} c^{4} e^{\left (6 \, d\right )} - 12 \, b^{3} c^{3} e e^{\left (6 \, d\right )} + 44 \, b^{2} c^{2} e^{2} e^{\left (6 \, d\right )} - 48 \, b c e^{3} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 3 \, {\left (b^{4} c^{4} e^{\left (4 \, d\right )} - 12 \, b^{3} c^{3} e e^{\left (4 \, d\right )} + 44 \, b^{2} c^{2} e^{2} e^{\left (4 \, d\right )} - 48 \, b c e^{3} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} - 3 \, {\left (b^{4} c^{4} e^{\left (2 \, d\right )} - 12 \, b^{3} c^{3} e e^{\left (2 \, d\right )} + 44 \, b^{2} c^{2} e^{2} e^{\left (2 \, d\right )} - 48 \, b c e^{3} 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)^3,x, algorithm="maxima")

[Out]

48*(b^2*c^2*e*e^(a*c) + 2*e^3*e^(a*c))*integrate(e^(b*c*x)/(b^3*c^3 - 12*b^2*c^2*e + 44*b*c*e^2 - 48*e^3 + (b^
3*c^3*e^(8*d) - 12*b^2*c^2*e*e^(8*d) + 44*b*c*e^2*e^(8*d) - 48*e^3*e^(8*d))*e^(8*e*x) - 4*(b^3*c^3*e^(6*d) - 1
2*b^2*c^2*e*e^(6*d) + 44*b*c*e^2*e^(6*d) - 48*e^3*e^(6*d))*e^(6*e*x) + 6*(b^3*c^3*e^(4*d) - 12*b^2*c^2*e*e^(4*
d) + 44*b*c*e^2*e^(4*d) - 48*e^3*e^(4*d))*e^(4*e*x) - 4*(b^3*c^3*e^(2*d) - 12*b^2*c^2*e*e^(2*d) + 44*b*c*e^2*e
^(2*d) - 48*e^3*e^(2*d))*e^(2*e*x)), x) - (b^3*c^3*e^(a*c) + 36*b^2*c^2*e*e^(a*c) + 44*b*c*e^2*e^(a*c) + 48*e^
3*e^(a*c) + (b^3*c^3*e^(a*c + 6*d) - 12*b^2*c^2*e*e^(a*c + 6*d) + 44*b*c*e^2*e^(a*c + 6*d) - 48*e^3*e^(a*c + 6
*d))*e^(6*e*x) + 3*(b^3*c^3*e^(a*c + 4*d) - 8*b^2*c^2*e*e^(a*c + 4*d) + 4*b*c*e^2*e^(a*c + 4*d) + 48*e^3*e^(a*
c + 4*d))*e^(4*e*x) + 3*(b^3*c^3*e^(a*c + 2*d) - 28*b*c*e^2*e^(a*c + 2*d) - 48*e^3*e^(a*c + 2*d))*e^(2*e*x))*e
^(b*c*x)/(b^4*c^4 - 12*b^3*c^3*e + 44*b^2*c^2*e^2 - 48*b*c*e^3 - (b^4*c^4*e^(6*d) - 12*b^3*c^3*e*e^(6*d) + 44*
b^2*c^2*e^2*e^(6*d) - 48*b*c*e^3*e^(6*d))*e^(6*e*x) + 3*(b^4*c^4*e^(4*d) - 12*b^3*c^3*e*e^(4*d) + 44*b^2*c^2*e
^2*e^(4*d) - 48*b*c*e^3*e^(4*d))*e^(4*e*x) - 3*(b^4*c^4*e^(2*d) - 12*b^3*c^3*e*e^(2*d) + 44*b^2*c^2*e^2*e^(2*d
) - 48*b*c*e^3*e^(2*d))*e^(2*e*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {coth}\left (d+e\,x\right )}^3\,{\mathrm {e}}^{c\,\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d + e*x)^3*exp(c*(a + b*x)),x)

[Out]

int(coth(d + e*x)^3*exp(c*(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int e^{b c x} \coth ^{3}{\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*coth(e*x+d)**3,x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*coth(d + e*x)**3, x)

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