3.10.65 \(\int e^{c+d x} \coth ^3(a+b x) \, dx\) [965]

Optimal. Leaf size=135 \[ \frac {e^{c+d x}}{d}-\frac {6 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}+\frac {12 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}-\frac {8 e^{c+d x} \, _2F_1\left (3,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d} \]

[Out]

exp(d*x+c)/d-6*exp(d*x+c)*hypergeom([1, 1/2*d/b],[1+1/2*d/b],exp(2*b*x+2*a))/d+12*exp(d*x+c)*hypergeom([2, 1/2
*d/b],[1+1/2*d/b],exp(2*b*x+2*a))/d-8*exp(d*x+c)*hypergeom([3, 1/2*d/b],[1+1/2*d/b],exp(2*b*x+2*a))/d

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Rubi [A]
time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5593, 2225, 2283} \begin {gather*} -\frac {6 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {12 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}-\frac {8 e^{c+d x} \, _2F_1\left (3,\frac {d}{2 b};\frac {d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac {e^{c+d x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(c + d*x)*Coth[a + b*x]^3,x]

[Out]

E^(c + d*x)/d - (6*E^(c + d*x)*Hypergeometric2F1[1, d/(2*b), 1 + d/(2*b), E^(2*(a + b*x))])/d + (12*E^(c + d*x
)*Hypergeometric2F1[2, d/(2*b), 1 + d/(2*b), E^(2*(a + b*x))])/d - (8*E^(c + d*x)*Hypergeometric2F1[3, d/(2*b)
, 1 + d/(2*b), E^(2*(a + b*x))])/d

Rule 2225

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 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 5593

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+d x} \coth ^3(a+b x) \, dx &=\int \left (e^{c+d x}+\frac {8 e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^3}+\frac {12 e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^2}+\frac {6 e^{c+d x}}{-1+e^{2 (a+b x)}}\right ) \, dx\\ &=6 \int \frac {e^{c+d x}}{-1+e^{2 (a+b x)}} \, dx+8 \int \frac {e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^3} \, dx+12 \int \frac {e^{c+d x}}{\left (-1+e^{2 (a+b x)}\right )^2} \, dx+\int e^{c+d x} \, dx\\ &=\frac {e^{c+d x}}{d}-\frac {6 e^{c+d x} \, _2F_1\left (1,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}+\frac {12 e^{c+d x} \, _2F_1\left (2,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}-\frac {8 e^{c+d x} \, _2F_1\left (3,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}\\ \end {align*}

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Mathematica [A]
time = 2.55, size = 176, normalized size = 1.30 \begin {gather*} \frac {1}{2} e^c \left (\frac {2 e^{d x} \coth (a)}{d}-\frac {e^{d x} \text {csch}^2(a+b x)}{b}-\frac {2 \left (2 b^2+d^2\right ) e^{2 a} \left (\frac {e^{d x} \, _2F_1\left (1,\frac {d}{2 b};1+\frac {d}{2 b};e^{2 (a+b x)}\right )}{d}-\frac {e^{(2 b+d) x} \, _2F_1\left (1,1+\frac {d}{2 b};2+\frac {d}{2 b};e^{2 (a+b x)}\right )}{2 b+d}\right )}{b^2 \left (-1+e^{2 a}\right )}+\frac {d e^{d x} \text {csch}(a) \text {csch}(a+b x) \sinh (b x)}{b^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^c*((2*E^(d*x)*Coth[a])/d - (E^(d*x)*Csch[a + b*x]^2)/b - (2*(2*b^2 + d^2)*E^(2*a)*((E^(d*x)*Hypergeometric2
F1[1, d/(2*b), 1 + d/(2*b), E^(2*(a + b*x))])/d - (E^((2*b + d)*x)*Hypergeometric2F1[1, 1 + d/(2*b), 2 + d/(2*
b), E^(2*(a + b*x))])/(2*b + d)))/(b^2*(-1 + E^(2*a))) + (d*E^(d*x)*Csch[a]*Csch[a + b*x]*Sinh[b*x])/b^2))/2

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Maple [F]
time = 3.45, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{d x +c} \left (\cosh ^{3}\left (b x +a \right )\right ) \mathrm {csch}\left (b x +a \right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(d*x+c)*cosh(b*x+a)^3*csch(b*x+a)^3,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)^3*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(d*x+c)*cosh(b*x+a)**3*csch(b*x+a)**3,x)

[Out]

Timed out

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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)^3*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^3*csch(b*x + a)^3*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 )}^3\,{\mathrm {e}}^{c+d\,x}}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^3*exp(c + d*x))/sinh(a + b*x)^3,x)

[Out]

int((cosh(a + b*x)^3*exp(c + d*x))/sinh(a + b*x)^3, x)

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