### 3.965 $$\int e^{c+d x} \coth ^3(a+b x) \, dx$$

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

[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

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Rubi [A]  time = 0.164018, antiderivative size = 135, normalized size of antiderivative = 1., 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 = {5485, 2194, 2251} $-\frac{6 e^{c+d x} \, _2F_1\left (1,\frac{d}{2 b};\frac{d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac{12 e^{c+d x} \, _2F_1\left (2,\frac{d}{2 b};\frac{d}{2 b}+1;e^{2 (a+b x)}\right )}{d}-\frac{8 e^{c+d x} \, _2F_1\left (3,\frac{d}{2 b};\frac{d}{2 b}+1;e^{2 (a+b x)}\right )}{d}+\frac{e^{c+d x}}{d}$

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 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]

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])

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*}

Mathematica [A]  time = 3.72413, size = 176, normalized size = 1.3 $\frac{1}{2} e^c \left (-\frac{2 e^{2 a} \left (2 b^2+d^2\right ) \left (\frac{e^{d x} \, _2F_1\left (1,\frac{d}{2 b};\frac{d}{2 b}+1;e^{2 (a+b x)}\right )}{d}-\frac{e^{x (2 b+d)} \, _2F_1\left (1,\frac{d}{2 b}+1;\frac{d}{2 b}+2;e^{2 (a+b x)}\right )}{2 b+d}\right )}{\left (e^{2 a}-1\right ) b^2}+\frac{d \text{csch}(a) e^{d x} \sinh (b x) \text{csch}(a+b x)}{b^2}-\frac{e^{d x} \text{csch}^2(a+b x)}{b}+\frac{2 \coth (a) e^{d x}}{d}\right )$

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 = 0.176, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{dx+c}} \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \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)^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., size = 0, normalized size = 0. \begin{align*} -48 \,{\left (2 \, b^{3} e^{c} + b d^{2} e^{c}\right )} \int \frac{e^{\left (d x\right )}}{48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3} +{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (8 \, b x + 8 \, a\right )} - 4 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 6 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 4 \,{\left (48 \, b^{3} - 44 \, b^{2} d + 12 \, b d^{2} - d^{3}\right )} e^{\left (2 \, b x + 2 \, a\right )}}\,{d x} + \frac{{\left (48 \, b^{3} e^{c} + 44 \, b^{2} d e^{c} + 36 \, b d^{2} e^{c} + d^{3} e^{c} -{\left (48 \, b^{3} e^{c} - 44 \, b^{2} d e^{c} + 12 \, b d^{2} e^{c} - d^{3} e^{c}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \,{\left (48 \, b^{3} e^{c} + 4 \, b^{2} d e^{c} - 8 \, b d^{2} e^{c} + d^{3} e^{c}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \,{\left (48 \, b^{3} e^{c} + 28 \, b^{2} d e^{c} - d^{3} e^{c}\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )} e^{\left (d x\right )}}{48 \, b^{3} d - 44 \, b^{2} d^{2} + 12 \, b d^{3} - d^{4} -{\left (48 \, b^{3} d - 44 \, b^{2} d^{2} + 12 \, b d^{3} - d^{4}\right )} e^{\left (6 \, b x + 6 \, a\right )} + 3 \,{\left (48 \, b^{3} d - 44 \, b^{2} d^{2} + 12 \, b d^{3} - d^{4}\right )} e^{\left (4 \, b x + 4 \, a\right )} - 3 \,{\left (48 \, b^{3} d - 44 \, b^{2} d^{2} + 12 \, b d^{3} - d^{4}\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)^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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cosh \left (b x + a\right )^{3} \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)^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)]  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)**3*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 )^{3} \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)^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)