3.2 \(\int (a+b \text{csch}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=74 \[ -\frac{b \left (3 a^2-3 a b+b^2\right ) \coth (c+d x)}{d}+a^3 x-\frac{b^2 (3 a-2 b) \coth ^3(c+d x)}{3 d}-\frac{b^3 \coth ^5(c+d x)}{5 d} \]

[Out]

a^3*x - (b*(3*a^2 - 3*a*b + b^2)*Coth[c + d*x])/d - ((3*a - 2*b)*b^2*Coth[c + d*x]^3)/(3*d) - (b^3*Coth[c + d*
x]^5)/(5*d)

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Rubi [A]  time = 0.047714, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4128, 390, 206} \[ -\frac{b \left (3 a^2-3 a b+b^2\right ) \coth (c+d x)}{d}+a^3 x-\frac{b^2 (3 a-2 b) \coth ^3(c+d x)}{3 d}-\frac{b^3 \coth ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x]^2)^3,x]

[Out]

a^3*x - (b*(3*a^2 - 3*a*b + b^2)*Coth[c + d*x])/d - ((3*a - 2*b)*b^2*Coth[c + d*x]^3)/(3*d) - (b^3*Coth[c + d*
x]^5)/(5*d)

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+b \text{csch}^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^3}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-b \left (3 a^2-3 a b+b^2\right )-(3 a-2 b) b^2 x^2-b^3 x^4+\frac{a^3}{1-x^2}\right ) \, dx,x,\coth (c+d x)\right )}{d}\\ &=-\frac{b \left (3 a^2-3 a b+b^2\right ) \coth (c+d x)}{d}-\frac{(3 a-2 b) b^2 \coth ^3(c+d x)}{3 d}-\frac{b^3 \coth ^5(c+d x)}{5 d}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=a^3 x-\frac{b \left (3 a^2-3 a b+b^2\right ) \coth (c+d x)}{d}-\frac{(3 a-2 b) b^2 \coth ^3(c+d x)}{3 d}-\frac{b^3 \coth ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 3.51963, size = 113, normalized size = 1.53 \[ \frac{8 \sinh ^6(c+d x) \left (a+b \text{csch}^2(c+d x)\right )^3 \left (15 a^3 (c+d x)-b \coth (c+d x) \left (45 a^2+b (15 a-4 b) \text{csch}^2(c+d x)-30 a b+3 b^2 \text{csch}^4(c+d x)+8 b^2\right )\right )}{15 d (a \cosh (2 (c+d x))-a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x]^2)^3,x]

[Out]

(8*(a + b*Csch[c + d*x]^2)^3*(15*a^3*(c + d*x) - b*Coth[c + d*x]*(45*a^2 - 30*a*b + 8*b^2 + (15*a - 4*b)*b*Csc
h[c + d*x]^2 + 3*b^2*Csch[c + d*x]^4))*Sinh[c + d*x]^6)/(15*d*(-a + 2*b + a*Cosh[2*(c + d*x)])^3)

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Maple [A]  time = 0.023, size = 83, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( dx+c \right ) -3\,{a}^{2}b{\rm coth} \left (dx+c\right )+3\,a{b}^{2} \left ( 2/3-1/3\, \left ({\rm csch} \left (dx+c\right ) \right ) ^{2} \right ){\rm coth} \left (dx+c\right )+{b}^{3} \left ( -{\frac{8}{15}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{15}} \right ){\rm coth} \left (dx+c\right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csch(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(d*x+c)-3*a^2*b*coth(d*x+c)+3*a*b^2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b^3*(-8/15-1/5*csch(d*x+c)^4+
4/15*csch(d*x+c)^2)*coth(d*x+c))

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Maxima [B]  time = 1.01154, size = 451, normalized size = 6.09 \begin{align*} a^{3} x - \frac{16}{15} \, b^{3}{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + 4 \, a b^{2}{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac{1}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac{6 \, a^{2} b}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

a^3*x - 16/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e
^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c)
+ 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 1/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*
x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))) + 4*a*b^2*(3*e^(-2*d*x - 2*c)/
(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)) - 1/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x
- 4*c) + e^(-6*d*x - 6*c) - 1))) + 6*a^2*b/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.6163, size = 1134, normalized size = 15.32 \begin{align*} -\frac{{\left (45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} -{\left (15 \, a^{3} d x + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} - 5 \,{\left (27 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \,{\left (15 \, a^{3} d x + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3} - 2 \,{\left (15 \, a^{3} d x + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 3 \,{\left (27 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (9 \, a^{2} b - 12 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) - 5 \,{\left (30 \, a^{3} d x +{\left (15 \, a^{3} d x + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 90 \, a^{2} b - 60 \, a b^{2} + 16 \, b^{3} - 3 \,{\left (15 \, a^{3} d x + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \sinh \left (d x + c\right )^{5} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((45*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x + c)^5 + 5*(45*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x + c)*sinh(d*x
+ c)^4 - (15*a^3*d*x + 45*a^2*b - 30*a*b^2 + 8*b^3)*sinh(d*x + c)^5 - 5*(27*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x
+ c)^3 + 5*(15*a^3*d*x + 45*a^2*b - 30*a*b^2 + 8*b^3 - 2*(15*a^3*d*x + 45*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x
+ c)^2)*sinh(d*x + c)^3 + 5*(2*(45*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x + c)^3 - 3*(27*a^2*b - 30*a*b^2 + 8*b^3)
*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(9*a^2*b - 12*a*b^2 + 8*b^3)*cosh(d*x + c) - 5*(30*a^3*d*x + (15*a^3*d*x
+ 45*a^2*b - 30*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + 90*a^2*b - 60*a*b^2 + 16*b^3 - 3*(15*a^3*d*x + 45*a^2*b - 30*
a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^
3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{csch}^{2}{\left (c + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)**2)**3,x)

[Out]

Integral((a + b*csch(c + d*x)**2)**3, x)

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Giac [B]  time = 1.19228, size = 246, normalized size = 3.32 \begin{align*} \frac{{\left (d x + c\right )} a^{3}}{d} - \frac{2 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 150 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 40 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b - 30 \, a b^{2} + 8 \, b^{3}\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)^2)^3,x, algorithm="giac")

[Out]

(d*x + c)*a^3/d - 2/15*(45*a^2*b*e^(8*d*x + 8*c) - 180*a^2*b*e^(6*d*x + 6*c) + 90*a*b^2*e^(6*d*x + 6*c) + 270*
a^2*b*e^(4*d*x + 4*c) - 210*a*b^2*e^(4*d*x + 4*c) + 80*b^3*e^(4*d*x + 4*c) - 180*a^2*b*e^(2*d*x + 2*c) + 150*a
*b^2*e^(2*d*x + 2*c) - 40*b^3*e^(2*d*x + 2*c) + 45*a^2*b - 30*a*b^2 + 8*b^3)/(d*(e^(2*d*x + 2*c) - 1)^5)