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

Optimal. Leaf size=43 \[ a^2 x-\frac{b (2 a-b) \coth (c+d x)}{d}-\frac{b^2 \coth ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0324473, antiderivative size = 43, 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} \[ a^2 x-\frac{b (2 a-b) \coth (c+d x)}{d}-\frac{b^2 \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x]^2)^2,x]

[Out]

a^2*x - ((2*a - b)*b*Coth[c + d*x])/d - (b^2*Coth[c + d*x]^3)/(3*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 )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^2}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-(2 a-b) b-b^2 x^2+\frac{a^2}{1-x^2}\right ) \, dx,x,\coth (c+d x)\right )}{d}\\ &=-\frac{(2 a-b) b \coth (c+d x)}{d}-\frac{b^2 \coth ^3(c+d x)}{3 d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=a^2 x-\frac{(2 a-b) b \coth (c+d x)}{d}-\frac{b^2 \coth ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.611587, size = 84, normalized size = 1.95 \[ \frac{4 \sinh ^4(c+d x) \left (a+b \text{csch}^2(c+d x)\right )^2 \left (3 a^2 (c+d x)-b \coth (c+d x) \left (6 a+b \text{csch}^2(c+d x)-2 b\right )\right )}{3 d (a (-\cosh (2 (c+d x)))+a-2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*Csch[c + d*x]^2)^2*(3*a^2*(c + d*x) - b*Coth[c + d*x]*(6*a - 2*b + b*Csch[c + d*x]^2))*Sinh[c + d*x]
^4)/(3*d*(a - 2*b - a*Cosh[2*(c + d*x)])^2)

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Maple [A]  time = 0.023, size = 47, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( dx+c \right ) -2\,ab{\rm coth} \left (dx+c\right )+{b}^{2} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \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)^2,x)

[Out]

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

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Maxima [B]  time = 1.0078, size = 163, normalized size = 3.79 \begin{align*} a^{2} x + \frac{4}{3} \, 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{4 \, a 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)^2,x, algorithm="maxima")

[Out]

a^2*x + 4/3*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))) + 4*a*b/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.70841, size = 429, normalized size = 9.98 \begin{align*} -\frac{2 \,{\left (3 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} + 6 \,{\left (3 \, a b - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (3 \, a^{2} d x + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 6 \,{\left (a b - b^{2}\right )} \cosh \left (d x + c\right ) + 3 \,{\left (3 \, a^{2} d x -{\left (3 \, a^{2} d x + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )}{3 \,{\left (d \sinh \left (d x + c\right )^{3} + 3 \,{\left (d \cosh \left (d x + c\right )^{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)^2,x, algorithm="fricas")

[Out]

-1/3*(2*(3*a*b - b^2)*cosh(d*x + c)^3 + 6*(3*a*b - b^2)*cosh(d*x + c)*sinh(d*x + c)^2 - (3*a^2*d*x + 6*a*b - 2
*b^2)*sinh(d*x + c)^3 - 6*(a*b - b^2)*cosh(d*x + c) + 3*(3*a^2*d*x - (3*a^2*d*x + 6*a*b - 2*b^2)*cosh(d*x + c)
^2 + 6*a*b - 2*b^2)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2411, size = 109, normalized size = 2.53 \begin{align*} \frac{{\left (d x + c\right )} a^{2}}{d} - \frac{4 \,{\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b - b^{2}\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)*a^2/d - 4/3*(3*a*b*e^(4*d*x + 4*c) - 6*a*b*e^(2*d*x + 2*c) + 3*b^2*e^(2*d*x + 2*c) + 3*a*b - b^2)/(d
*(e^(2*d*x + 2*c) - 1)^3)