3.4 \(\int (a+b \coth ^2(c+d x))^2 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0331715, 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 = {3661, 390, 206} \[ -\frac{b (2 a+b) \coth (c+d x)}{d}+x (a+b)^2-\frac{b^2 \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x]^2)^2,x]

[Out]

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

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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

Mathematica [A]  time = 0.468242, size = 65, normalized size = 1.51 \[ -\frac{\coth (c+d x) \left (b \left (6 a+b \coth ^2(c+d x)+3 b\right )-3 (a+b)^2 \tanh ^{-1}\left (\sqrt{\tanh ^2(c+d x)}\right ) \sqrt{\tanh ^2(c+d x)}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Coth[c + d*x]*(b*(6*a + 3*b + b*Coth[c + d*x]^2) - 3*(a + b)^2*ArcTanh[Sqrt[Tanh[c + d*x]^2]]*Sqrt[Tanh[c +
d*x]^2]))/(3*d)

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Maple [B]  time = 0.003, size = 144, normalized size = 3.4 \begin{align*} -{\frac{{b}^{2} \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}}{3\,d}}-2\,{\frac{{\rm coth} \left (dx+c\right )ab}{d}}-{\frac{{b}^{2}{\rm coth} \left (dx+c\right )}{d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{2}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) ab}{d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){b}^{2}}{2\,d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{2}}{2\,d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) ab}{d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){b}^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.25198, size = 154, normalized size = 3.58 \begin{align*} \frac{1}{3} \, b^{2}{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\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 )}}\right )} + 2 \, a b{\left (x + \frac{c}{d} + \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.8694, size = 489, normalized size = 11.37 \begin{align*} -\frac{2 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 6 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 6 \, a b + 4 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 6 \, a b \cosh \left (d x + c\right ) + 3 \,{\left (3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x -{\left (3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 6 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a b + 4 \, 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*coth(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-1/3*(2*(3*a*b + 2*b^2)*cosh(d*x + c)^3 + 6*(3*a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 - (3*(a^2 + 2*a*b +
b^2)*d*x + 6*a*b + 4*b^2)*sinh(d*x + c)^3 - 6*a*b*cosh(d*x + c) + 3*(3*(a^2 + 2*a*b + b^2)*d*x - (3*(a^2 + 2*a
*b + b^2)*d*x + 6*a*b + 4*b^2)*cosh(d*x + c)^2 + 6*a*b + 4*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 [A]  time = 13.1673, size = 102, normalized size = 2.37 \begin{align*} \begin{cases} a^{2} x + \tilde{\infty } a b x + \tilde{\infty } b^{2} x & \text{for}\: c = \log{\left (- e^{- d x} \right )} \vee c = \log{\left (e^{- d x} \right )} \\x \left (a + b \coth ^{2}{\left (c \right )}\right )^{2} & \text{for}\: d = 0 \\a^{2} x + 2 a b x - \frac{2 a b}{d \tanh{\left (c + d x \right )}} + b^{2} x - \frac{b^{2}}{d \tanh{\left (c + d x \right )}} - \frac{b^{2}}{3 d \tanh ^{3}{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x + zoo*a*b*x + zoo*b**2*x, Eq(c, log(exp(-d*x))) | Eq(c, log(-exp(-d*x)))), (x*(a + b*coth(c)
**2)**2, Eq(d, 0)), (a**2*x + 2*a*b*x - 2*a*b/(d*tanh(c + d*x)) + b**2*x - b**2/(d*tanh(c + d*x)) - b**2/(3*d*
tanh(c + d*x)**3), True))

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Giac [B]  time = 1.16184, size = 139, normalized size = 3.23 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} - \frac{4 \,{\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} 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 + 2 \, 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*coth(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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