### 3.106 $$\int \cosh ^2(a+b x) \coth ^2(a+b x) \, dx$$

Optimal. Leaf size=40 $-\frac{3 \coth (a+b x)}{2 b}+\frac{\cosh ^2(a+b x) \coth (a+b x)}{2 b}+\frac{3 x}{2}$

[Out]

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

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Rubi [A]  time = 0.0411313, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {2591, 288, 321, 206} $-\frac{3 \coth (a+b x)}{2 b}+\frac{\cosh ^2(a+b x) \coth (a+b x)}{2 b}+\frac{3 x}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^2*Coth[a + b*x]^2,x]

[Out]

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

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [A]  time = 0.10684, size = 31, normalized size = 0.78 $\frac{6 (a+b x)+\sinh (2 (a+b x))-4 \coth (a+b x)}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^2*Coth[a + b*x]^2,x]

[Out]

(6*(a + b*x) - 4*Coth[a + b*x] + Sinh[2*(a + b*x)])/(4*b)

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Maple [A]  time = 0.018, size = 39, normalized size = 1. \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{2\,\sinh \left ( bx+a \right ) }}+{\frac{3\,bx}{2}}+{\frac{3\,a}{2}}-{\frac{3\,{\rm coth} \left (bx+a\right )}{2}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/2/sinh(b*x+a)*cosh(b*x+a)^3+3/2*b*x+3/2*a-3/2*coth(b*x+a))

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Maxima [A]  time = 1.12189, size = 89, normalized size = 2.22 \begin{align*} \frac{3 \,{\left (b x + a\right )}}{2 \, b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} - \frac{17 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1}{8 \, b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(b*x + a)/b - 1/8*e^(-2*b*x - 2*a)/b - 1/8*(17*e^(-2*b*x - 2*a) - 1)/(b*(e^(-2*b*x - 2*a) - e^(-4*b*x - 4*
a)))

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Fricas [A]  time = 1.72865, size = 166, normalized size = 4.15 \begin{align*} \frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 4 \,{\left (3 \, b x + 2\right )} \sinh \left (b x + a\right ) - 9 \, \cosh \left (b x + a\right )}{8 \, b \sinh \left (b x + a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*coth(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + 4*(3*b*x + 2)*sinh(b*x + a) - 9*cosh(b*x + a))/(b*sin
h(b*x + a))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2*coth(b*x+a)**2,x)

[Out]

Integral(cosh(a + b*x)**2*coth(a + b*x)**2, x)

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Giac [A]  time = 1.25637, size = 90, normalized size = 2.25 \begin{align*} \frac{12 \, b x + \frac{{\left (3 \, e^{\left (4 \, b x + 4 \, a\right )} + 14 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, a\right )}}{e^{\left (2 \, b x\right )} - e^{\left (4 \, b x + 2 \, a\right )}} + e^{\left (2 \, b x + 2 \, a\right )}}{8 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*coth(b*x+a)^2,x, algorithm="giac")

[Out]

1/8*(12*b*x + (3*e^(4*b*x + 4*a) + 14*e^(2*b*x + 2*a) - 1)*e^(-2*a)/(e^(2*b*x) - e^(4*b*x + 2*a)) + e^(2*b*x +
2*a))/b