### 3.108 $$\int \cosh ^3(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=38 $\frac{\cosh ^3(a+b x)}{3 b}+\frac{\cosh (a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

[Out]

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

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Rubi [A]  time = 0.0300179, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2592, 302, 206} $\frac{\cosh ^3(a+b x)}{3 b}+\frac{\cosh (a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^3*Coth[a + b*x],x]

[Out]

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

Rule 2592

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.0293825, size = 44, normalized size = 1.16 $\frac{5 \cosh (a+b x)}{4 b}+\frac{\cosh (3 (a+b x))}{12 b}+\frac{\log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^3*Coth[a + b*x],x]

[Out]

(5*Cosh[a + b*x])/(4*b) + Cosh[3*(a + b*x)]/(12*b) + Log[Tanh[(a + b*x)/2]]/b

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Maple [A]  time = 0.017, size = 31, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{3}}+\cosh \left ( bx+a \right ) -2\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*coth(b*x+a),x)

[Out]

1/b*(1/3*cosh(b*x+a)^3+cosh(b*x+a)-2*arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.04711, size = 117, normalized size = 3.08 \begin{align*} \frac{{\left (15 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{24 \, b} + \frac{15 \, e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*coth(b*x+a),x, algorithm="maxima")

[Out]

1/24*(15*e^(-2*b*x - 2*a) + 1)*e^(3*b*x + 3*a)/b + 1/24*(15*e^(-b*x - a) + e^(-3*b*x - 3*a))/b - log(e^(-b*x -
a) + 1)/b + log(e^(-b*x - a) - 1)/b

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Fricas [B]  time = 1.89726, size = 1037, normalized size = 27.29 \begin{align*} \frac{\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 15 \,{\left (\cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 15 \, \cosh \left (b x + a\right )^{4} + 20 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 15 \,{\left (\cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 15 \, \cosh \left (b x + a\right )^{2} - 24 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 24 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 6 \,{\left (\cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{24 \,{\left (b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*coth(b*x+a),x, algorithm="fricas")

[Out]

1/24*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 15*(cosh(b*x + a)^2 + 1)*sinh(b*x
+ a)^4 + 15*cosh(b*x + a)^4 + 20*(cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 15*(cosh(b*x + a)^4 + 6
*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 15*cosh(b*x + a)^2 - 24*(cosh(b*x + a)^3 + 3*cosh(b*x + a)^2*sinh(b*x
+ a) + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 24*(cosh(b*
x + a)^3 + 3*cosh(b*x + a)^2*sinh(b*x + a) + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3)*log(cosh(b*x +
a) + sinh(b*x + a) - 1) + 6*(cosh(b*x + a)^5 + 10*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a) + 1)/(b*co
sh(b*x + a)^3 + 3*b*cosh(b*x + a)^2*sinh(b*x + a) + 3*b*cosh(b*x + a)*sinh(b*x + a)^2 + b*sinh(b*x + a)^3)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**3*coth(b*x+a),x)

[Out]

Integral(cosh(a + b*x)**3*coth(a + b*x), x)

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Giac [B]  time = 1.19812, size = 104, normalized size = 2.74 \begin{align*} \frac{{\left (15 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} +{\left (e^{\left (3 \, b x + 18 \, a\right )} + 15 \, e^{\left (b x + 16 \, a\right )}\right )} e^{\left (-15 \, a\right )} - 24 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 24 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{24 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*coth(b*x+a),x, algorithm="giac")

[Out]

1/24*((15*e^(2*b*x + 2*a) + 1)*e^(-3*b*x - 3*a) + (e^(3*b*x + 18*a) + 15*e^(b*x + 16*a))*e^(-15*a) - 24*log(e^
(b*x + a) + 1) + 24*log(abs(e^(b*x + a) - 1)))/b