### 3.103 $$\int \cosh (a+b x) \coth ^3(a+b x) \, dx$$

Optimal. Leaf size=49 $\frac{3 \cosh (a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{\cosh (a+b x) \coth ^2(a+b x)}{2 b}$

[Out]

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

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Rubi [A]  time = 0.0359131, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.267, Rules used = {2592, 288, 321, 206} $\frac{3 \cosh (a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{\cosh (a+b x) \coth ^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]*Coth[a + b*x]^3,x]

[Out]

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

Mathematica [A]  time = 0.0319093, size = 67, normalized size = 1.37 $\frac{\cosh (a+b x)}{b}-\frac{\text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{3 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Cosh[a + b*x]/b - Csch[(a + b*x)/2]^2/(8*b) + (3*Log[Tanh[(a + b*x)/2]])/(2*b) - Sech[(a + b*x)/2]^2/(8*b)

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Maple [A]  time = 0.02, size = 62, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}-3\,{\frac{\cosh \left ( bx+a \right ) }{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3\,{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}-3\,{\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)*coth(b*x+a)^3,x)

[Out]

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

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Maxima [B]  time = 1.21084, size = 146, normalized size = 2.98 \begin{align*} \frac{e^{\left (-b x - a\right )}}{2 \, b} - \frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} - \frac{4 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} - 1}{2 \, b{\left (e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^(-b*x - a)/b - 3/2*log(e^(-b*x - a) + 1)/b + 3/2*log(e^(-b*x - a) - 1)/b - 1/2*(4*e^(-2*b*x - 2*a) + e^(
-4*b*x - 4*a) - 1)/(b*(e^(-b*x - a) - 2*e^(-3*b*x - 3*a) + e^(-5*b*x - 5*a)))

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Fricas [B]  time = 2.02582, size = 1710, normalized size = 34.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x
+ a)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 -
6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 3*cosh(b*x + a)^2 - 3*(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a
)^4 + sinh(b*x + a)^5 + 2*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^3 - 2*cosh(b*x + a)^3 + 2*(5*cosh(b*x + a)^3 -
3*cosh(b*x + a))*sinh(b*x + a)^2 + (5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))
*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 3*(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a)^4 + sinh(b*x + a)
^5 + 2*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^3 - 2*cosh(b*x + a)^3 + 2*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*s
inh(b*x + a)^2 + (5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))*log(cosh(b*x + a)
+ sinh(b*x + a) - 1) + 6*(cosh(b*x + a)^5 - 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)/(b*cosh(b*x
+ a)^5 + 5*b*cosh(b*x + a)*sinh(b*x + a)^4 + b*sinh(b*x + a)^5 - 2*b*cosh(b*x + a)^3 + 2*(5*b*cosh(b*x + a)^2
- b)*sinh(b*x + a)^3 + 2*(5*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^2 + b*cosh(b*x + a) + (5*b*co
sh(b*x + a)^4 - 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28521, size = 107, normalized size = 2.18 \begin{align*} -\frac{\frac{2 \,{\left (e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - e^{\left (b x + a\right )} - e^{\left (-b x - a\right )} + 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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