3.442 \(\int \cosh (a+b x) \coth ^2(a+b x) \, dx\)

Optimal. Leaf size=22 \[ \frac{\sinh (a+b x)}{b}-\frac{\text{csch}(a+b x)}{b} \]

[Out]

-(Csch[a + b*x]/b) + Sinh[a + b*x]/b

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Rubi [A]  time = 0.0221304, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2590, 14} \[ \frac{\sinh (a+b x)}{b}-\frac{\text{csch}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]*Coth[a + b*x]^2,x]

[Out]

-(Csch[a + b*x]/b) + Sinh[a + b*x]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0129803, size = 22, normalized size = 1. \[ \frac{\sinh (a+b x)}{b}-\frac{\text{csch}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csch[a + b*x]/b) + Sinh[a + b*x]/b

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Maple [A]  time = 0.013, size = 32, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{\sinh \left ( bx+a \right ) }}+2\,\sinh \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*csch(b*x+a)^2,x)

[Out]

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

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Maxima [B]  time = 1.11023, size = 76, normalized size = 3.45 \begin{align*} -\frac{e^{\left (-b x - a\right )}}{2 \, b} - \frac{5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1}{2 \, b{\left (e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.19848, size = 85, normalized size = 3.86 \begin{align*} \frac{\cosh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{2} - 3}{2 \, 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)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**3*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.16081, size = 66, normalized size = 3. \begin{align*} \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} - \frac{2}{b{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="giac")

[Out]

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