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

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

[Out]

-csch(b*x+a)/b+sinh(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, 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 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]]

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]

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*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ \frac {\sinh (a+b x)}{b}-\frac {\text {csch}(a+b x)}{b} \]

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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fricas [A]  time = 0.49, size = 31, normalized size = 1.41 \[ \frac {\cosh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{2} - 3}{2 \, b \sinh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*coth(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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giac [B]  time = 0.14, size = 50, normalized size = 2.27 \[ -\frac {\frac {{\left (5 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-a\right )}}{e^{\left (3 \, b x + 2 \, a\right )} - e^{\left (b x\right )}} - e^{\left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 33, normalized size = 1.50 \[ \frac {\frac {\cosh ^{2}\left (b x +a \right )}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 56, normalized size = 2.55 \[ -\frac {e^{\left (-b x - a\right )}}{2 \, b} - \frac {5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1}{2 \, b {\left (e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*coth(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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mupad [B]  time = 1.43, size = 49, normalized size = 2.23 \[ \frac {{\mathrm {e}}^{-a-b\,x}\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-6\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}{2\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*coth(a + b*x)^2,x)

[Out]

(exp(- a - b*x)*(exp(4*a + 4*b*x) - 6*exp(2*a + 2*b*x) + 1))/(2*b*(exp(2*a + 2*b*x) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh {\left (a + b x \right )} \coth ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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