3.26 \(\int \text{csch}(a+b x) \text{sech}^3(a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac{\log (\tanh (a+b x))}{b}-\frac{\tanh ^2(a+b x)}{2 b} \]

[Out]

Log[Tanh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0235788, antiderivative size = 27, 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 = {2620, 14} \[ \frac{\log (\tanh (a+b x))}{b}-\frac{\tanh ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]*Sech[a + b*x]^3,x]

[Out]

Log[Tanh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b)

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/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 \text{csch}(a+b x) \text{sech}^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x}+x\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{\log (\tanh (a+b x))}{b}-\frac{\tanh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0362133, size = 36, normalized size = 1.33 \[ -\frac{-\text{sech}^2(a+b x)-2 \log (\sinh (a+b x))+2 \log (\cosh (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[a + b*x]*Sech[a + b*x]^3,x]

[Out]

-(2*Log[Cosh[a + b*x]] - 2*Log[Sinh[a + b*x]] - Sech[a + b*x]^2)/(2*b)

________________________________________________________________________________________

Maple [A]  time = 0.018, size = 26, normalized size = 1. \begin{align*}{\frac{1}{2\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{\ln \left ( \tanh \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b/cosh(b*x+a)^2+ln(tanh(b*x+a))/b

________________________________________________________________________________________

Maxima [B]  time = 1.60227, size = 119, normalized size = 4.41 \begin{align*} \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac{\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac{2 \, e^{\left (-2 \, b x - 2 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.21561, size = 1035, normalized size = 38.33 \begin{align*} \frac{2 \, \cosh \left (b x + a\right )^{2} -{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac{2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) +{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )^{2}}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)*sech(b*x+a)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)*sech(b*x+a)**3,x)

[Out]

Integral(csch(a + b*x)*sech(a + b*x)**3, x)

________________________________________________________________________________________

Giac [B]  time = 1.21783, size = 135, normalized size = 5. \begin{align*} -\frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}{2 \, b} + \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{2 \, b} + \frac{e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 6}{2 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2)/b + 1/2*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) - 2)/b + 1/2*(
e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 6)/(b*(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2))