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

Optimal. Leaf size=28 \[ -\frac {\coth ^2(a+b x)}{2 b}-\frac {\log (\tanh (a+b x))}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 28, 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 = {2620, 14} \[ -\frac {\coth ^2(a+b x)}{2 b}-\frac {\log (\tanh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]^3*Sech[a + b*x],x]

[Out]

-Coth[a + b*x]^2/(2*b) - Log[Tanh[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 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]

Rubi steps

\begin {align*} \int \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{x^3} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{x^3}+\frac {1}{x}\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=-\frac {\coth ^2(a+b x)}{2 b}-\frac {\log (\tanh (a+b x))}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 34, normalized size = 1.21 \[ -\frac {\text {csch}^2(a+b x)+2 \log (\sinh (a+b x))-2 \log (\cosh (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Csch[a + b*x]^2 - 2*Log[Cosh[a + b*x]] + 2*Log[Sinh[a + b*x]])/b

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 379, normalized size = 13.54 \[ -\frac {2 \, \cosh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )^{2}}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^3*sech(b*x+a),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*co
sh(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))*si
nh(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)

________________________________________________________________________________________

giac [B]  time = 0.15, size = 93, normalized size = 3.32 \[ \frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 6}{e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2} + \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right ) - \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.18, size = 27, normalized size = 0.96 \[ -\frac {1}{2 b \sinh \left (b x +a \right )^{2}}-\frac {\ln \left (\tanh \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 1.18, size = 91, normalized size = 3.25 \[ -\frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac {2 \, e^{\left (-2 \, b x - 2 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [B]  time = 1.45, size = 78, normalized size = 2.79 \[ \frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)*sinh(a + b*x)^3),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________