### 3.497 $$\int x \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx$$

Optimal. Leaf size=30 $\frac{\log (\sinh (2 a+2 b x))}{b^2}-\frac{2 x \coth (2 a+2 b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.0568041, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {5461, 4184, 3475} $\frac{\log (\sinh (2 a+2 b x))}{b^2}-\frac{2 x \coth (2 a+2 b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx &=4 \int x \text{csch}^2(2 a+2 b x) \, dx\\ &=-\frac{2 x \coth (2 a+2 b x)}{b}+\frac{2 \int \coth (2 a+2 b x) \, dx}{b}\\ &=-\frac{2 x \coth (2 a+2 b x)}{b}+\frac{\log (\sinh (2 a+2 b x))}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.145231, size = 26, normalized size = 0.87 $\frac{\log (\sinh (2 (a+b x)))-2 b x \coth (2 (a+b x))}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

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

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Maple [B]  time = 0.037, size = 62, normalized size = 2.1 \begin{align*} -4\,{\frac{x}{b}}-4\,{\frac{a}{{b}^{2}}}-4\,{\frac{x}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+{\frac{\ln \left ({{\rm e}^{4\,bx+4\,a}}-1 \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.11937, size = 117, normalized size = 3.9 \begin{align*} -\frac{4 \, x e^{\left (4 \, b x + 4 \, a\right )}}{b e^{\left (4 \, b x + 4 \, a\right )} - b} + \frac{\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{2}} + \frac{\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{2}} + \frac{\log \left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, a\right )}\right )}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.2761, size = 788, normalized size = 26.27 \begin{align*} -\frac{4 \, b x \cosh \left (b x + a\right )^{4} + 16 \, b x \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 24 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 4 \, b x \sinh \left (b x + a\right )^{4} -{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} - 1\right )} \log \left (\frac{4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2}}\right )}{b^{2} \cosh \left (b x + a\right )^{4} + 4 \, b^{2} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, b^{2} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b^{2} \sinh \left (b x + a\right )^{4} - b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(b*x+a)^2*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

-(4*b*x*cosh(b*x + a)^4 + 16*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 24*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 16*b
*x*cosh(b*x + a)*sinh(b*x + a)^3 + 4*b*x*sinh(b*x + a)^4 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a)
+ 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*log(4*cosh(b*x +
a)*sinh(b*x + a)/(cosh(b*x + a)^2 - 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)))/(b^2*cosh(b*x + a)^4 +
4*b^2*cosh(b*x + a)^3*sinh(b*x + a) + 6*b^2*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b^2*cosh(b*x + a)*sinh(b*x + a
)^3 + b^2*sinh(b*x + a)^4 - b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}^{2}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(b*x+a)**2*sech(b*x+a)**2,x)

[Out]

Integral(x*csch(a + b*x)**2*sech(a + b*x)**2, x)

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Giac [B]  time = 1.16235, size = 97, normalized size = 3.23 \begin{align*} -\frac{4 \, b x e^{\left (4 \, b x + 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} \log \left (e^{\left (4 \, b x + 4 \, a\right )} - 1\right ) + \log \left (e^{\left (4 \, b x + 4 \, a\right )} - 1\right )}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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