### 3.427 $$\int x \coth (a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=25 $-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}$

[Out]

-(ArcTanh[Cosh[a + b*x]]/b^2) - (x*Csch[a + b*x])/b

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Rubi [A]  time = 0.018915, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {5419, 3770} $-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

-(ArcTanh[Cosh[a + b*x]]/b^2) - (x*Csch[a + b*x])/b

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 3770

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

Rubi steps

\begin{align*} \int x \coth (a+b x) \text{csch}(a+b x) \, dx &=-\frac{x \text{csch}(a+b x)}{b}+\frac{\int \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac{x \text{csch}(a+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 0.0483079, size = 114, normalized size = 4.56 $\frac{\log \left (\sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b^2}-\frac{\log \left (\cosh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b^2}-\frac{x \text{csch}(a)}{b}+\frac{x \text{csch}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{csch}\left (\frac{a}{2}+\frac{b x}{2}\right )}{2 b}+\frac{x \text{sech}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{sech}\left (\frac{a}{2}+\frac{b x}{2}\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.029, size = 54, normalized size = 2.2 \begin{align*} -2\,{\frac{x{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+{\frac{\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.31727, size = 86, normalized size = 3.44 \begin{align*} -\frac{2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, 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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.05509, size = 483, normalized size = 19.32 \begin{align*} -\frac{2 \, b x \cosh \left (b x + a\right ) + 2 \, b x \sinh \left (b x + a\right ) +{\left (\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} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) -{\left (\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} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right )}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} - b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.27432, size = 126, normalized size = 5.04 \begin{align*} -\frac{2 \, b x e^{\left (b x + a\right )} + e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) - e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (b x + a\right )} - 1\right ) - \log \left (e^{\left (b x + a\right )} + 1\right ) + \log \left (e^{\left (b x + a\right )} - 1\right )}{b^{2} e^{\left (2 \, b x + 2 \, a\right )} - b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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