### 3.407 $$\int x \cosh (a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=66 $-\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac{\sinh (a+b x)}{b^2}+\frac{x \cosh (a+b x)}{b}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

(-2*x*ArcTanh[E^(a + b*x)])/b + (x*Cosh[a + b*x])/b - PolyLog[2, -E^(a + b*x)]/b^2 + PolyLog[2, E^(a + b*x)]/b
^2 - Sinh[a + b*x]/b^2

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Rubi [A]  time = 0.0611381, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.429, Rules used = {5450, 3296, 2637, 4182, 2279, 2391} $-\frac{\text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{\text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac{\sinh (a+b x)}{b^2}+\frac{x \cosh (a+b x)}{b}-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Cosh[a + b*x]*Coth[a + b*x],x]

[Out]

(-2*x*ArcTanh[E^(a + b*x)])/b + (x*Cosh[a + b*x])/b - PolyLog[2, -E^(a + b*x)]/b^2 + PolyLog[2, E^(a + b*x)]/b
^2 - Sinh[a + b*x]/b^2

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x \cosh (a+b x) \coth (a+b x) \, dx &=\int x \text{csch}(a+b x) \, dx+\int x \sinh (a+b x) \, dx\\ &=-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \cosh (a+b x)}{b}-\frac{\int \cosh (a+b x) \, dx}{b}-\frac{\int \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac{\int \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \cosh (a+b x)}{b}-\frac{\sinh (a+b x)}{b^2}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}\\ &=-\frac{2 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \cosh (a+b x)}{b}-\frac{\text{Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac{\text{Li}_2\left (e^{a+b x}\right )}{b^2}-\frac{\sinh (a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.132542, size = 131, normalized size = 1.98 $-\frac{-\text{PolyLog}\left (2,-e^{-a-b x}\right )+\text{PolyLog}\left (2,e^{-a-b x}\right )-a \log \left (1-e^{-a-b x}\right )-b x \log \left (1-e^{-a-b x}\right )+a \log \left (e^{-a-b x}+1\right )+b x \log \left (e^{-a-b x}+1\right )+\sinh (a+b x)-b x \cosh (a+b x)+a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((-(b*x*Cosh[a + b*x]) - a*Log[1 - E^(-a - b*x)] - b*x*Log[1 - E^(-a - b*x)] + a*Log[1 + E^(-a - b*x)] + b*x*
Log[1 + E^(-a - b*x)] + a*Log[Tanh[(a + b*x)/2]] - PolyLog[2, -E^(-a - b*x)] + PolyLog[2, E^(-a - b*x)] + Sinh
[a + b*x])/b^2)

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Maple [B]  time = 0.061, size = 139, normalized size = 2.1 \begin{align*}{\frac{ \left ( bx-1 \right ){{\rm e}^{bx+a}}}{2\,{b}^{2}}}+{\frac{ \left ( bx+1 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{a{\it Artanh} \left ({{\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)^2*csch(b*x+a),x)

[Out]

1/2*(b*x-1)/b^2*exp(b*x+a)+1/2*(b*x+1)/b^2*exp(-b*x-a)-1/b*ln(1+exp(b*x+a))*x-1/b^2*ln(1+exp(b*x+a))*a-1/b^2*p
olylog(2,-exp(b*x+a))+1/b*ln(1-exp(b*x+a))*x+1/b^2*ln(1-exp(b*x+a))*a+1/b^2*polylog(2,exp(b*x+a))+2/b^2*a*arct
anh(exp(b*x+a))

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Maxima [A]  time = 1.29479, size = 127, normalized size = 1.92 \begin{align*} \frac{{\left ({\left (b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} +{\left (b x + 1\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{2}} - \frac{b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac{b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.16253, size = 753, normalized size = 11.41 \begin{align*} \frac{{\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x - 1\right )} \sinh \left (b x + a\right )^{2} + b x + 2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 2 \,{\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\left (a \cosh \left (b x + a\right ) + a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left ({\left (b x + a\right )} \cosh \left (b x + a\right ) +{\left (b x + a\right )} \sinh \left (b x + a\right )\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 1}{2 \,{\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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