3.287 \(\int x \coth ^{-1}(e^{a+b x}) \, dx\)

Optimal. Leaf size=83 \[ \frac{\text{PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac{\text{PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2}+\frac{x \text{PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac{x \text{PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]

[Out]

(x*PolyLog[2, -E^(-a - b*x)])/(2*b) - (x*PolyLog[2, E^(-a - b*x)])/(2*b) + PolyLog[3, -E^(-a - b*x)]/(2*b^2) -
PolyLog[3, E^(-a - b*x)]/(2*b^2)

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Rubi [A]  time = 0.0642786, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6214, 2531, 2282, 6589} \[ \frac{\text{PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac{\text{PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2}+\frac{x \text{PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac{x \text{PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[E^(a + b*x)],x]

[Out]

(x*PolyLog[2, -E^(-a - b*x)])/(2*b) - (x*PolyLog[2, E^(-a - b*x)])/(2*b) + PolyLog[3, -E^(-a - b*x)]/(2*b^2) -
PolyLog[3, E^(-a - b*x)]/(2*b^2)

Rule 6214

Int[ArcCoth[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + 1/(a +
b*f^(c + d*x))], x], x] - Dist[1/2, Int[x^m*Log[1 - 1/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f},
x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac{1}{2} \int x \log \left (1-e^{-a-b x}\right ) \, dx\right )+\frac{1}{2} \int x \log \left (1+e^{-a-b x}\right ) \, dx\\ &=\frac{x \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x \text{Li}_2\left (e^{-a-b x}\right )}{2 b}-\frac{\int \text{Li}_2\left (-e^{-a-b x}\right ) \, dx}{2 b}+\frac{\int \text{Li}_2\left (e^{-a-b x}\right ) \, dx}{2 b}\\ &=\frac{x \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x \text{Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}\\ &=\frac{x \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x \text{Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac{\text{Li}_3\left (-e^{-a-b x}\right )}{2 b^2}-\frac{\text{Li}_3\left (e^{-a-b x}\right )}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0309224, size = 113, normalized size = 1.36 \[ \frac{-2 b x \text{PolyLog}\left (2,-e^{a+b x}\right )+2 b x \text{PolyLog}\left (2,e^{a+b x}\right )+2 \text{PolyLog}\left (3,-e^{a+b x}\right )-2 \text{PolyLog}\left (3,e^{a+b x}\right )+b^2 x^2 \log \left (1-e^{a+b x}\right )-b^2 x^2 \log \left (e^{a+b x}+1\right )+2 b^2 x^2 \coth ^{-1}\left (e^{a+b x}\right )}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[E^(a + b*x)],x]

[Out]

(2*b^2*x^2*ArcCoth[E^(a + b*x)] + b^2*x^2*Log[1 - E^(a + b*x)] - b^2*x^2*Log[1 + E^(a + b*x)] - 2*b*x*PolyLog[
2, -E^(a + b*x)] + 2*b*x*PolyLog[2, E^(a + b*x)] + 2*PolyLog[3, -E^(a + b*x)] - 2*PolyLog[3, E^(a + b*x)])/(4*
b^2)

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Maple [B]  time = 0.07, size = 153, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left ({{\rm e}^{bx+a}}\right )}{2}}-{\frac{\ln \left ({{\rm e}^{bx+a}}+1 \right ){x}^{2}}{4}}+{\frac{\ln \left ({{\rm e}^{bx+a}}+1 \right ){a}^{2}}{4\,{b}^{2}}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) x}{2\,b}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{4}}-{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{4\,{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) x}{2\,b}}-{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}-{\frac{{a}^{2}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(exp(b*x+a)),x)

[Out]

1/2*x^2*arccoth(exp(b*x+a))-1/4*ln(exp(b*x+a)+1)*x^2+1/4/b^2*ln(exp(b*x+a)+1)*a^2-1/2/b*polylog(2,-exp(b*x+a))
*x+1/2/b^2*polylog(3,-exp(b*x+a))+1/4*ln(1-exp(b*x+a))*x^2-1/4/b^2*ln(1-exp(b*x+a))*a^2+1/2/b*polylog(2,exp(b*
x+a))*x-1/2/b^2*polylog(3,exp(b*x+a))-1/2/b^2*a^2*arctanh(exp(b*x+a))

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Maxima [A]  time = 1.1348, size = 146, normalized size = 1.76 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (e^{\left (b x + a\right )}\right ) - \frac{1}{4} \, b{\left (\frac{b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} - \frac{b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(b*x+a)),x, algorithm="maxima")

[Out]

1/2*x^2*arccoth(e^(b*x + a)) - 1/4*b*((b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylog(3,
-e^(b*x + a)))/b^3 - (b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b
^3)

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Fricas [C]  time = 1.61181, size = 583, normalized size = 7.02 \begin{align*} \frac{b^{2} x^{2} \log \left (\frac{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, b x{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, b x{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) +{\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \,{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \,{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(b*x+a)),x, algorithm="fricas")

[Out]

1/4*(b^2*x^2*log((cosh(b*x + a) + sinh(b*x + a) + 1)/(cosh(b*x + a) + sinh(b*x + a) - 1)) - b^2*x^2*log(cosh(b
*x + a) + sinh(b*x + a) + 1) + 2*b*x*dilog(cosh(b*x + a) + sinh(b*x + a)) - 2*b*x*dilog(-cosh(b*x + a) - sinh(
b*x + a)) + a^2*log(cosh(b*x + a) + sinh(b*x + a) - 1) + (b^2*x^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) +
1) - 2*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 2*polylog(3, -cosh(b*x + a) - sinh(b*x + a)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(exp(b*x+a)),x)

[Out]

Integral(x*acoth(exp(a)*exp(b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arccoth(e^(b*x + a)), x)