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

Optimal. Leaf size=76 \[ -\frac{\text{PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} x \log \left (1-(1-d) e^{2 a+2 b x}\right )+x \coth ^{-1}(d (-\coth (a+b x))-d+1)+\frac{b x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.144469, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6234, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} x \log \left (1-(1-d) e^{2 a+2 b x}\right )+x \coth ^{-1}(d (-\coth (a+b x))-d+1)+\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[1 - d - d*Coth[a + b*x]],x]

[Out]

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

Rule 6234

Int[ArcCoth[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcCoth[c + d*Coth[a + b*x]], x] + Di
st[b, Int[x/(c - d - c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c - d)^2, 1]

Rule 2184

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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 \coth ^{-1}(1-d-d \coth (a+b x)) \, dx &=x \coth ^{-1}(1-d-d \coth (a+b x))+b \int \frac{x}{1+(-1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^2}{2}+x \coth ^{-1}(1-d-d \coth (a+b x))+(b (1-d)) \int \frac{e^{2 a+2 b x} x}{1+(-1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^2}{2}+x \coth ^{-1}(1-d-d \coth (a+b x))-\frac{1}{2} x \log \left (1-(1-d) e^{2 a+2 b x}\right )+\frac{1}{2} \int \log \left (1+(-1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^2}{2}+x \coth ^{-1}(1-d-d \coth (a+b x))-\frac{1}{2} x \log \left (1-(1-d) e^{2 a+2 b x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\log (1+(-1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=\frac{b x^2}{2}+x \coth ^{-1}(1-d-d \coth (a+b x))-\frac{1}{2} x \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac{\text{Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}\\ \end{align*}

Mathematica [B]  time = 0.749857, size = 208, normalized size = 2.74 \[ \frac{-2 \text{PolyLog}\left (2,-\sqrt{1-d} e^{a+b x}\right )-2 \text{PolyLog}\left (2,\sqrt{1-d} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (1-\sqrt{1-d} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (\sqrt{1-d} e^{a+b x}+1\right )+2 \log \left (e^{a+b x}\right ) \log \left (e^{-a-b x} \left ((d-1) e^{2 (a+b x)}+1\right )\right )-2 b x \log ((d-2) \sinh (a+b x)+d \cosh (a+b x))+\log ^2\left (e^{a+b x}\right )+b^2 x^2}{4 b}+x \coth ^{-1}(d (-\coth (a+b x))-d+1) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[1 - d - d*Coth[a + b*x]],x]

[Out]

x*ArcCoth[1 - d - d*Coth[a + b*x]] + (b^2*x^2 + Log[E^(a + b*x)]^2 - 2*Log[E^(a + b*x)]*Log[1 - Sqrt[1 - d]*E^
(a + b*x)] - 2*Log[E^(a + b*x)]*Log[1 + Sqrt[1 - d]*E^(a + b*x)] + 2*Log[E^(a + b*x)]*Log[E^(-a - b*x)*(1 + (-
1 + d)*E^(2*(a + b*x)))] - 2*b*x*Log[d*Cosh[a + b*x] + (-2 + d)*Sinh[a + b*x]] - 2*PolyLog[2, -(Sqrt[1 - d]*E^
(a + b*x))] - 2*PolyLog[2, Sqrt[1 - d]*E^(a + b*x)])/(4*b)

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Maple [B]  time = 0.175, size = 271, normalized size = 3.6 \begin{align*}{\frac{{\rm arccoth} \left (1-d-d{\rm coth} \left (bx+a\right )\right )\ln \left ( -d{\rm coth} \left (bx+a\right )-d \right ) }{2\,b}}-{\frac{{\rm arccoth} \left (1-d-d{\rm coth} \left (bx+a\right )\right )\ln \left ( -d{\rm coth} \left (bx+a\right )+d \right ) }{2\,b}}+{\frac{ \left ( \ln \left ( -d{\rm coth} \left (bx+a\right )-d \right ) \right ) ^{2}}{8\,b}}-{\frac{1}{4\,b}{\it dilog} \left ( 1-{\frac{d{\rm coth} \left (bx+a\right )}{2}}-{\frac{d}{2}} \right ) }-{\frac{\ln \left ( -d{\rm coth} \left (bx+a\right )-d \right ) }{4\,b}\ln \left ( 1-{\frac{d{\rm coth} \left (bx+a\right )}{2}}-{\frac{d}{2}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ( -{\frac{-d{\rm coth} \left (bx+a\right )-d}{2\,d}} \right ) }-{\frac{\ln \left ( -d{\rm coth} \left (bx+a\right )+d \right ) }{4\,b}\ln \left ( -{\frac{-d{\rm coth} \left (bx+a\right )-d}{2\,d}} \right ) }+{\frac{1}{4\,b}{\it dilog} \left ({\frac{-d{\rm coth} \left (bx+a\right )-d+2}{-2\,d+2}} \right ) }+{\frac{\ln \left ( -d{\rm coth} \left (bx+a\right )+d \right ) }{4\,b}\ln \left ({\frac{-d{\rm coth} \left (bx+a\right )-d+2}{-2\,d+2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1-d-d*coth(b*x+a)),x)

[Out]

1/2/b*arccoth(1-d-d*coth(b*x+a))*ln(-d*coth(b*x+a)-d)-1/2/b*arccoth(1-d-d*coth(b*x+a))*ln(-d*coth(b*x+a)+d)+1/
8/b*ln(-d*coth(b*x+a)-d)^2-1/4/b*dilog(1-1/2*d*coth(b*x+a)-1/2*d)-1/4/b*ln(-d*coth(b*x+a)-d)*ln(1-1/2*d*coth(b
*x+a)-1/2*d)-1/4/b*dilog(-1/2*(-d*coth(b*x+a)-d)/d)-1/4/b*ln(-d*coth(b*x+a)+d)*ln(-1/2*(-d*coth(b*x+a)-d)/d)+1
/4/b*dilog((-d*coth(b*x+a)-d+2)/(-2*d+2))+1/4/b*ln(-d*coth(b*x+a)+d)*ln((-d*coth(b*x+a)-d+2)/(-2*d+2))

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Maxima [A]  time = 3.52798, size = 99, normalized size = 1.3 \begin{align*} \frac{1}{4} \, b d{\left (\frac{2 \, x^{2}}{d} - \frac{2 \, b x \log \left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} - x \operatorname{arcoth}\left (d \coth \left (b x + a\right ) + d - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-d-d*coth(b*x+a)),x, algorithm="maxima")

[Out]

1/4*b*d*(2*x^2/d - (2*b*x*log((d - 1)*e^(2*b*x + 2*a) + 1) + dilog(-(d - 1)*e^(2*b*x + 2*a)))/(b^2*d)) - x*arc
coth(d*coth(b*x + a) + d - 1)

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Fricas [B]  time = 1.74687, size = 707, normalized size = 9.3 \begin{align*} \frac{b^{2} x^{2} - b x \log \left (\frac{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) +{\left (d - 2\right )} \sinh \left (b x + a\right )}\right ) + a \log \left (2 \,{\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d - 1\right )} \sinh \left (b x + a\right ) + \sqrt{-4 \, d + 4}\right ) + a \log \left (2 \,{\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d - 1\right )} \sinh \left (b x + a\right ) - \sqrt{-4 \, d + 4}\right ) -{\left (b x + a\right )} \log \left (\frac{1}{2} \, \sqrt{-4 \, d + 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) -{\left (b x + a\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 \, d + 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) -{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 \, d + 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) -{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 \, d + 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-d-d*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/2*(b^2*x^2 - b*x*log((d*cosh(b*x + a) + d*sinh(b*x + a))/(d*cosh(b*x + a) + (d - 2)*sinh(b*x + a))) + a*log(
2*(d - 1)*cosh(b*x + a) + 2*(d - 1)*sinh(b*x + a) + sqrt(-4*d + 4)) + a*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 1
)*sinh(b*x + a) - sqrt(-4*d + 4)) - (b*x + a)*log(1/2*sqrt(-4*d + 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b
*x + a)*log(-1/2*sqrt(-4*d + 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - dilog(1/2*sqrt(-4*d + 4)*(cosh(b*x + a)
+ sinh(b*x + a))) - dilog(-1/2*sqrt(-4*d + 4)*(cosh(b*x + a) + sinh(b*x + a))))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{acoth}{\left (- d \coth{\left (a + b x \right )} - d + 1 \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(1-d-d*coth(b*x+a)),x)

[Out]

Integral(acoth(-d*coth(a + b*x) - d + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arcoth}\left (-d \coth \left (b x + a\right ) - d + 1\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-d-d*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(-d*coth(b*x + a) - d + 1), x)