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

Optimal. Leaf size=150 \[ \frac{\text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{2} x \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \coth ^{-1}(d \tanh (a+b x)+c) \]

[Out]

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

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Rubi [A]  time = 0.231966, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6236, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{2} x \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \coth ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[c + d*Tanh[a + b*x]],x]

[Out]

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

Rule 6236

Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCoth[c + d*Tanh[a + b*x]], x] + (D
ist[b*(1 - c - d), Int[(x*E^(2*a + 2*b*x))/(1 - c + d + (1 - c - d)*E^(2*a + 2*b*x)), x], x] - Dist[b*(1 + c +
d), Int[(x*E^(2*a + 2*b*x))/(1 + c - d + (1 + c + d)*E^(2*a + 2*b*x)), x], x]) /; FreeQ[{a, b, c, d}, x] && N
eQ[(c - d)^2, 1]

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}(c+d \tanh (a+b x)) \, dx &=x \coth ^{-1}(c+d \tanh (a+b x))+(b (1-c-d)) \int \frac{e^{2 a+2 b x} x}{1-c+d+(1-c-d) e^{2 a+2 b x}} \, dx-(b (1+c+d)) \int \frac{e^{2 a+2 b x} x}{1+c-d+(1+c+d) e^{2 a+2 b x}} \, dx\\ &=x \coth ^{-1}(c+d \tanh (a+b x))+\frac{1}{2} x \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )-\frac{1}{2} \int \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx+\frac{1}{2} \int \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx\\ &=x \coth ^{-1}(c+d \tanh (a+b x))+\frac{1}{2} x \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{(1-c-d) x}{1-c+d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}+\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{(1+c+d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=x \coth ^{-1}(c+d \tanh (a+b x))+\frac{1}{2} x \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{\text{Li}_2\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{\text{Li}_2\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 1.24638, size = 131, normalized size = 0.87 \[ \frac{\text{PolyLog}\left (2,-\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}\right )-\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}\right )+2 b x \left (\log \left (\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}+1\right )-\log \left (\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}+1\right )\right )}{4 b}+x \coth ^{-1}(d \tanh (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[c + d*Tanh[a + b*x]],x]

[Out]

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

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Maple [B]  time = 0.164, size = 306, normalized size = 2. \begin{align*} -{\frac{{\rm arccoth} \left (c+d\tanh \left ( bx+a \right ) \right )\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{2\,b}}+{\frac{{\rm arccoth} \left (c+d\tanh \left ( bx+a \right ) \right )\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{2\,b}}+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c+d}} \right ) }+{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c+d}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c+d-1}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c+d-1}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c-d}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c-d}} \right ) }+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c-d-1}} \right ) }+{\frac{\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c-d-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(c+d*tanh(b*x+a)),x)

[Out]

-1/2/b*arccoth(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)-d)+1/2/b*arccoth(c+d*tanh(b*x+a))*ln(d*tanh(b*x+a)+d)+1/4/b*d
ilog((d*tanh(b*x+a)+c+1)/(1+c+d))+1/4/b*ln(d*tanh(b*x+a)-d)*ln((d*tanh(b*x+a)+c+1)/(1+c+d))-1/4/b*dilog((d*tan
h(b*x+a)+c-1)/(c+d-1))-1/4/b*ln(d*tanh(b*x+a)-d)*ln((d*tanh(b*x+a)+c-1)/(c+d-1))-1/4/b*dilog((d*tanh(b*x+a)+c+
1)/(1+c-d))-1/4/b*ln(d*tanh(b*x+a)+d)*ln((d*tanh(b*x+a)+c+1)/(1+c-d))+1/4/b*dilog((d*tanh(b*x+a)+c-1)/(c-d-1))
+1/4/b*ln(d*tanh(b*x+a)+d)*ln((d*tanh(b*x+a)+c-1)/(c-d-1))

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Maxima [A]  time = 2.1111, size = 192, normalized size = 1.28 \begin{align*} -\frac{1}{4} \, b d{\left (\frac{2 \, b x \log \left (\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) +{\rm Li}_2\left (-\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right )}{b^{2} d} - \frac{2 \, b x \log \left (\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) +{\rm Li}_2\left (-\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right )}{b^{2} d}\right )} + x \operatorname{arcoth}\left (d \tanh \left (b x + a\right ) + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(c+d*tanh(b*x+a)),x, algorithm="maxima")

[Out]

-1/4*b*d*((2*b*x*log((c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1) + 1) + dilog(-(c + d + 1)*e^(2*b*x + 2*a)/(c - d
+ 1)))/(b^2*d) - (2*b*x*log((c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1) + 1) + dilog(-(c + d - 1)*e^(2*b*x + 2*a)/
(c - d - 1)))/(b^2*d)) + x*arccoth(d*tanh(b*x + a) + c)

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Fricas [B]  time = 1.85118, size = 1601, normalized size = 10.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(c+d*tanh(b*x+a)),x, algorithm="fricas")

[Out]

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

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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(acoth(c+d*tanh(b*x+a)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(c+d*tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccoth(d*tanh(b*x + a) + c), x)