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

Optimal. Leaf size=76 \[ -\frac {\text {Li}_2\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{2} x \log \left ((1-d) e^{2 a+2 b x}+1\right )+x \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 76, normalized size of antiderivative = 1.00, 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 = {6232, 2184, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-(1-d) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{2} x \log \left ((1-d) e^{2 a+2 b x}+1\right )+x \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^2)/2 + x*ArcCoth[1 - d - d*Tanh[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 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]

Rule 6232

Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCoth[c + d*Tanh[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]

Rubi steps

\begin {align*} \int \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx &=x \coth ^{-1}(1-d-d \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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*}

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Mathematica [B]  time = 0.98, size = 200, normalized size = 2.63 \[ \frac {-2 \text {Li}_2\left (-\sqrt {d-1} e^{a+b x}\right )-2 \text {Li}_2\left (\sqrt {d-1} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (1-\sqrt {d-1} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (\sqrt {d-1} e^{a+b x}+1\right )+2 \log \left (e^{a+b x}\right ) \log \left (e^{-a-b x} \left ((d-1) e^{2 (a+b x)}-1\right )\right )-2 b x \log (d \sinh (a+b x)+(d-2) \cosh (a+b x))+\log ^2\left (e^{a+b x}\right )+b^2 x^2}{4 b}+x \coth ^{-1}(d (-\tanh (a+b x))-d+1) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x*ArcCoth[1 - d - d*Tanh[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[(-2 + d)*Cosh[a + b*x] + 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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fricas [B]  time = 0.80, size = 227, normalized size = 2.99 \[ \frac {b^{2} x^{2} - b x \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) + a \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) - {\left (b x + a\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b x + a\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1-d-d*tanh(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 - 2)*cosh(b*x + a) + d*sinh(b*x + a))) + a*log(
2*(d - 1)*cosh(b*x + a) + 2*(d - 1)*sinh(b*x + a) + 2*sqrt(d - 1)) + a*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 1)
*sinh(b*x + a) - 2*sqrt(d - 1)) - (b*x + a)*log(sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b*x + a)*l
og(-sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - dilog(sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - di
log(-sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (-d \tanh \left (b x + a\right ) - d + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 271, normalized size = 3.57 \[ -\frac {\mathrm {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) \ln \left (-d \tanh \left (b x +a \right )+d \right )}{2 b}+\frac {\mathrm {arccoth}\left (1-d -d \tanh \left (b x +a \right )\right ) \ln \left (-d \tanh \left (b x +a \right )-d \right )}{2 b}+\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right )^{2}}{8 b}-\frac {\dilog \left (1-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}\right )}{4 b}-\frac {\ln \left (-d \tanh \left (b x +a \right )-d \right ) \ln \left (1-\frac {d \tanh \left (b x +a \right )}{2}-\frac {d}{2}\right )}{4 b}+\frac {\dilog \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{4 b}+\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (\frac {-d \tanh \left (b x +a \right )-d +2}{-2 d +2}\right )}{4 b}-\frac {\dilog \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{4 b}-\frac {\ln \left (-d \tanh \left (b x +a \right )+d \right ) \ln \left (-\frac {-d \tanh \left (b x +a \right )-d}{2 d}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.10, size = 73, normalized size = 0.96 \[ \frac {1}{4} \, b d {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, b x \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} - x \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-acoth(d + d*tanh(a + b*x) - 1),x)

[Out]

int(-acoth(d + d*tanh(a + b*x) - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {acoth}{\left (- d \tanh {\left (a + b x \right )} - d + 1 \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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