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

Optimal. Leaf size=41 \[ \frac {\text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {\text {Li}_2\left (e^{-a-b x}\right )}{2 b} \]

[Out]

1/2*polylog(2,-exp(-b*x-a))/b-1/2*polylog(2,exp(-b*x-a))/b

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2282, 5913} \[ \frac {\text {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {\text {PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[E^(a + b*x)],x]

[Out]

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

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 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 68, normalized size = 1.66 \[ \frac {-\text {Li}_2\left (-e^{a+b x}\right )+\text {Li}_2\left (e^{a+b x}\right )+b x \left (\log \left (1-e^{a+b x}\right )-\log \left (e^{a+b x}+1\right )+2 \coth ^{-1}\left (e^{a+b x}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[E^(a + b*x)],x]

[Out]

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

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fricas [B]  time = 0.63, size = 137, normalized size = 3.34 \[ \frac {b x \log \left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b x \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b x + a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(e^(b*x + a)), x)

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maple [A]  time = 0.07, size = 67, normalized size = 1.63 \[ \frac {\ln \left ({\mathrm e}^{b x +a}\right ) \mathrm {arccoth}\left ({\mathrm e}^{b x +a}\right )}{b}-\frac {\dilog \left ({\mathrm e}^{b x +a}\right )}{2 b}-\frac {\dilog \left ({\mathrm e}^{b x +a}+1\right )}{2 b}-\frac {\ln \left ({\mathrm e}^{b x +a}\right ) \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*ln(exp(b*x+a))*arccoth(exp(b*x+a))-1/2/b*dilog(exp(b*x+a))-1/2/b*dilog(exp(b*x+a)+1)-1/2/b*ln(exp(b*x+a))*
ln(exp(b*x+a)+1)

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maxima [B]  time = 0.33, size = 107, normalized size = 2.61 \[ \frac {{\left (b x + a\right )} \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right )}{b} - \frac {{\left (b x + a\right )} {\left (\log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left (e^{\left (b x + a\right )} - 1\right )\right )} - \log \left (-e^{\left (b x + a\right )}\right ) \log \left (e^{\left (b x + a\right )} + 1\right ) + {\left (b x + a\right )} \log \left (e^{\left (b x + a\right )} - 1\right ) - {\rm Li}_2\left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )} + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {acoth}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(exp(a + b*x)),x)

[Out]

int(acoth(exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(exp(a + b*x)), x)

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