\(\int x \coth ^{-1}(e^{a+b x}) \, dx\) [287]

Optimal result

Integrand size = 10, antiderivative size = 83 \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )}{2 b}+\frac {\operatorname {PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac {\operatorname {PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2} \]

[Out]

1/2*x*polylog(2,-exp(-b*x-a))/b-1/2*x*polylog(2,exp(-b*x-a))/b+1/2*polylog(3,-exp(-b*x-a))/b^2-1/2*polylog(3,e
xp(-b*x-a))/b^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6349, 2611, 2320, 6724} \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {\operatorname {PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac {\operatorname {PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2}+\frac {x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]

[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 2320

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 2611

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 6349

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 6724

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*} \text {integral}& = -\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 \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (2,-e^{-a-b x}\right ) \, dx}{2 b}+\frac {\int \operatorname {PolyLog}\left (2,e^{-a-b x}\right ) \, dx}{2 b} \\ & = \frac {x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2}-\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,x)}{x} \, dx,x,e^{-a-b x}\right )}{2 b^2} \\ & = \frac {x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )}{2 b}+\frac {\operatorname {PolyLog}\left (3,-e^{-a-b x}\right )}{2 b^2}-\frac {\operatorname {PolyLog}\left (3,e^{-a-b x}\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {2 b^2 x^2 \coth ^{-1}\left (e^{a+b x}\right )+b^2 x^2 \log \left (1-e^{a+b x}\right )-b^2 x^2 \log \left (1+e^{a+b x}\right )-2 b x \operatorname {PolyLog}\left (2,-e^{a+b x}\right )+2 b x \operatorname {PolyLog}\left (2,e^{a+b x}\right )+2 \operatorname {PolyLog}\left (3,-e^{a+b x}\right )-2 \operatorname {PolyLog}\left (3,e^{a+b x}\right )}{4 b^2} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(71)=142\).

Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.16

method	result	size
default	\(\frac {x^{2} \operatorname {arccoth}\left ({\mathrm e}^{b x +a}\right )}{2}+\frac {-a^{2} \operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\left (b x +a \right )^{2} \ln \left (1-{\mathrm e}^{b x +a}\right )}{2}+\left (b x +a \right ) \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )-\operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )-\frac {\left (b x +a \right )^{2} \ln \left ({\mathrm e}^{b x +a}+1\right )}{2}-\left (b x +a \right ) \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )+\operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )-a \left (b x +a \right ) \ln \left (1-{\mathrm e}^{b x +a}\right )+a \left (b x +a \right ) \ln \left ({\mathrm e}^{b x +a}+1\right )-a \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )+a \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}\)	\(179\)
parts	\(\frac {x^{2} \operatorname {arccoth}\left ({\mathrm e}^{b x +a}\right )}{2}+\frac {-a^{2} \operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\left (b x +a \right )^{2} \ln \left (1-{\mathrm e}^{b x +a}\right )}{2}+\left (b x +a \right ) \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )-\operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )-\frac {\left (b x +a \right )^{2} \ln \left ({\mathrm e}^{b x +a}+1\right )}{2}-\left (b x +a \right ) \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )+\operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )-a \left (b x +a \right ) \ln \left (1-{\mathrm e}^{b x +a}\right )+a \left (b x +a \right ) \ln \left ({\mathrm e}^{b x +a}+1\right )-a \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )+a \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}\)	\(179\)
risch	\(-\frac {x^{2} \ln \left (-1+{\mathrm e}^{b x +a}\right )}{4}+\frac {x^{2} \ln \left (1-{\mathrm e}^{b x +a}\right )}{4}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a x}{2 b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{4 b^{2}}+\frac {x \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{2 b}+\frac {a^{2} \ln \left (-1+{\mathrm e}^{b x +a}\right )}{4 b^{2}}+\frac {\operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {a \operatorname {dilog}\left ({\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {\operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{2 b}+\frac {\operatorname {dilog}\left ({\mathrm e}^{b x +a}+1\right ) a}{2 b^{2}}-\frac {\operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{2 b^{2}}\)	\(202\)

[In]

int(x*arccoth(exp(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (69) = 138\).

Time = 0.25 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.39 \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {b^{2} x^{2} \log \left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{2} x^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 2 \, b x {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, b x {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{2}} \]

[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

Sympy [F]

\[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\int x \operatorname {acoth}{\left (e^{a} e^{b x} \right )}\, dx \]

[In]

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

[Out]

Integral(x*acoth(exp(a)*exp(b*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.30 \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {1}{2} \, x^{2} \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right ) - \frac {1}{4} \, b {\left (\frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} - \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}}\right )} \]

[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)

Giac [F]

\[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\int { x \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right ) \,d x } \]

[In]

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

[Out]

integrate(x*arccoth(e^(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int x \coth ^{-1}\left (e^{a+b x}\right ) \, dx=\int x\,\mathrm {acoth}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]

[In]

int(x*acoth(exp(a + b*x)),x)

[Out]

int(x*acoth(exp(a + b*x)), x)