3.285 \(\int x^2 \coth ^{-1}(e^x) \, dx\)

Optimal. Leaf size=70 \[ \frac {1}{2} x^2 \text {Li}_2\left (-e^{-x}\right )-\frac {1}{2} x^2 \text {Li}_2\left (e^{-x}\right )+x \text {Li}_3\left (-e^{-x}\right )-x \text {Li}_3\left (e^{-x}\right )+\text {Li}_4\left (-e^{-x}\right )-\text {Li}_4\left (e^{-x}\right ) \]

[Out]

1/2*x^2*polylog(2,-1/exp(x))-1/2*x^2*polylog(2,exp(-x))+x*polylog(3,-1/exp(x))-x*polylog(3,exp(-x))+polylog(4,
-1/exp(x))-polylog(4,exp(-x))

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Rubi [A]  time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6214, 2531, 6609, 2282, 6589} \[ \frac {1}{2} x^2 \text {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x^2 \text {PolyLog}\left (2,e^{-x}\right )+x \text {PolyLog}\left (3,-e^{-x}\right )-x \text {PolyLog}\left (3,e^{-x}\right )+\text {PolyLog}\left (4,-e^{-x}\right )-\text {PolyLog}\left (4,e^{-x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[E^x],x]

[Out]

(x^2*PolyLog[2, -E^(-x)])/2 - (x^2*PolyLog[2, E^(-x)])/2 + x*PolyLog[3, -E^(-x)] - x*PolyLog[3, E^(-x)] + Poly
Log[4, -E^(-x)] - PolyLog[4, E^(-x)]

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 2531

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 6214

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 6589

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]

Rule 6609

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

Rubi steps

\begin {align*} \int x^2 \coth ^{-1}\left (e^x\right ) \, dx &=-\left (\frac {1}{2} \int x^2 \log \left (1-e^{-x}\right ) \, dx\right )+\frac {1}{2} \int x^2 \log \left (1+e^{-x}\right ) \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2\left (-e^{-x}\right )-\frac {1}{2} x^2 \text {Li}_2\left (e^{-x}\right )-\int x \text {Li}_2\left (-e^{-x}\right ) \, dx+\int x \text {Li}_2\left (e^{-x}\right ) \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2\left (-e^{-x}\right )-\frac {1}{2} x^2 \text {Li}_2\left (e^{-x}\right )+x \text {Li}_3\left (-e^{-x}\right )-x \text {Li}_3\left (e^{-x}\right )-\int \text {Li}_3\left (-e^{-x}\right ) \, dx+\int \text {Li}_3\left (e^{-x}\right ) \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2\left (-e^{-x}\right )-\frac {1}{2} x^2 \text {Li}_2\left (e^{-x}\right )+x \text {Li}_3\left (-e^{-x}\right )-x \text {Li}_3\left (e^{-x}\right )+\operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-x}\right )-\operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{-x}\right )\\ &=\frac {1}{2} x^2 \text {Li}_2\left (-e^{-x}\right )-\frac {1}{2} x^2 \text {Li}_2\left (e^{-x}\right )+x \text {Li}_3\left (-e^{-x}\right )-x \text {Li}_3\left (e^{-x}\right )+\text {Li}_4\left (-e^{-x}\right )-\text {Li}_4\left (e^{-x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 93, normalized size = 1.33 \[ \frac {1}{6} \left (-3 x^2 \text {Li}_2\left (-e^x\right )+3 x^2 \text {Li}_2\left (e^x\right )+6 x \text {Li}_3\left (-e^x\right )-6 x \text {Li}_3\left (e^x\right )-6 \text {Li}_4\left (-e^x\right )+6 \text {Li}_4\left (e^x\right )+x^3 \log \left (1-e^x\right )-x^3 \log \left (e^x+1\right )+2 x^3 \coth ^{-1}\left (e^x\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[E^x],x]

[Out]

(2*x^3*ArcCoth[E^x] + x^3*Log[1 - E^x] - x^3*Log[1 + E^x] - 3*x^2*PolyLog[2, -E^x] + 3*x^2*PolyLog[2, E^x] + 6
*x*PolyLog[3, -E^x] - 6*x*PolyLog[3, E^x] - 6*PolyLog[4, -E^x] + 6*PolyLog[4, E^x])/6

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fricas [C]  time = 0.69, size = 119, normalized size = 1.70 \[ \frac {1}{6} \, x^{3} \log \left (\frac {\cosh \relax (x) + \sinh \relax (x) + 1}{\cosh \relax (x) + \sinh \relax (x) - 1}\right ) - \frac {1}{6} \, x^{3} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + \frac {1}{6} \, x^{3} \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right ) + \frac {1}{2} \, x^{2} {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) - \frac {1}{2} \, x^{2} {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) - x {\rm polylog}\left (3, \cosh \relax (x) + \sinh \relax (x)\right ) + x {\rm polylog}\left (3, -\cosh \relax (x) - \sinh \relax (x)\right ) + {\rm polylog}\left (4, \cosh \relax (x) + \sinh \relax (x)\right ) - {\rm polylog}\left (4, -\cosh \relax (x) - \sinh \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(x)),x, algorithm="fricas")

[Out]

1/6*x^3*log((cosh(x) + sinh(x) + 1)/(cosh(x) + sinh(x) - 1)) - 1/6*x^3*log(cosh(x) + sinh(x) + 1) + 1/6*x^3*lo
g(-cosh(x) - sinh(x) + 1) + 1/2*x^2*dilog(cosh(x) + sinh(x)) - 1/2*x^2*dilog(-cosh(x) - sinh(x)) - x*polylog(3
, cosh(x) + sinh(x)) + x*polylog(3, -cosh(x) - sinh(x)) + polylog(4, cosh(x) + sinh(x)) - polylog(4, -cosh(x)
- sinh(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arcoth}\left (e^{x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(x)),x, algorithm="giac")

[Out]

integrate(x^2*arccoth(e^x), x)

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maple [A]  time = 0.06, size = 79, normalized size = 1.13 \[ \frac {x^{3} \mathrm {arccoth}\left ({\mathrm e}^{x}\right )}{3}-\frac {x^{3} \ln \left ({\mathrm e}^{x}+1\right )}{6}-\frac {x^{2} \polylog \left (2, -{\mathrm e}^{x}\right )}{2}+x \polylog \left (3, -{\mathrm e}^{x}\right )-\polylog \left (4, -{\mathrm e}^{x}\right )+\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{6}+\frac {x^{2} \polylog \left (2, {\mathrm e}^{x}\right )}{2}-x \polylog \left (3, {\mathrm e}^{x}\right )+\polylog \left (4, {\mathrm e}^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(exp(x)),x)

[Out]

1/3*x^3*arccoth(exp(x))-1/6*x^3*ln(exp(x)+1)-1/2*x^2*polylog(2,-exp(x))+x*polylog(3,-exp(x))-polylog(4,-exp(x)
)+1/6*x^3*ln(1-exp(x))+1/2*x^2*polylog(2,exp(x))-x*polylog(3,exp(x))+polylog(4,exp(x))

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maxima [A]  time = 0.32, size = 76, normalized size = 1.09 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (e^{x}\right ) - \frac {1}{6} \, x^{3} \log \left (e^{x} + 1\right ) + \frac {1}{6} \, x^{3} \log \left (-e^{x} + 1\right ) - \frac {1}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) + \frac {1}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) + x {\rm Li}_{3}(-e^{x}) - x {\rm Li}_{3}(e^{x}) - {\rm Li}_{4}(-e^{x}) + {\rm Li}_{4}(e^{x}) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(x)),x, algorithm="maxima")

[Out]

1/3*x^3*arccoth(e^x) - 1/6*x^3*log(e^x + 1) + 1/6*x^3*log(-e^x + 1) - 1/2*x^2*dilog(-e^x) + 1/2*x^2*dilog(e^x)
+ x*polylog(3, -e^x) - x*polylog(3, e^x) - polylog(4, -e^x) + polylog(4, e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {acoth}\left ({\mathrm {e}}^x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acoth(exp(x)),x)

[Out]

int(x^2*acoth(exp(x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {acoth}{\left (e^{x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(exp(x)),x)

[Out]

Integral(x**2*acoth(exp(x)), x)

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