3.42 \(\int \frac{e^x x^3}{1-e^{2 x}} \, dx\)

Optimal. Leaf size=69 \[ \frac{3}{2} x^2 \text{PolyLog}\left (2,-e^x\right )-\frac{3}{2} x^2 \text{PolyLog}\left (2,e^x\right )-3 x \text{PolyLog}\left (3,-e^x\right )+3 x \text{PolyLog}\left (3,e^x\right )+3 \text{PolyLog}\left (4,-e^x\right )-3 \text{PolyLog}\left (4,e^x\right )+x^3 \tanh ^{-1}\left (e^x\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.121056, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {2249, 206, 2245, 6213, 2531, 6609, 2282, 6589} \[ \frac{3}{2} x^2 \text{PolyLog}\left (2,-e^x\right )-\frac{3}{2} x^2 \text{PolyLog}\left (2,e^x\right )-3 x \text{PolyLog}\left (3,-e^x\right )+3 x \text{PolyLog}\left (3,e^x\right )+3 \text{PolyLog}\left (4,-e^x\right )-3 \text{PolyLog}\left (4,e^x\right )+x^3 \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^x*x^3)/(1 - E^(2*x)),x]

[Out]

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

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2245

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^(m_.), x_Symbol] :> With[{u = IntHid
e[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Dist[x^m, u, x] - Dist[m, Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c,
d, e}, x] && EqQ[v, 2*e*(c + d*x)] && GtQ[m, 0] && ILtQ[p, 0]

Rule 6213

Int[ArcTanh[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + a + b*
f^(c + d*x)], x], x] - Dist[1/2, Int[x^m*Log[1 - a - b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] && IG
tQ[m, 0]

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 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]

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 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]

Rubi steps

\begin{align*} \int \frac{e^x x^3}{1-e^{2 x}} \, dx &=x^3 \tanh ^{-1}\left (e^x\right )-3 \int x^2 \tanh ^{-1}\left (e^x\right ) \, dx\\ &=x^3 \tanh ^{-1}\left (e^x\right )+\frac{3}{2} \int x^2 \log \left (1-e^x\right ) \, dx-\frac{3}{2} \int x^2 \log \left (1+e^x\right ) \, dx\\ &=x^3 \tanh ^{-1}\left (e^x\right )+\frac{3}{2} x^2 \text{Li}_2\left (-e^x\right )-\frac{3}{2} x^2 \text{Li}_2\left (e^x\right )-3 \int x \text{Li}_2\left (-e^x\right ) \, dx+3 \int x \text{Li}_2\left (e^x\right ) \, dx\\ &=x^3 \tanh ^{-1}\left (e^x\right )+\frac{3}{2} x^2 \text{Li}_2\left (-e^x\right )-\frac{3}{2} x^2 \text{Li}_2\left (e^x\right )-3 x \text{Li}_3\left (-e^x\right )+3 x \text{Li}_3\left (e^x\right )+3 \int \text{Li}_3\left (-e^x\right ) \, dx-3 \int \text{Li}_3\left (e^x\right ) \, dx\\ &=x^3 \tanh ^{-1}\left (e^x\right )+\frac{3}{2} x^2 \text{Li}_2\left (-e^x\right )-\frac{3}{2} x^2 \text{Li}_2\left (e^x\right )-3 x \text{Li}_3\left (-e^x\right )+3 x \text{Li}_3\left (e^x\right )+3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^x\right )-3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^x\right )\\ &=x^3 \tanh ^{-1}\left (e^x\right )+\frac{3}{2} x^2 \text{Li}_2\left (-e^x\right )-\frac{3}{2} x^2 \text{Li}_2\left (e^x\right )-3 x \text{Li}_3\left (-e^x\right )+3 x \text{Li}_3\left (e^x\right )+3 \text{Li}_4\left (-e^x\right )-3 \text{Li}_4\left (e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0384728, size = 89, normalized size = 1.29 \[ \frac{3}{2} x^2 \text{PolyLog}\left (2,-e^x\right )-\frac{3}{2} x^2 \text{PolyLog}\left (2,e^x\right )-3 x \text{PolyLog}\left (3,-e^x\right )+3 x \text{PolyLog}\left (3,e^x\right )+3 \text{PolyLog}\left (4,-e^x\right )-3 \text{PolyLog}\left (4,e^x\right )-\frac{1}{2} x^3 \log \left (1-e^x\right )+\frac{1}{2} x^3 \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x^3)/(1 - E^(2*x)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 74, normalized size = 1.1 \begin{align*}{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}}+{\frac{3\,{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{2}}-3\,x{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) +3\,{\it polylog} \left ( 4,-{{\rm e}^{x}} \right ) -{\frac{{x}^{3}\ln \left ( 1-{{\rm e}^{x}} \right ) }{2}}-{\frac{3\,{x}^{2}{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{2}}+3\,x{\it polylog} \left ( 3,{{\rm e}^{x}} \right ) -3\,{\it polylog} \left ( 4,{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*x^3/(1-exp(2*x)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.20274, size = 96, normalized size = 1.39 \begin{align*} \frac{1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac{1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac{3}{2} \, x^{2}{\rm Li}_2\left (-e^{x}\right ) - \frac{3}{2} \, x^{2}{\rm Li}_2\left (e^{x}\right ) - 3 \, x{\rm Li}_{3}(-e^{x}) + 3 \, x{\rm Li}_{3}(e^{x}) + 3 \,{\rm Li}_{4}(-e^{x}) - 3 \,{\rm Li}_{4}(e^{x}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C]  time = 1.49903, size = 236, normalized size = 3.42 \begin{align*} \frac{1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac{1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac{3}{2} \, x^{2}{\rm Li}_2\left (-e^{x}\right ) - \frac{3}{2} \, x^{2}{\rm Li}_2\left (e^{x}\right ) - 3 \, x{\rm polylog}\left (3, -e^{x}\right ) + 3 \, x{\rm polylog}\left (3, e^{x}\right ) + 3 \,{\rm polylog}\left (4, -e^{x}\right ) - 3 \,{\rm polylog}\left (4, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3} e^{x}}{e^{2 x} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x**3/(1-exp(2*x)),x)

[Out]

-Integral(x**3*exp(x)/(exp(2*x) - 1), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{3} e^{x}}{e^{\left (2 \, x\right )} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="giac")

[Out]

integrate(-x^3*e^x/(e^(2*x) - 1), x)