Integrand size = 18, antiderivative size = 40 \[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=x^2 \text {arctanh}\left (e^x\right )+x \operatorname {PolyLog}\left (2,-e^x\right )-x \operatorname {PolyLog}\left (2,e^x\right )-\operatorname {PolyLog}\left (3,-e^x\right )+\operatorname {PolyLog}\left (3,e^x\right ) \]
x^2*arctanh(exp(x))+x*polylog(2,-exp(x))-x*polylog(2,exp(x))-polylog(3,-ex p(x))+polylog(3,exp(x))
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.50 \[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=-\frac {1}{2} x^2 \log \left (1-e^x\right )+\frac {1}{2} x^2 \log \left (1+e^x\right )+x \operatorname {PolyLog}\left (2,-e^x\right )-x \operatorname {PolyLog}\left (2,e^x\right )-\operatorname {PolyLog}\left (3,-e^x\right )+\operatorname {PolyLog}\left (3,e^x\right ) \]
-1/2*(x^2*Log[1 - E^x]) + (x^2*Log[1 + E^x])/2 + x*PolyLog[2, -E^x] - x*Po lyLog[2, E^x] - PolyLog[3, -E^x] + PolyLog[3, E^x]
Time = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.32, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2675, 6767, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^x x^2}{1-e^{2 x}} \, dx\) |
\(\Big \downarrow \) 2675 |
\(\displaystyle x^2 \text {arctanh}\left (e^x\right )-2 \int x \text {arctanh}\left (e^x\right )dx\) |
\(\Big \downarrow \) 6767 |
\(\displaystyle x^2 \text {arctanh}\left (e^x\right )-2 \left (\frac {1}{2} \int x \log \left (1+e^x\right )dx-\frac {1}{2} \int x \log \left (1-e^x\right )dx\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle x^2 \text {arctanh}\left (e^x\right )-2 \left (\frac {1}{2} \left (\int \operatorname {PolyLog}\left (2,-e^x\right )dx-x \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x \operatorname {PolyLog}\left (2,e^x\right )-\int \operatorname {PolyLog}\left (2,e^x\right )dx\right )\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle x^2 \text {arctanh}\left (e^x\right )-2 \left (\frac {1}{2} \left (\int e^{-x} \operatorname {PolyLog}\left (2,-e^x\right )de^x-x \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x \operatorname {PolyLog}\left (2,e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (2,e^x\right )de^x\right )\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle x^2 \text {arctanh}\left (e^x\right )-2 \left (\frac {1}{2} \left (\operatorname {PolyLog}\left (3,-e^x\right )-x \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x \operatorname {PolyLog}\left (2,e^x\right )-\operatorname {PolyLog}\left (3,e^x\right )\right )\right )\) |
x^2*ArcTanh[E^x] - 2*((-(x*PolyLog[2, -E^x]) + PolyLog[3, -E^x])/2 + (x*Po lyLog[2, E^x] - PolyLog[3, E^x])/2)
3.1.41.3.1 Defintions of rubi rules used
Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^( m_.), x_Symbol] :> With[{u = IntHide[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Si mp[x^m u, x] - Simp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[ArcTanh[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Simp[1/2 Int[x^m*Log[1 + a + b*f^(c + d*x)], x], x] - Simp[1/2 Int[x ^m*Log[1 - a - b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] && IGtQ[ m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.28
method | result | size |
default | \(\frac {x^{2} \ln \left (1+{\mathrm e}^{x}\right )}{2}+x \,\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )-\operatorname {Li}_{3}\left (-{\mathrm e}^{x}\right )-\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{2}-x \,\operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )+\operatorname {Li}_{3}\left ({\mathrm e}^{x}\right )\) | \(51\) |
risch | \(\frac {x^{2} \ln \left (1+{\mathrm e}^{x}\right )}{2}+x \,\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )-\operatorname {Li}_{3}\left (-{\mathrm e}^{x}\right )-\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{2}-x \,\operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )+\operatorname {Li}_{3}\left ({\mathrm e}^{x}\right )\) | \(51\) |
1/2*x^2*ln(1+exp(x))+x*polylog(2,-exp(x))-polylog(3,-exp(x))-1/2*x^2*ln(1- exp(x))-x*polylog(2,exp(x))+polylog(3,exp(x))
Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{2} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{2} \log \left (-e^{x} + 1\right ) + x {\rm Li}_2\left (-e^{x}\right ) - x {\rm Li}_2\left (e^{x}\right ) - {\rm polylog}\left (3, -e^{x}\right ) + {\rm polylog}\left (3, e^{x}\right ) \]
1/2*x^2*log(e^x + 1) - 1/2*x^2*log(-e^x + 1) + x*dilog(-e^x) - x*dilog(e^x ) - polylog(3, -e^x) + polylog(3, e^x)
\[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=- \int \frac {x^{2} e^{x}}{e^{2 x} - 1}\, dx \]
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{2} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{2} \log \left (-e^{x} + 1\right ) + x {\rm Li}_2\left (-e^{x}\right ) - x {\rm Li}_2\left (e^{x}\right ) - {\rm Li}_{3}(-e^{x}) + {\rm Li}_{3}(e^{x}) \]
1/2*x^2*log(e^x + 1) - 1/2*x^2*log(-e^x + 1) + x*dilog(-e^x) - x*dilog(e^x ) - polylog(3, -e^x) + polylog(3, e^x)
\[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=\int { -\frac {x^{2} e^{x}}{e^{\left (2 \, x\right )} - 1} \,d x } \]
Timed out. \[ \int \frac {e^x x^2}{1-e^{2 x}} \, dx=-\int \frac {x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1} \,d x \]