Integrand size = 18, antiderivative size = 69 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=x^3 \text {arctanh}\left (e^x\right )+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^x\right )-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^x\right )-3 x \operatorname {PolyLog}\left (3,-e^x\right )+3 x \operatorname {PolyLog}\left (3,e^x\right )+3 \operatorname {PolyLog}\left (4,-e^x\right )-3 \operatorname {PolyLog}\left (4,e^x\right ) \] Output:
x^3*arctanh(exp(x))+3/2*x^2*polylog(2,-exp(x))-3/2*x^2*polylog(2,exp(x))-3 *x*polylog(3,-exp(x))+3*x*polylog(3,exp(x))+3*polylog(4,-exp(x))-3*polylog (4,exp(x))
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=-\frac {1}{2} x^3 \log \left (1-e^x\right )+\frac {1}{2} x^3 \log \left (1+e^x\right )+\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,-e^x\right )-\frac {3}{2} x^2 \operatorname {PolyLog}\left (2,e^x\right )-3 x \operatorname {PolyLog}\left (3,-e^x\right )+3 x \operatorname {PolyLog}\left (3,e^x\right )+3 \operatorname {PolyLog}\left (4,-e^x\right )-3 \operatorname {PolyLog}\left (4,e^x\right ) \] Input:
Integrate[(E^x*x^3)/(1 - E^(2*x)),x]
Output:
-1/2*(x^3*Log[1 - E^x]) + (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*PolyLog[3, -E^x] + 3*x*PolyLog[3, E^x] + 3*PolyLog[4, -E^x] - 3*PolyLog[4, E^x]
Time = 0.87 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2675, 6767, 3011, 7163, 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.
\(\displaystyle \int \frac {e^x x^3}{1-e^{2 x}} \, dx\) |
\(\Big \downarrow \) 2675 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \int x^2 \text {arctanh}\left (e^x\right )dx\) |
\(\Big \downarrow \) 6767 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \left (\frac {1}{2} \int x^2 \log \left (1+e^x\right )dx-\frac {1}{2} \int x^2 \log \left (1-e^x\right )dx\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \left (\frac {1}{2} \left (2 \int x \operatorname {PolyLog}\left (2,-e^x\right )dx-x^2 \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x^2 \operatorname {PolyLog}\left (2,e^x\right )-2 \int x \operatorname {PolyLog}\left (2,e^x\right )dx\right )\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \left (\frac {1}{2} \left (2 \left (x \operatorname {PolyLog}\left (3,-e^x\right )-\int \operatorname {PolyLog}\left (3,-e^x\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x^2 \operatorname {PolyLog}\left (2,e^x\right )-2 \left (x \operatorname {PolyLog}\left (3,e^x\right )-\int \operatorname {PolyLog}\left (3,e^x\right )dx\right )\right )\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \left (\frac {1}{2} \left (2 \left (x \operatorname {PolyLog}\left (3,-e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,-e^x\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x^2 \operatorname {PolyLog}\left (2,e^x\right )-2 \left (x \operatorname {PolyLog}\left (3,e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,e^x\right )de^x\right )\right )\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle x^3 \text {arctanh}\left (e^x\right )-3 \left (\frac {1}{2} \left (2 \left (x \operatorname {PolyLog}\left (3,-e^x\right )-\operatorname {PolyLog}\left (4,-e^x\right )\right )-x^2 \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {1}{2} \left (x^2 \operatorname {PolyLog}\left (2,e^x\right )-2 \left (x \operatorname {PolyLog}\left (3,e^x\right )-\operatorname {PolyLog}\left (4,e^x\right )\right )\right )\right )\) |
Input:
Int[(E^x*x^3)/(1 - E^(2*x)),x]
Output:
x^3*ArcTanh[E^x] - 3*((-(x^2*PolyLog[2, -E^x]) + 2*(x*PolyLog[3, -E^x] - P olyLog[4, -E^x]))/2 + (x^2*PolyLog[2, E^x] - 2*(x*PolyLog[3, E^x] - PolyLo g[4, E^x]))/2)
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]
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] - Simp[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]
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {x^{3} \ln \left ({\mathrm e}^{x}+1\right )}{2}+\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}-3 x \operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )+3 \operatorname {polylog}\left (4, -{\mathrm e}^{x}\right )-\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {3 x^{2} \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}+3 x \operatorname {polylog}\left (3, {\mathrm e}^{x}\right )-3 \operatorname {polylog}\left (4, {\mathrm e}^{x}\right )\) | \(74\) |
risch | \(\frac {x^{3} \ln \left ({\mathrm e}^{x}+1\right )}{2}+\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}-3 x \operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )+3 \operatorname {polylog}\left (4, -{\mathrm e}^{x}\right )-\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {3 x^{2} \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}+3 x \operatorname {polylog}\left (3, {\mathrm e}^{x}\right )-3 \operatorname {polylog}\left (4, {\mathrm e}^{x}\right )\) | \(74\) |
Input:
int(exp(x)*x^3/(1-exp(2*x)),x,method=_RETURNVERBOSE)
Output:
1/2*x^3*ln(exp(x)+1)+3/2*x^2*polylog(2,-exp(x))-3*x*polylog(3,-exp(x))+3*p olylog(4,-exp(x))-1/2*x^3*ln(1-exp(x))-3/2*x^2*polylog(2,exp(x))+3*x*polyl og(3,exp(x))-3*polylog(4,exp(x))
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac {3}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) - \frac {3}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) - 3 \, x {\rm polylog}\left (3, -e^{x}\right ) + 3 \, x {\rm polylog}\left (3, e^{x}\right ) + 3 \, {\rm polylog}\left (4, -e^{x}\right ) - 3 \, {\rm polylog}\left (4, e^{x}\right ) \] Input:
integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="fricas")
Output:
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)
\[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=- \int \frac {x^{3} e^{x}}{e^{2 x} - 1}\, dx \] Input:
integrate(exp(x)*x**3/(1-exp(2*x)),x)
Output:
-Integral(x**3*exp(x)/(exp(2*x) - 1), x)
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\frac {1}{2} \, x^{3} \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x^{3} \log \left (-e^{x} + 1\right ) + \frac {3}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) - \frac {3}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) - 3 \, x {\rm Li}_{3}(-e^{x}) + 3 \, x {\rm Li}_{3}(e^{x}) + 3 \, {\rm Li}_{4}(-e^{x}) - 3 \, {\rm Li}_{4}(e^{x}) \] Input:
integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="maxima")
Output:
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)
\[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=\int { -\frac {x^{3} e^{x}}{e^{\left (2 \, x\right )} - 1} \,d x } \] Input:
integrate(exp(x)*x^3/(1-exp(2*x)),x, algorithm="giac")
Output:
integrate(-x^3*e^x/(e^(2*x) - 1), x)
Timed out. \[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=-\int \frac {x^3\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1} \,d x \] Input:
int(-(x^3*exp(x))/(exp(2*x) - 1),x)
Output:
-int((x^3*exp(x))/(exp(2*x) - 1), x)
\[ \int \frac {e^x x^3}{1-e^{2 x}} \, dx=-\left (\int \frac {e^{x} x^{3}}{e^{2 x}-1}d x \right ) \] Input:
int(exp(x)*x^3/(1-exp(2*x)),x)
Output:
- int((e**x*x**3)/(e**(2*x) - 1),x)