3.34.28 \(\int \frac {-20-16 x-15 x^2-4 x^3+e^{3 x} (20+24 x+36 x^2+16 x^3)+e^{2 x} (-60-64 x-88 x^2-40 x^3-4 x^4)+e^x (60+56 x+67 x^2+26 x^3+2 x^4)}{-16+48 e^x-48 e^{2 x}+16 e^{3 x}} \, dx\)

Optimal. Leaf size=33 \[ x+x (1+x) \left (-x+\frac {1}{4} \left (-2+2 x+\frac {x}{1-e^x}\right )\right )^2 \]

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Rubi [B]  time = 2.80, antiderivative size = 107, normalized size of antiderivative = 3.24, number of steps used = 128, number of rules used = 13, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6741, 12, 6742, 2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2279, 2391} \begin {gather*} -\frac {x^4}{4 \left (1-e^x\right )}+\frac {x^4}{16 \left (1-e^x\right )^2}+\frac {x^4}{4}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^3}{16 \left (1-e^x\right )^2}+\frac {3 x^3}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^2}{4}+\frac {5 x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 16*x - 15*x^2 - 4*x^3 + E^(3*x)*(20 + 24*x + 36*x^2 + 16*x^3) + E^(2*x)*(-60 - 64*x - 88*x^2 - 40*x
^3 - 4*x^4) + E^x*(60 + 56*x + 67*x^2 + 26*x^3 + 2*x^4))/(-16 + 48*E^x - 48*E^(2*x) + 16*E^(3*x)),x]

[Out]

(5*x)/4 + (3*x^2)/4 - x^2/(4*(1 - E^x)) + (3*x^3)/4 + x^3/(16*(1 - E^x)^2) - x^3/(2*(1 - E^x)) + x^4/4 + x^4/(
16*(1 - E^x)^2) - x^4/(4*(1 - E^x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2184

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

Rule 2185

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

Rule 2190

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

Rule 2191

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20+16 x+15 x^2+4 x^3-e^{3 x} \left (20+24 x+36 x^2+16 x^3\right )-e^{2 x} \left (-60-64 x-88 x^2-40 x^3-4 x^4\right )-e^x \left (60+56 x+67 x^2+26 x^3+2 x^4\right )}{16 \left (1-e^x\right )^3} \, dx\\ &=\frac {1}{16} \int \frac {20+16 x+15 x^2+4 x^3-e^{3 x} \left (20+24 x+36 x^2+16 x^3\right )-e^{2 x} \left (-60-64 x-88 x^2-40 x^3-4 x^4\right )-e^x \left (60+56 x+67 x^2+26 x^3+2 x^4\right )}{\left (1-e^x\right )^3} \, dx\\ &=\frac {1}{16} \int \left (-\frac {2 x^3 (1+x)}{\left (-1+e^x\right )^3}-\frac {x^2 \left (1+6 x+6 x^2\right )}{\left (-1+e^x\right )^2}-\frac {4 x \left (-2-5 x-2 x^2+x^3\right )}{-1+e^x}+4 \left (5+6 x+9 x^2+4 x^3\right )\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {x^2 \left (1+6 x+6 x^2\right )}{\left (-1+e^x\right )^2} \, dx\right )-\frac {1}{8} \int \frac {x^3 (1+x)}{\left (-1+e^x\right )^3} \, dx-\frac {1}{4} \int \frac {x \left (-2-5 x-2 x^2+x^3\right )}{-1+e^x} \, dx+\frac {1}{4} \int \left (5+6 x+9 x^2+4 x^3\right ) \, dx\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}+\frac {3 x^3}{4}+\frac {x^4}{4}-\frac {1}{16} \int \left (\frac {x^2}{\left (-1+e^x\right )^2}+\frac {6 x^3}{\left (-1+e^x\right )^2}+\frac {6 x^4}{\left (-1+e^x\right )^2}\right ) \, dx-\frac {1}{8} \int \left (\frac {x^3}{\left (-1+e^x\right )^3}+\frac {x^4}{\left (-1+e^x\right )^3}\right ) \, dx-\frac {1}{4} \int \left (-\frac {2 x}{-1+e^x}-\frac {5 x^2}{-1+e^x}-\frac {2 x^3}{-1+e^x}+\frac {x^4}{-1+e^x}\right ) \, dx\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}+\frac {3 x^3}{4}+\frac {x^4}{4}-\frac {1}{16} \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx-\frac {1}{8} \int \frac {x^3}{\left (-1+e^x\right )^3} \, dx-\frac {1}{8} \int \frac {x^4}{\left (-1+e^x\right )^3} \, dx-\frac {1}{4} \int \frac {x^4}{-1+e^x} \, dx-\frac {3}{8} \int \frac {x^3}{\left (-1+e^x\right )^2} \, dx-\frac {3}{8} \int \frac {x^4}{\left (-1+e^x\right )^2} \, dx+\frac {1}{2} \int \frac {x}{-1+e^x} \, dx+\frac {1}{2} \int \frac {x^3}{-1+e^x} \, dx+\frac {5}{4} \int \frac {x^2}{-1+e^x} \, dx\\ &=\frac {5 x}{4}+\frac {x^2}{2}+\frac {x^3}{3}+\frac {x^4}{8}+\frac {x^5}{20}-\frac {1}{16} \int \frac {e^x x^2}{\left (-1+e^x\right )^2} \, dx+\frac {1}{16} \int \frac {x^2}{-1+e^x} \, dx-\frac {1}{8} \int \frac {e^x x^3}{\left (-1+e^x\right )^3} \, dx+\frac {1}{8} \int \frac {x^3}{\left (-1+e^x\right )^2} \, dx-\frac {1}{8} \int \frac {e^x x^4}{\left (-1+e^x\right )^3} \, dx+\frac {1}{8} \int \frac {x^4}{\left (-1+e^x\right )^2} \, dx-\frac {1}{4} \int \frac {e^x x^4}{-1+e^x} \, dx-\frac {3}{8} \int \frac {e^x x^3}{\left (-1+e^x\right )^2} \, dx+\frac {3}{8} \int \frac {x^3}{-1+e^x} \, dx-\frac {3}{8} \int \frac {e^x x^4}{\left (-1+e^x\right )^2} \, dx+\frac {3}{8} \int \frac {x^4}{-1+e^x} \, dx+\frac {1}{2} \int \frac {e^x x}{-1+e^x} \, dx+\frac {1}{2} \int \frac {e^x x^3}{-1+e^x} \, dx+\frac {5}{4} \int \frac {e^x x^2}{-1+e^x} \, dx\\ &=\frac {5 x}{4}+\frac {x^2}{2}-\frac {x^2}{16 \left (1-e^x\right )}+\frac {5 x^3}{16}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {3 x^3}{8 \left (1-e^x\right )}+\frac {x^4}{32}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {3 x^4}{8 \left (1-e^x\right )}-\frac {x^5}{40}+\frac {1}{2} x \log \left (1-e^x\right )+\frac {5}{4} x^2 \log \left (1-e^x\right )+\frac {1}{2} x^3 \log \left (1-e^x\right )-\frac {1}{4} x^4 \log \left (1-e^x\right )+\frac {1}{16} \int \frac {e^x x^2}{-1+e^x} \, dx-\frac {1}{8} \int \frac {x}{-1+e^x} \, dx+\frac {1}{8} \int \frac {e^x x^3}{\left (-1+e^x\right )^2} \, dx-\frac {1}{8} \int \frac {x^3}{-1+e^x} \, dx+\frac {1}{8} \int \frac {e^x x^4}{\left (-1+e^x\right )^2} \, dx-\frac {1}{8} \int \frac {x^4}{-1+e^x} \, dx-\frac {3}{16} \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx-\frac {1}{4} \int \frac {x^3}{\left (-1+e^x\right )^2} \, dx+\frac {3}{8} \int \frac {e^x x^3}{-1+e^x} \, dx+\frac {3}{8} \int \frac {e^x x^4}{-1+e^x} \, dx-\frac {1}{2} \int \log \left (1-e^x\right ) \, dx-\frac {9}{8} \int \frac {x^2}{-1+e^x} \, dx-\frac {3}{2} \int \frac {x^3}{-1+e^x} \, dx-\frac {3}{2} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {5}{2} \int x \log \left (1-e^x\right ) \, dx+\int x^3 \log \left (1-e^x\right ) \, dx\\ &=\frac {5 x}{4}+\frac {9 x^2}{16}-\frac {x^2}{16 \left (1-e^x\right )}+\frac {11 x^3}{16}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{4 \left (1-e^x\right )}+\frac {7 x^4}{16}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}+\frac {1}{2} x \log \left (1-e^x\right )+\frac {21}{16} x^2 \log \left (1-e^x\right )+\frac {7}{8} x^3 \log \left (1-e^x\right )+\frac {1}{8} x^4 \log \left (1-e^x\right )+\frac {5 x \text {Li}_2\left (e^x\right )}{2}+\frac {3}{2} x^2 \text {Li}_2\left (e^x\right )-x^3 \text {Li}_2\left (e^x\right )-\frac {1}{8} \int \frac {e^x x}{-1+e^x} \, dx-\frac {1}{8} \int \frac {e^x x^3}{-1+e^x} \, dx-\frac {1}{8} \int \frac {e^x x^4}{-1+e^x} \, dx-\frac {1}{8} \int x \log \left (1-e^x\right ) \, dx-\frac {3}{16} \int \frac {e^x x^2}{\left (-1+e^x\right )^2} \, dx+\frac {3}{16} \int \frac {x^2}{-1+e^x} \, dx-\frac {1}{4} \int \frac {e^x x^3}{\left (-1+e^x\right )^2} \, dx+\frac {1}{4} \int \frac {x^3}{-1+e^x} \, dx+\frac {3}{8} \int \frac {x^2}{-1+e^x} \, dx+\frac {1}{2} \int \frac {x^3}{-1+e^x} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^x\right )-\frac {9}{8} \int \frac {e^x x^2}{-1+e^x} \, dx-\frac {9}{8} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {3}{2} \int \frac {e^x x^3}{-1+e^x} \, dx-\frac {3}{2} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {5}{2} \int \text {Li}_2\left (e^x\right ) \, dx-3 \int x \text {Li}_2\left (e^x\right ) \, dx+3 \int x^2 \text {Li}_2\left (e^x\right ) \, dx\\ &=\frac {5 x}{4}+\frac {9 x^2}{16}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {x^3}{2}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}+\frac {3}{8} x \log \left (1-e^x\right )+\frac {3}{16} x^2 \log \left (1-e^x\right )-\frac {3}{4} x^3 \log \left (1-e^x\right )+\frac {\text {Li}_2\left (e^x\right )}{2}+\frac {21 x \text {Li}_2\left (e^x\right )}{8}+\frac {21}{8} x^2 \text {Li}_2\left (e^x\right )+\frac {1}{2} x^3 \text {Li}_2\left (e^x\right )-3 x \text {Li}_3\left (e^x\right )+3 x^2 \text {Li}_3\left (e^x\right )+\frac {1}{8} \int \log \left (1-e^x\right ) \, dx-\frac {1}{8} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {3}{16} \int \frac {e^x x^2}{-1+e^x} \, dx+\frac {1}{4} \int \frac {e^x x^3}{-1+e^x} \, dx-\frac {3}{8} \int \frac {x}{-1+e^x} \, dx+\frac {3}{8} \int \frac {e^x x^2}{-1+e^x} \, dx+\frac {3}{8} \int x^2 \log \left (1-e^x\right ) \, dx+\frac {1}{2} \int \frac {e^x x^3}{-1+e^x} \, dx+\frac {1}{2} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {3}{4} \int \frac {x^2}{-1+e^x} \, dx+\frac {9}{4} \int x \log \left (1-e^x\right ) \, dx-\frac {9}{4} \int x \text {Li}_2\left (e^x\right ) \, dx-\frac {5}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )+3 \int \text {Li}_3\left (e^x\right ) \, dx+\frac {9}{2} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {9}{2} \int x^2 \text {Li}_2\left (e^x\right ) \, dx-6 \int x \text {Li}_3\left (e^x\right ) \, dx\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^3}{4}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}+\frac {3}{8} x \log \left (1-e^x\right )+\frac {3}{4} x^2 \log \left (1-e^x\right )+\frac {\text {Li}_2\left (e^x\right )}{2}+\frac {3 x \text {Li}_2\left (e^x\right )}{8}-\frac {9}{4} x^2 \text {Li}_2\left (e^x\right )-\frac {5 \text {Li}_3\left (e^x\right )}{2}-\frac {21 x \text {Li}_3\left (e^x\right )}{4}-\frac {3}{2} x^2 \text {Li}_3\left (e^x\right )-6 x \text {Li}_4\left (e^x\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )-\frac {3}{8} \int \frac {e^x x}{-1+e^x} \, dx-\frac {3}{8} \int x \log \left (1-e^x\right ) \, dx-\frac {3}{4} \int \frac {e^x x^2}{-1+e^x} \, dx-\frac {3}{4} \int x \log \left (1-e^x\right ) \, dx-\frac {3}{4} \int x^2 \log \left (1-e^x\right ) \, dx+\frac {3}{4} \int x \text {Li}_2\left (e^x\right ) \, dx-\frac {3}{2} \int x^2 \log \left (1-e^x\right ) \, dx+\frac {3}{2} \int x^2 \text {Li}_2\left (e^x\right ) \, dx+\frac {9}{4} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {9}{4} \int \text {Li}_3\left (e^x\right ) \, dx+3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )+6 \int \text {Li}_4\left (e^x\right ) \, dx+9 \int x \text {Li}_2\left (e^x\right ) \, dx+9 \int x \text {Li}_3\left (e^x\right ) \, dx\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^3}{4}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}+\frac {3 \text {Li}_2\left (e^x\right )}{8}+\frac {3 x \text {Li}_2\left (e^x\right )}{2}-\frac {21 \text {Li}_3\left (e^x\right )}{8}+\frac {9 x \text {Li}_3\left (e^x\right )}{2}+3 \text {Li}_4\left (e^x\right )+3 x \text {Li}_4\left (e^x\right )+\frac {3}{8} \int \log \left (1-e^x\right ) \, dx-\frac {3}{8} \int \text {Li}_2\left (e^x\right ) \, dx-\frac {3}{4} \int \text {Li}_2\left (e^x\right ) \, dx-\frac {3}{4} \int \text {Li}_3\left (e^x\right ) \, dx+\frac {3}{2} \int x \log \left (1-e^x\right ) \, dx-\frac {3}{2} \int x \text {Li}_2\left (e^x\right ) \, dx+\frac {9}{4} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )+\frac {9}{4} \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )-3 \int x \text {Li}_2\left (e^x\right ) \, dx-3 \int x \text {Li}_3\left (e^x\right ) \, dx+6 \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^x\right )-9 \int \text {Li}_3\left (e^x\right ) \, dx-9 \int \text {Li}_4\left (e^x\right ) \, dx\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^3}{4}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}+\frac {3 \text {Li}_2\left (e^x\right )}{8}-\frac {3 \text {Li}_3\left (e^x\right )}{8}+\frac {21 \text {Li}_4\left (e^x\right )}{4}+6 \text {Li}_5\left (e^x\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^x\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )-\frac {3}{4} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )-\frac {3}{4} \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )+\frac {3}{2} \int \text {Li}_2\left (e^x\right ) \, dx+\frac {3}{2} \int \text {Li}_3\left (e^x\right ) \, dx+3 \int \text {Li}_3\left (e^x\right ) \, dx+3 \int \text {Li}_4\left (e^x\right ) \, dx-9 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^x\right )-9 \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^x\right )\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^3}{4}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}-\frac {3 \text {Li}_3\left (e^x\right )}{2}-\frac {9 \text {Li}_4\left (e^x\right )}{2}-3 \text {Li}_5\left (e^x\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^x\right )+\frac {3}{2} \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 )+3 \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^x\right )\\ &=\frac {5 x}{4}+\frac {3 x^2}{4}-\frac {x^2}{4 \left (1-e^x\right )}+\frac {3 x^3}{4}+\frac {x^3}{16 \left (1-e^x\right )^2}-\frac {x^3}{2 \left (1-e^x\right )}+\frac {x^4}{4}+\frac {x^4}{16 \left (1-e^x\right )^2}-\frac {x^4}{4 \left (1-e^x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 64, normalized size = 1.94 \begin {gather*} \frac {1}{16} \left (20 x+12 x^2+12 x^3+4 x^4-\frac {-x^3-x^4}{\left (-1+e^x\right )^2}+\frac {4 \left (x^2+2 x^3+x^4\right )}{-1+e^x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 16*x - 15*x^2 - 4*x^3 + E^(3*x)*(20 + 24*x + 36*x^2 + 16*x^3) + E^(2*x)*(-60 - 64*x - 88*x^2
- 40*x^3 - 4*x^4) + E^x*(60 + 56*x + 67*x^2 + 26*x^3 + 2*x^4))/(-16 + 48*E^x - 48*E^(2*x) + 16*E^(3*x)),x]

[Out]

(20*x + 12*x^2 + 12*x^3 + 4*x^4 - (-x^3 - x^4)/(-1 + E^x)^2 + (4*(x^2 + 2*x^3 + x^4))/(-1 + E^x))/16

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fricas [B]  time = 0.70, size = 75, normalized size = 2.27 \begin {gather*} \frac {x^{4} + 5 \, x^{3} + 8 \, x^{2} + 4 \, {\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{4} + 4 \, x^{3} + 5 \, x^{2} + 10 \, x\right )} e^{x} + 20 \, x}{16 \, {\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^3+36*x^2+24*x+20)*exp(x)^3+(-4*x^4-40*x^3-88*x^2-64*x-60)*exp(x)^2+(2*x^4+26*x^3+67*x^2+56*x+
60)*exp(x)-4*x^3-15*x^2-16*x-20)/(16*exp(x)^3-48*exp(x)^2+48*exp(x)-16),x, algorithm="fricas")

[Out]

1/16*(x^4 + 5*x^3 + 8*x^2 + 4*(x^4 + 3*x^3 + 3*x^2 + 5*x)*e^(2*x) - 4*(x^4 + 4*x^3 + 5*x^2 + 10*x)*e^x + 20*x)
/(e^(2*x) - 2*e^x + 1)

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giac [B]  time = 0.24, size = 91, normalized size = 2.76 \begin {gather*} \frac {4 \, x^{4} e^{\left (2 \, x\right )} - 4 \, x^{4} e^{x} + x^{4} + 12 \, x^{3} e^{\left (2 \, x\right )} - 16 \, x^{3} e^{x} + 5 \, x^{3} + 12 \, x^{2} e^{\left (2 \, x\right )} - 20 \, x^{2} e^{x} + 8 \, x^{2} + 20 \, x e^{\left (2 \, x\right )} - 40 \, x e^{x} + 20 \, x}{16 \, {\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^3+36*x^2+24*x+20)*exp(x)^3+(-4*x^4-40*x^3-88*x^2-64*x-60)*exp(x)^2+(2*x^4+26*x^3+67*x^2+56*x+
60)*exp(x)-4*x^3-15*x^2-16*x-20)/(16*exp(x)^3-48*exp(x)^2+48*exp(x)-16),x, algorithm="giac")

[Out]

1/16*(4*x^4*e^(2*x) - 4*x^4*e^x + x^4 + 12*x^3*e^(2*x) - 16*x^3*e^x + 5*x^3 + 12*x^2*e^(2*x) - 20*x^2*e^x + 8*
x^2 + 20*x*e^(2*x) - 40*x*e^x + 20*x)/(e^(2*x) - 2*e^x + 1)

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maple [B]  time = 0.22, size = 57, normalized size = 1.73

method	result	size
risch	\(\frac {x^{4}}{4}+\frac {3 x^{3}}{4}+\frac {3 x^{2}}{4}+\frac {5 x}{4}+\frac {x^{2} \left (4 \,{\mathrm e}^{x} x^{2}-3 x^{2}+8 \,{\mathrm e}^{x} x -7 x +4 \,{\mathrm e}^{x}-4\right )}{16 \left ({\mathrm e}^{x}-1\right )^{2}}\)	\(57\)
norman	\(\frac {\frac {5 x}{4}+\frac {x^{2}}{2}+\frac {5 x^{3}}{16}+\frac {x^{4}}{16}+\frac {5 x \,{\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{x} x}{2}-\frac {5 \,{\mathrm e}^{x} x^{2}}{4}-{\mathrm e}^{x} x^{3}-\frac {{\mathrm e}^{x} x^{4}}{4}+\frac {3 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {3 \,{\mathrm e}^{2 x} x^{3}}{4}+\frac {{\mathrm e}^{2 x} x^{4}}{4}}{\left ({\mathrm e}^{x}-1\right )^{2}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^3+36*x^2+24*x+20)*exp(x)^3+(-4*x^4-40*x^3-88*x^2-64*x-60)*exp(x)^2+(2*x^4+26*x^3+67*x^2+56*x+60)*ex
p(x)-4*x^3-15*x^2-16*x-20)/(16*exp(x)^3-48*exp(x)^2+48*exp(x)-16),x,method=_RETURNVERBOSE)

[Out]

1/4*x^4+3/4*x^3+3/4*x^2+5/4*x+1/16*x^2*(4*exp(x)*x^2-3*x^2+8*exp(x)*x-7*x+4*exp(x)-4)/(exp(x)-1)^2

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maxima [B]  time = 0.67, size = 92, normalized size = 2.79 \begin {gather*} \frac {5}{4} \, x + \frac {x^{4} + 5 \, x^{3} + 8 \, x^{2} + 4 \, {\left (x^{4} + 3 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{4} + 4 \, x^{3} + 5 \, x^{2} - 5\right )} e^{x} - 30}{16 \, {\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}} - \frac {5 \, {\left (2 \, e^{x} - 3\right )}}{8 \, {\left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^3+36*x^2+24*x+20)*exp(x)^3+(-4*x^4-40*x^3-88*x^2-64*x-60)*exp(x)^2+(2*x^4+26*x^3+67*x^2+56*x+
60)*exp(x)-4*x^3-15*x^2-16*x-20)/(16*exp(x)^3-48*exp(x)^2+48*exp(x)-16),x, algorithm="maxima")

[Out]

5/4*x + 1/16*(x^4 + 5*x^3 + 8*x^2 + 4*(x^4 + 3*x^3 + 3*x^2)*e^(2*x) - 4*(x^4 + 4*x^3 + 5*x^2 - 5)*e^x - 30)/(e
^(2*x) - 2*e^x + 1) - 5/8*(2*e^x - 3)/(e^(2*x) - 2*e^x + 1)

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mupad [B]  time = 0.17, size = 92, normalized size = 2.79 \begin {gather*} \frac {20\,x+20\,x\,{\mathrm {e}}^{2\,x}-20\,x^2\,{\mathrm {e}}^x-16\,x^3\,{\mathrm {e}}^x-4\,x^4\,{\mathrm {e}}^x+12\,x^2\,{\mathrm {e}}^{2\,x}+12\,x^3\,{\mathrm {e}}^{2\,x}+4\,x^4\,{\mathrm {e}}^{2\,x}-40\,x\,{\mathrm {e}}^x+8\,x^2+5\,x^3+x^4}{16\,{\mathrm {e}}^{2\,x}-32\,{\mathrm {e}}^x+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - exp(x)*(56*x + 67*x^2 + 26*x^3 + 2*x^4 + 60) - exp(3*x)*(24*x + 36*x^2 + 16*x^3 + 20) + exp(2*x)*(
64*x + 88*x^2 + 40*x^3 + 4*x^4 + 60) + 15*x^2 + 4*x^3 + 20)/(48*exp(2*x) - 16*exp(3*x) - 48*exp(x) + 16),x)

[Out]

(20*x + 20*x*exp(2*x) - 20*x^2*exp(x) - 16*x^3*exp(x) - 4*x^4*exp(x) + 12*x^2*exp(2*x) + 12*x^3*exp(2*x) + 4*x
^4*exp(2*x) - 40*x*exp(x) + 8*x^2 + 5*x^3 + x^4)/(16*exp(2*x) - 32*exp(x) + 16)

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sympy [B]  time = 0.18, size = 70, normalized size = 2.12 \begin {gather*} \frac {x^{4}}{4} + \frac {3 x^{3}}{4} + \frac {3 x^{2}}{4} + \frac {5 x}{4} + \frac {- 3 x^{4} - 7 x^{3} - 4 x^{2} + \left (4 x^{4} + 8 x^{3} + 4 x^{2}\right ) e^{x}}{16 e^{2 x} - 32 e^{x} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**3+36*x**2+24*x+20)*exp(x)**3+(-4*x**4-40*x**3-88*x**2-64*x-60)*exp(x)**2+(2*x**4+26*x**3+67*
x**2+56*x+60)*exp(x)-4*x**3-15*x**2-16*x-20)/(16*exp(x)**3-48*exp(x)**2+48*exp(x)-16),x)

[Out]

x**4/4 + 3*x**3/4 + 3*x**2/4 + 5*x/4 + (-3*x**4 - 7*x**3 - 4*x**2 + (4*x**4 + 8*x**3 + 4*x**2)*exp(x))/(16*exp
(2*x) - 32*exp(x) + 16)

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