3.49.16 \(\int \frac {64 e x-64 e^4 x-16 e^2 x^3+e^{2 x} (e (4 x-2 x^2)+e^4 (-4 x+2 x^2))+e^x (e^4 (32 x-8 x^2)+e (-32 x+8 x^2)+e^2 (4 x^3-2 x^4))}{-64+48 e^x-12 e^{2 x}+e^{3 x}} \, dx\)

Optimal. Leaf size=21 \[ \left (-1+e^3+\frac {e x^2}{4-e^x}\right )^2 \]

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Rubi [A]  time = 2.46, antiderivative size = 39, normalized size of antiderivative = 1.86, number of steps used = 87, number of rules used = 14, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6, 6688, 12, 6742, 2184, 2190, 2279, 2391, 2531, 2282, 6589, 2185, 6609, 2191} \begin {gather*} \frac {e^2 x^4}{\left (4-e^x\right )^2}-\frac {2 e \left (1-e^3\right ) x^2}{4-e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64*E*x - 64*E^4*x - 16*E^2*x^3 + E^(2*x)*(E*(4*x - 2*x^2) + E^4*(-4*x + 2*x^2)) + E^x*(E^4*(32*x - 8*x^2)
+ E*(-32*x + 8*x^2) + E^2*(4*x^3 - 2*x^4)))/(-64 + 48*E^x - 12*E^(2*x) + E^(3*x)),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6688

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

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 {\left (64 e-64 e^4\right ) x-16 e^2 x^3+e^{2 x} \left (e \left (4 x-2 x^2\right )+e^4 \left (-4 x+2 x^2\right )\right )+e^x \left (e^4 \left (32 x-8 x^2\right )+e \left (-32 x+8 x^2\right )+e^2 \left (4 x^3-2 x^4\right )\right )}{-64+48 e^x-12 e^{2 x}+e^{3 x}} \, dx\\ &=\int \frac {2 e x \left (-32 \left (1-e^3\right )-4 e^x \left (1-e^3\right ) (-4+x)+e^{2 x} \left (1-e^3\right ) (-2+x)+8 e x^2+e^{1+x} (-2+x) x^2\right )}{\left (4-e^x\right )^3} \, dx\\ &=(2 e) \int \frac {x \left (-32 \left (1-e^3\right )-4 e^x \left (1-e^3\right ) (-4+x)+e^{2 x} \left (1-e^3\right ) (-2+x)+8 e x^2+e^{1+x} (-2+x) x^2\right )}{\left (4-e^x\right )^3} \, dx\\ &=(2 e) \int \left (\frac {\left (-1+e^3\right ) (-2+x) x}{-4+e^x}-\frac {4 e x^4}{\left (-4+e^x\right )^3}+\frac {x^2 \left (-4 \left (1-e^3\right )+2 e x-e x^2\right )}{\left (4-e^x\right )^2}\right ) \, dx\\ &=(2 e) \int \frac {x^2 \left (-4 \left (1-e^3\right )+2 e x-e x^2\right )}{\left (4-e^x\right )^2} \, dx-\left (8 e^2\right ) \int \frac {x^4}{\left (-4+e^x\right )^3} \, dx-\left (2 e \left (1-e^3\right )\right ) \int \frac {(-2+x) x}{-4+e^x} \, dx\\ &=(2 e) \int \left (\frac {4 (-1+e) \left (1+e+e^2\right ) x^2}{\left (-4+e^x\right )^2}+\frac {2 e x^3}{\left (-4+e^x\right )^2}-\frac {e x^4}{\left (-4+e^x\right )^2}\right ) \, dx-\left (2 e^2\right ) \int \frac {e^x x^4}{\left (-4+e^x\right )^3} \, dx+\left (2 e^2\right ) \int \frac {x^4}{\left (-4+e^x\right )^2} \, dx-\left (2 e \left (1-e^3\right )\right ) \int \left (-\frac {2 x}{-4+e^x}+\frac {x^2}{-4+e^x}\right ) \, dx\\ &=\frac {e^2 x^4}{\left (4-e^x\right )^2}+\frac {1}{2} e^2 \int \frac {e^x x^4}{\left (-4+e^x\right )^2} \, dx-\frac {1}{2} e^2 \int \frac {x^4}{-4+e^x} \, dx-\left (2 e^2\right ) \int \frac {x^4}{\left (-4+e^x\right )^2} \, dx+\left (8 (-1+e) e \left (1+e+e^2\right )\right ) \int \frac {x^2}{\left (-4+e^x\right )^2} \, dx-\left (2 e \left (1-e^3\right )\right ) \int \frac {x^2}{-4+e^x} \, dx+\left (4 e \left (1-e^3\right )\right ) \int \frac {x}{-4+e^x} \, dx\\ &=-\frac {1}{2} e \left (1-e^3\right ) x^2+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}+\frac {e^2 x^4}{2 \left (4-e^x\right )}+\frac {e^2 x^5}{40}-\frac {1}{8} e^2 \int \frac {e^x x^4}{-4+e^x} \, dx-\frac {1}{2} e^2 \int \frac {e^x x^4}{\left (-4+e^x\right )^2} \, dx+\frac {1}{2} e^2 \int \frac {x^4}{-4+e^x} \, dx+\left (2 e^2\right ) \int \frac {x^3}{-4+e^x} \, dx+\left (2 (-1+e) e \left (1+e+e^2\right )\right ) \int \frac {e^x x^2}{\left (-4+e^x\right )^2} \, dx-\left (2 (-1+e) e \left (1+e+e^2\right )\right ) \int \frac {x^2}{-4+e^x} \, dx-\frac {1}{2} \left (e \left (1-e^3\right )\right ) \int \frac {e^x x^2}{-4+e^x} \, dx+\left (e \left (1-e^3\right )\right ) \int \frac {e^x x}{-4+e^x} \, dx\\ &=-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3-\frac {e^2 x^4}{8}+\frac {e^2 x^4}{\left (4-e^x\right )^2}+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{8} e^2 x^4 \log \left (1-\frac {e^x}{4}\right )+\frac {1}{8} e^2 \int \frac {e^x x^4}{-4+e^x} \, dx+\frac {1}{2} e^2 \int \frac {e^x x^3}{-4+e^x} \, dx+\frac {1}{2} e^2 \int x^3 \log \left (1-\frac {e^x}{4}\right ) \, dx-\left (2 e^2\right ) \int \frac {x^3}{-4+e^x} \, dx-\frac {1}{2} \left ((-1+e) e \left (1+e+e^2\right )\right ) \int \frac {e^x x^2}{-4+e^x} \, dx+\left (4 (-1+e) e \left (1+e+e^2\right )\right ) \int \frac {x}{-4+e^x} \, dx-\left (e \left (1-e^3\right )\right ) \int \log \left (1-\frac {e^x}{4}\right ) \, dx+\left (e \left (1-e^3\right )\right ) \int x \log \left (1-\frac {e^x}{4}\right ) \, dx\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} e^2 x^3 \log \left (1-\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-\frac {1}{2} e^2 x^3 \text {Li}_2\left (\frac {e^x}{4}\right )-\frac {1}{2} e^2 \int \frac {e^x x^3}{-4+e^x} \, dx-\frac {1}{2} e^2 \int x^3 \log \left (1-\frac {e^x}{4}\right ) \, dx-\frac {1}{2} \left (3 e^2\right ) \int x^2 \log \left (1-\frac {e^x}{4}\right ) \, dx+\frac {1}{2} \left (3 e^2\right ) \int x^2 \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx+\left ((-1+e) e \left (1+e+e^2\right )\right ) \int \frac {e^x x}{-4+e^x} \, dx+\left ((-1+e) e \left (1+e+e^2\right )\right ) \int x \log \left (1-\frac {e^x}{4}\right ) \, dx+\left (e \left (1-e^3\right )\right ) \int \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx-\left (e \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{4}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}-(1-e) e \left (1+e+e^2\right ) x \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+(1-e) e \left (1+e+e^2\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )+\frac {3}{2} e^2 x^2 \text {Li}_2\left (\frac {e^x}{4}\right )+\frac {3}{2} e^2 x^2 \text {Li}_3\left (\frac {e^x}{4}\right )+\frac {1}{2} \left (3 e^2\right ) \int x^2 \log \left (1-\frac {e^x}{4}\right ) \, dx-\frac {1}{2} \left (3 e^2\right ) \int x^2 \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx-\left (3 e^2\right ) \int x \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx-\left (3 e^2\right ) \int x \text {Li}_3\left (\frac {e^x}{4}\right ) \, dx-\left ((-1+e) e \left (1+e+e^2\right )\right ) \int \log \left (1-\frac {e^x}{4}\right ) \, dx+\left ((-1+e) e \left (1+e+e^2\right )\right ) \int \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx+\left (e \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}-(1-e) e \left (1+e+e^2\right ) x \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+(1-e) e \left (1+e+e^2\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_3\left (\frac {e^x}{4}\right )-3 e^2 x \text {Li}_3\left (\frac {e^x}{4}\right )-3 e^2 x \text {Li}_4\left (\frac {e^x}{4}\right )+\left (3 e^2\right ) \int x \text {Li}_2\left (\frac {e^x}{4}\right ) \, dx+\left (3 e^2\right ) \int \text {Li}_3\left (\frac {e^x}{4}\right ) \, dx+\left (3 e^2\right ) \int x \text {Li}_3\left (\frac {e^x}{4}\right ) \, dx+\left (3 e^2\right ) \int \text {Li}_4\left (\frac {e^x}{4}\right ) \, dx-\left ((-1+e) e \left (1+e+e^2\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{4}\right )}{x} \, dx,x,e^x\right )+\left ((-1+e) e \left (1+e+e^2\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}-(1-e) e \left (1+e+e^2\right ) x \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+(1-e) e \left (1+e+e^2\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_3\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_3\left (\frac {e^x}{4}\right )-\left (3 e^2\right ) \int \text {Li}_3\left (\frac {e^x}{4}\right ) \, dx-\left (3 e^2\right ) \int \text {Li}_4\left (\frac {e^x}{4}\right ) \, dx+\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )+\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}-(1-e) e \left (1+e+e^2\right ) x \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+(1-e) e \left (1+e+e^2\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_3\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_3\left (\frac {e^x}{4}\right )+3 e^2 \text {Li}_4\left (\frac {e^x}{4}\right )+3 e^2 \text {Li}_5\left (\frac {e^x}{4}\right )-\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )-\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4\left (\frac {x}{4}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2-\frac {1}{2} e \left (1-e^3\right ) x^2-\frac {2 (1-e) e \left (1+e+e^2\right ) x^2}{4-e^x}-\frac {1}{6} (1-e) e \left (1+e+e^2\right ) x^3+\frac {1}{6} e \left (1-e^3\right ) x^3+\frac {e^2 x^4}{\left (4-e^x\right )^2}-(1-e) e \left (1+e+e^2\right ) x \log \left (1-\frac {e^x}{4}\right )+e \left (1-e^3\right ) x \log \left (1-\frac {e^x}{4}\right )+\frac {1}{2} (1-e) e \left (1+e+e^2\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-\frac {1}{2} e \left (1-e^3\right ) x^2 \log \left (1-\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_2\left (\frac {e^x}{4}\right )+(1-e) e \left (1+e+e^2\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-e \left (1-e^3\right ) x \text {Li}_2\left (\frac {e^x}{4}\right )-(1-e) e \left (1+e+e^2\right ) \text {Li}_3\left (\frac {e^x}{4}\right )+e \left (1-e^3\right ) \text {Li}_3\left (\frac {e^x}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 30, normalized size = 1.43 \begin {gather*} \frac {e \left (-2 \left (-1+e^3\right ) \left (-4+e^x\right ) x^2+e x^4\right )}{\left (-4+e^x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64*E*x - 64*E^4*x - 16*E^2*x^3 + E^(2*x)*(E*(4*x - 2*x^2) + E^4*(-4*x + 2*x^2)) + E^x*(E^4*(32*x -
8*x^2) + E*(-32*x + 8*x^2) + E^2*(4*x^3 - 2*x^4)))/(-64 + 48*E^x - 12*E^(2*x) + E^(3*x)),x]

[Out]

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

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fricas [B]  time = 0.79, size = 52, normalized size = 2.48 \begin {gather*} \frac {x^{4} e^{2} + 8 \, x^{2} e^{4} - 8 \, x^{2} e - 2 \, {\left (x^{2} e^{4} - x^{2} e\right )} e^{x}}{e^{\left (2 \, x\right )} - 8 \, e^{x} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-4*x)*exp(1)*exp(3)+(-2*x^2+4*x)*exp(1))*exp(x)^2+((-8*x^2+32*x)*exp(1)*exp(3)+(-2*x^4+4*x^3
)*exp(1)^2+(8*x^2-32*x)*exp(1))*exp(x)-64*x*exp(1)*exp(3)-16*x^3*exp(1)^2+64*x*exp(1))/(exp(x)^3-12*exp(x)^2+4
8*exp(x)-64),x, algorithm="fricas")

[Out]

(x^4*e^2 + 8*x^2*e^4 - 8*x^2*e - 2*(x^2*e^4 - x^2*e)*e^x)/(e^(2*x) - 8*e^x + 16)

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giac [B]  time = 0.25, size = 52, normalized size = 2.48 \begin {gather*} \frac {x^{4} e^{2} + 8 \, x^{2} e^{4} - 8 \, x^{2} e - 2 \, x^{2} e^{\left (x + 4\right )} + 2 \, x^{2} e^{\left (x + 1\right )}}{e^{\left (2 \, x\right )} - 8 \, e^{x} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-4*x)*exp(1)*exp(3)+(-2*x^2+4*x)*exp(1))*exp(x)^2+((-8*x^2+32*x)*exp(1)*exp(3)+(-2*x^4+4*x^3
)*exp(1)^2+(8*x^2-32*x)*exp(1))*exp(x)-64*x*exp(1)*exp(3)-16*x^3*exp(1)^2+64*x*exp(1))/(exp(x)^3-12*exp(x)^2+4
8*exp(x)-64),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 37, normalized size = 1.76

method	result	size
risch	\(-\frac {{\mathrm e} x^{2} \left (2 \,{\mathrm e}^{3+x}-8 \,{\mathrm e}^{3}-x^{2} {\mathrm e}-2 \,{\mathrm e}^{x}+8\right )}{\left ({\mathrm e}^{x}-4\right )^{2}}\)	\(37\)
norman	\(\frac {x^{4} {\mathrm e}^{2}+\left (8 \,{\mathrm e} \,{\mathrm e}^{3}-8 \,{\mathrm e}\right ) x^{2}+\left (-2 \,{\mathrm e} \,{\mathrm e}^{3}+2 \,{\mathrm e}\right ) x^{2} {\mathrm e}^{x}}{\left ({\mathrm e}^{x}-4\right )^{2}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-4*x)*exp(1)*exp(3)+(-2*x^2+4*x)*exp(1))*exp(x)^2+((-8*x^2+32*x)*exp(1)*exp(3)+(-2*x^4+4*x^3)*exp(
1)^2+(8*x^2-32*x)*exp(1))*exp(x)-64*x*exp(1)*exp(3)-16*x^3*exp(1)^2+64*x*exp(1))/(exp(x)^3-12*exp(x)^2+48*exp(
x)-64),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.40, size = 46, normalized size = 2.19 \begin {gather*} \frac {x^{4} e^{2} - 2 \, x^{2} {\left (e^{4} - e\right )} e^{x} + 8 \, x^{2} {\left (e^{4} - e\right )}}{e^{\left (2 \, x\right )} - 8 \, e^{x} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-4*x)*exp(1)*exp(3)+(-2*x^2+4*x)*exp(1))*exp(x)^2+((-8*x^2+32*x)*exp(1)*exp(3)+(-2*x^4+4*x^3
)*exp(1)^2+(8*x^2-32*x)*exp(1))*exp(x)-64*x*exp(1)*exp(3)-16*x^3*exp(1)^2+64*x*exp(1))/(exp(x)^3-12*exp(x)^2+4
8*exp(x)-64),x, algorithm="maxima")

[Out]

(x^4*e^2 - 2*x^2*(e^4 - e)*e^x + 8*x^2*(e^4 - e))/(e^(2*x) - 8*e^x + 16)

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mupad [B]  time = 3.37, size = 44, normalized size = 2.10 \begin {gather*} \frac {x^4\,{\mathrm {e}}^2-2\,x^2\,{\mathrm {e}}^{x+1}\,\left ({\mathrm {e}}^3-1\right )+8\,x^2\,\mathrm {e}\,\left ({\mathrm {e}}^3-1\right )}{{\mathrm {e}}^{2\,x}-8\,{\mathrm {e}}^x+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*x*exp(1) - 64*x*exp(4) + exp(2*x)*(exp(1)*(4*x - 2*x^2) - exp(4)*(4*x - 2*x^2)) + exp(x)*(exp(4)*(32*
x - 8*x^2) - exp(1)*(32*x - 8*x^2) + exp(2)*(4*x^3 - 2*x^4)) - 16*x^3*exp(2))/(12*exp(2*x) - exp(3*x) - 48*exp
(x) + 64),x)

[Out]

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

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sympy [B]  time = 0.16, size = 54, normalized size = 2.57 \begin {gather*} \frac {x^{4} e^{2} - 8 e x^{2} + 8 x^{2} e^{4} + \left (- 2 x^{2} e^{4} + 2 e x^{2}\right ) e^{x}}{e^{2 x} - 8 e^{x} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-4*x)*exp(1)*exp(3)+(-2*x**2+4*x)*exp(1))*exp(x)**2+((-8*x**2+32*x)*exp(1)*exp(3)+(-2*x**4+
4*x**3)*exp(1)**2+(8*x**2-32*x)*exp(1))*exp(x)-64*x*exp(1)*exp(3)-16*x**3*exp(1)**2+64*x*exp(1))/(exp(x)**3-12
*exp(x)**2+48*exp(x)-64),x)

[Out]

(x**4*exp(2) - 8*E*x**2 + 8*x**2*exp(4) + (-2*x**2*exp(4) + 2*E*x**2)*exp(x))/(exp(2*x) - 8*exp(x) + 16)

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