3.20.7 \(\int \frac {2 e^{5 x}-2 x+e^{4 x} (-6-8 \log (3))-8 \log (3)+e^{3-x} (-2+2 x+8 \log (3)) \log (5)+2 e^{6-2 x} \log ^2(5)+e^x (2+4 x-2 x^2+(24-8 x) \log (3)+e^{3-x} (4-6 x-24 \log (3)) \log (5)-6 e^{6-2 x} \log ^2(5))+e^{3 x} (8+24 \log (3)-8 e^{3-x} \log (3) \log (5)-2 e^{6-2 x} \log ^2(5))+e^{2 x} (-6-2 x+(-32+8 x) \log (3)+e^{3-x} (-2+4 x+24 \log (3)) \log (5)+6 e^{6-2 x} \log ^2(5))}{-1+3 e^x-3 e^{2 x}+e^{3 x}} \, dx\)

Optimal. Leaf size=33 \[ \left (-e^x+\frac {x}{1-e^x}+4 \log (3)+e^{3-x} \log (5)\right )^2 \]

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Rubi [C]  time = 8.36, antiderivative size = 582, normalized size of antiderivative = 17.64, number of steps used = 135, number of rules used = 20, integrand size = 224, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6688, 12, 6742, 2194, 2176, 2254, 2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391, 36, 31, 29, 2248, 44} \begin {gather*} 2 \text {Li}_2\left (e^x\right )-2 \left (1-e^3 \log (125)-\log (531441)\right ) \text {Li}_2\left (e^x\right )-6 \left (e^3 \log (5)+\log (81)\right ) \text {Li}_2\left (e^x\right )+\frac {x^2}{\left (1-e^x\right )^2}-x^2+x^2 \left (1-e^3 \log (125)-\log (531441)\right )+3 x^2 \left (e^3 \log (5)+\log (81)\right )-\frac {2 x}{1-e^x}+4 x+e^{2 x}-e^{-2 x} \left (1-e^6 \log ^2(5)-e^3 \log (5)-\log (81)\right )-2 x \left (1-e^3 \log (125)-\log (531441)\right ) \log \left (1-e^x\right )-6 x \left (e^3 \log (5)+\log (81)\right ) \log \left (1-e^x\right )+2 x \log \left (1-e^x\right )-e^{-2 x} x \left (1-e^3 \log (125)-\log (531441)\right )-2 e^{-x} x \left (1-e^3 \log (125)-\log (531441)\right )+e^{-2 x} x \left (1-e^3 \log (25)-\log (6561)\right )-e^{-2 x} x \left (e^3 \log (5)+\log (81)\right )-4 e^{-x} x \left (e^3 \log (5)+\log (81)\right )+\frac {2 x \left (e^3 \log (5)+\log (81)\right )}{1-e^x}-2 x \left (e^3 \log (5)+\log (81)\right )-2 x \left (1-e^3 \log (5)-\log (81)\right )-2 e^{-x} \left (-(x (1-\log (81)))+1-e^3 \log (5) \log (81)-\log (81)\right )+2 \left (e^3 \log (5)+\log (81)\right ) \log \left (1-e^x\right )+2 \left (1-e^3 \log (5)-\log (81)\right ) \log \left (1-e^x\right )-2 \log \left (1-e^x\right )-\frac {1}{2} e^{-2 x} \left (1-e^3 \log (125)-\log (531441)\right )-2 e^{-x} \left (1-e^3 \log (125)-\log (531441)\right )+\frac {1}{2} e^{-2 x} \left (1-e^3 \log (25)-\log (6561)\right )-\frac {1}{2} e^{-2 x} \left (e^3 \log (5)+\log (81)\right )-4 e^{-x} \left (e^3 \log (5)+\log (81)\right )-2 e^x \log (81)+e^{-2 x} \left (1-e^3 \log (5)-\log (81)\right )+2 e^{-x} \left (1-e^3 \log (5)-\log (81)\right )+2 e^{-x} (1-\log (81)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(5*x) - 2*x + E^(4*x)*(-6 - 8*Log[3]) - 8*Log[3] + E^(3 - x)*(-2 + 2*x + 8*Log[3])*Log[5] + 2*E^(6 -
2*x)*Log[5]^2 + E^x*(2 + 4*x - 2*x^2 + (24 - 8*x)*Log[3] + E^(3 - x)*(4 - 6*x - 24*Log[3])*Log[5] - 6*E^(6 - 2
*x)*Log[5]^2) + E^(3*x)*(8 + 24*Log[3] - 8*E^(3 - x)*Log[3]*Log[5] - 2*E^(6 - 2*x)*Log[5]^2) + E^(2*x)*(-6 - 2
*x + (-32 + 8*x)*Log[3] + E^(3 - x)*(-2 + 4*x + 24*Log[3])*Log[5] + 6*E^(6 - 2*x)*Log[5]^2))/(-1 + 3*E^x - 3*E
^(2*x) + E^(3*x)),x]

[Out]

E^(2*x) + 4*x - (2*x)/(1 - E^x) - x^2 + x^2/(1 - E^x)^2 + (2*(1 - Log[81]))/E^x + (1 - E^3*Log[5] - Log[81])/E
^(2*x) + (2*(1 - E^3*Log[5] - Log[81]))/E^x - 2*x*(1 - E^3*Log[5] - Log[81]) - (1 - E^3*Log[5] - E^6*Log[5]^2
- Log[81])/E^(2*x) - 2*E^x*Log[81] - (E^3*Log[5] + Log[81])/(2*E^(2*x)) - (4*(E^3*Log[5] + Log[81]))/E^x - 2*x
*(E^3*Log[5] + Log[81]) - (x*(E^3*Log[5] + Log[81]))/E^(2*x) - (4*x*(E^3*Log[5] + Log[81]))/E^x + (2*x*(E^3*Lo
g[5] + Log[81]))/(1 - E^x) + 3*x^2*(E^3*Log[5] + Log[81]) - (2*(1 - x*(1 - Log[81]) - Log[81] - E^3*Log[5]*Log
[81]))/E^x + (1 - E^3*Log[25] - Log[6561])/(2*E^(2*x)) + (x*(1 - E^3*Log[25] - Log[6561]))/E^(2*x) - (1 - E^3*
Log[125] - Log[531441])/(2*E^(2*x)) - (2*(1 - E^3*Log[125] - Log[531441]))/E^x - (x*(1 - E^3*Log[125] - Log[53
1441]))/E^(2*x) - (2*x*(1 - E^3*Log[125] - Log[531441]))/E^x + x^2*(1 - E^3*Log[125] - Log[531441]) - 2*Log[1
- E^x] + 2*x*Log[1 - E^x] + 2*(1 - E^3*Log[5] - Log[81])*Log[1 - E^x] + 2*(E^3*Log[5] + Log[81])*Log[1 - E^x]
- 6*x*(E^3*Log[5] + Log[81])*Log[1 - E^x] - 2*x*(1 - E^3*Log[125] - Log[531441])*Log[1 - E^x] + 2*PolyLog[2, E
^x] - 6*(E^3*Log[5] + Log[81])*PolyLog[2, E^x] - 2*(1 - E^3*Log[125] - Log[531441])*PolyLog[2, 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2248

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

Rule 2254

Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> W
ith[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a,
b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simplify[u/v]]

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 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 {2 e^{-2 x} \left (2 e^{3 x}-e^{4 x}+e^{2 x} (-2+x)-e^3 \log (5)-e^{3+2 x} \log (5)+e^x \left (1+e^3 \log (25)\right )\right ) \left (e^{3 x}+e^3 \log (5)-e^{3+x} \log (5)-e^{2 x} (1+\log (81))+e^x (x+\log (81))\right )}{\left (1-e^x\right )^3} \, dx\\ &=2 \int \frac {e^{-2 x} \left (2 e^{3 x}-e^{4 x}+e^{2 x} (-2+x)-e^3 \log (5)-e^{3+2 x} \log (5)+e^x \left (1+e^3 \log (25)\right )\right ) \left (e^{3 x}+e^3 \log (5)-e^{3+x} \log (5)-e^{2 x} (1+\log (81))+e^x (x+\log (81))\right )}{\left (1-e^x\right )^3} \, dx\\ &=2 \int \left (1+e^{2 x}-e^{-2 x} x^2-\frac {e^{-2 x} x^2}{\left (-1+e^x\right )^3}-e^{-2 x} x \left (1-e^3 \log (25)-2 \log (81)\right )-\frac {e^{-2 x} x \left (3 x-e^3 \log (5)-\log (81)\right )}{\left (-1+e^x\right )^2}-e^x \log (81)+e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right )+e^{-2 x} \left (1+\log (5) \left (-e^3-e^6 \log (5)-\frac {3 \log (81)}{\log (125)}\right )\right )+\frac {e^{-2 x} \left (-1+3 x^2+e^3 \log (5)+\log (81)+x \left (1-e^3 \log (125)-\log (531441)\right )\right )}{1-e^x}\right ) \, dx\\ &=2 x+2 \int e^{2 x} \, dx-2 \int e^{-2 x} x^2 \, dx-2 \int \frac {e^{-2 x} x^2}{\left (-1+e^x\right )^3} \, dx-2 \int \frac {e^{-2 x} x \left (3 x-e^3 \log (5)-\log (81)\right )}{\left (-1+e^x\right )^2} \, dx+2 \int e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right ) \, dx+2 \int \frac {e^{-2 x} \left (-1+3 x^2+e^3 \log (5)+\log (81)+x \left (1-e^3 \log (125)-\log (531441)\right )\right )}{1-e^x} \, dx+\left (2 \left (1-e^3 \log (5)-e^6 \log ^2(5)-\log (81)\right )\right ) \int e^{-2 x} \, dx-(2 \log (81)) \int e^x \, dx-\left (2 \left (1-e^3 \log (25)-\log (6561)\right )\right ) \int e^{-2 x} x \, dx\\ &=e^{2 x}+2 x+e^{-2 x} x^2-e^{-2 x} \left (1-e^3 \log (5)-e^6 \log ^2(5)-\log (81)\right )-2 e^x \log (81)-2 e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right )+e^{-2 x} x \left (1-e^3 \log (25)-\log (6561)\right )-2 \int e^{-2 x} x \, dx-2 \int \left (-e^{-2 x} x^2-3 e^{-x} x^2+\frac {x^2}{\left (-1+e^x\right )^3}-\frac {2 x^2}{\left (-1+e^x\right )^2}+\frac {3 x^2}{-1+e^x}\right ) \, dx-2 \int \left (\frac {3 e^{-2 x} x^2}{\left (-1+e^x\right )^2}-\frac {e^{-2 x} x \left (e^3 \log (5)+\log (81)\right )}{\left (-1+e^x\right )^2}\right ) \, dx+2 \int \left (-\frac {3 e^{-2 x} x^2}{-1+e^x}+\frac {e^{-2 x} \left (1-e^3 \log (5)-\log (81)\right )}{-1+e^x}+\frac {e^{-2 x} x \left (-1+e^3 \log (125)+\log (531441)\right )}{-1+e^x}\right ) \, dx-(2 (1-\log (81))) \int e^{-x} \, dx-\left (1-e^3 \log (25)-\log (6561)\right ) \int e^{-2 x} \, dx\\ &=e^{2 x}+2 x+e^{-2 x} x+e^{-2 x} x^2+2 e^{-x} (1-\log (81))-e^{-2 x} \left (1-e^3 \log (5)-e^6 \log ^2(5)-\log (81)\right )-2 e^x \log (81)-2 e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right )+\frac {1}{2} e^{-2 x} \left (1-e^3 \log (25)-\log (6561)\right )+e^{-2 x} x \left (1-e^3 \log (25)-\log (6561)\right )+2 \int e^{-2 x} x^2 \, dx-2 \int \frac {x^2}{\left (-1+e^x\right )^3} \, dx+4 \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx+6 \int e^{-x} x^2 \, dx-6 \int \frac {e^{-2 x} x^2}{\left (-1+e^x\right )^2} \, dx-6 \int \frac {x^2}{-1+e^x} \, dx-6 \int \frac {e^{-2 x} x^2}{-1+e^x} \, dx+\left (2 \left (1-e^3 \log (5)-\log (81)\right )\right ) \int \frac {e^{-2 x}}{-1+e^x} \, dx+\left (2 \left (e^3 \log (5)+\log (81)\right )\right ) \int \frac {e^{-2 x} x}{\left (-1+e^x\right )^2} \, dx-\left (2 \left (1-e^3 \log (125)-\log (531441)\right )\right ) \int \frac {e^{-2 x} x}{-1+e^x} \, dx-\int e^{-2 x} \, dx\\ &=\frac {e^{-2 x}}{2}+e^{2 x}+2 x+e^{-2 x} x-6 e^{-x} x^2+2 x^3+2 e^{-x} (1-\log (81))-e^{-2 x} \left (1-e^3 \log (5)-e^6 \log ^2(5)-\log (81)\right )-2 e^x \log (81)-2 e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right )+\frac {1}{2} e^{-2 x} \left (1-e^3 \log (25)-\log (6561)\right )+e^{-2 x} x \left (1-e^3 \log (25)-\log (6561)\right )+2 \int e^{-2 x} x \, dx-2 \int \frac {e^x x^2}{\left (-1+e^x\right )^3} \, dx+2 \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx+4 \int \frac {e^x x^2}{\left (-1+e^x\right )^2} \, dx-4 \int \frac {x^2}{-1+e^x} \, dx-6 \int \frac {e^x x^2}{-1+e^x} \, dx-6 \int \left (e^{-2 x} x^2+2 e^{-x} x^2+\frac {x^2}{\left (-1+e^x\right )^2}-\frac {2 x^2}{-1+e^x}\right ) \, dx-6 \int \left (-e^{-2 x} x^2-e^{-x} x^2+\frac {x^2}{-1+e^x}\right ) \, dx+12 \int e^{-x} x \, dx+\left (2 \left (1-e^3 \log (5)-\log (81)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) x^3} \, dx,x,e^x\right )+\left (2 \left (e^3 \log (5)+\log (81)\right )\right ) \int \left (e^{-2 x} x+2 e^{-x} x+\frac {x}{\left (-1+e^x\right )^2}-\frac {2 x}{-1+e^x}\right ) \, dx-\left (2 \left (1-e^3 \log (125)-\log (531441)\right )\right ) \int \left (-e^{-2 x} x-e^{-x} x+\frac {x}{-1+e^x}\right ) \, dx\\ &=\frac {e^{-2 x}}{2}+e^{2 x}+2 x-12 e^{-x} x-6 e^{-x} x^2+\frac {x^2}{\left (1-e^x\right )^2}+\frac {4 x^2}{1-e^x}+\frac {10 x^3}{3}+2 e^{-x} (1-\log (81))-e^{-2 x} \left (1-e^3 \log (5)-e^6 \log ^2(5)-\log (81)\right )-2 e^x \log (81)-2 e^{-x} \left (1-x (1-\log (81))-\log (81)-e^3 \log (5) \log (81)\right )+\frac {1}{2} e^{-2 x} \left (1-e^3 \log (25)-\log (6561)\right )+e^{-2 x} x \left (1-e^3 \log (25)-\log (6561)\right )-6 x^2 \log \left (1-e^x\right )-2 \int \frac {x}{\left (-1+e^x\right )^2} \, dx+2 \int \frac {e^x x^2}{\left (-1+e^x\right )^2} \, dx-2 \int \frac {x^2}{-1+e^x} \, dx-4 \int \frac {e^x x^2}{-1+e^x} \, dx+6 \int e^{-x} x^2 \, dx-6 \int \frac {x^2}{\left (-1+e^x\right )^2} \, dx-6 \int \frac {x^2}{-1+e^x} \, dx+8 \int \frac {x}{-1+e^x} \, dx+12 \int e^{-x} \, dx-12 \int e^{-x} x^2 \, dx+12 \int \frac {x^2}{-1+e^x} \, dx+12 \int x \log \left (1-e^x\right ) \, dx+\left (2 \left (1-e^3 \log (5)-\log (81)\right )\right ) \operatorname {Subst}\left (\int \left (\frac {1}{-1+x}-\frac {1}{x^3}-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx,x,e^x\right )+\left (2 \left (e^3 \log (5)+\log (81)\right )\right ) \int e^{-2 x} x \, dx+\left (2 \left (e^3 \log (5)+\log (81)\right )\right ) \int \frac {x}{\left (-1+e^x\right )^2} \, dx+\left (4 \left (e^3 \log (5)+\log (81)\right )\right ) \int e^{-x} x \, dx-\left (4 \left (e^3 \log (5)+\log (81)\right )\right ) \int \frac {x}{-1+e^x} \, dx+\left (2 \left (1-e^3 \log (125)-\log (531441)\right )\right ) \int e^{-2 x} x \, dx+\left (2 \left (1-e^3 \log (125)-\log (531441)\right )\right ) \int e^{-x} x \, dx-\left (2 \left (1-e^3 \log (125)-\log (531441)\right )\right ) \int \frac {x}{-1+e^x} \, dx+\int e^{-2 x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.25, size = 92, normalized size = 2.79 \begin {gather*} e^{2 x}+x^2+\frac {x^2}{\left (-1+e^{-x}\right )^2}+e^{6-2 x} \log ^2(5)-2 e^x \log (81)+2 e^{3-x} \log (5) (x+\log (81))+2 x \left (e^3 \log (5)+\log (81)\right )-\frac {2 e^x x \left (-1+x+e^3 \log (5)+\log (81)\right )}{-1+e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(5*x) - 2*x + E^(4*x)*(-6 - 8*Log[3]) - 8*Log[3] + E^(3 - x)*(-2 + 2*x + 8*Log[3])*Log[5] + 2*E
^(6 - 2*x)*Log[5]^2 + E^x*(2 + 4*x - 2*x^2 + (24 - 8*x)*Log[3] + E^(3 - x)*(4 - 6*x - 24*Log[3])*Log[5] - 6*E^
(6 - 2*x)*Log[5]^2) + E^(3*x)*(8 + 24*Log[3] - 8*E^(3 - x)*Log[3]*Log[5] - 2*E^(6 - 2*x)*Log[5]^2) + E^(2*x)*(
-6 - 2*x + (-32 + 8*x)*Log[3] + E^(3 - x)*(-2 + 4*x + 24*Log[3])*Log[5] + 6*E^(6 - 2*x)*Log[5]^2))/(-1 + 3*E^x
- 3*E^(2*x) + E^(3*x)),x]

[Out]

E^(2*x) + x^2 + x^2/(-1 + E^(-x))^2 + E^(6 - 2*x)*Log[5]^2 - 2*E^x*Log[81] + 2*E^(3 - x)*Log[5]*(x + Log[81])
+ 2*x*(E^3*Log[5] + Log[81]) - (2*E^x*x*(-1 + x + E^3*Log[5] + Log[81]))/(-1 + E^x)

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fricas [B]  time = 0.75, size = 144, normalized size = 4.36 \begin {gather*} \frac {e^{6} \log \relax (5)^{2} - 2 \, {\left (4 \, \log \relax (3) + 1\right )} e^{\left (5 \, x\right )} + {\left (2 \, x + 16 \, \log \relax (3) + 1\right )} e^{\left (4 \, x\right )} + 2 \, {\left (4 \, e^{3} \log \relax (5) \log \relax (3) - 4 \, {\left (x + 1\right )} \log \relax (3) - x\right )} e^{\left (3 \, x\right )} + {\left (e^{6} \log \relax (5)^{2} + x^{2} - 2 \, {\left (x e^{3} + 8 \, e^{3} \log \relax (3)\right )} \log \relax (5) + 8 \, x \log \relax (3)\right )} e^{\left (2 \, x\right )} - 2 \, {\left (e^{6} \log \relax (5)^{2} - {\left (x e^{3} + 4 \, e^{3} \log \relax (3)\right )} \log \relax (5)\right )} e^{x} + e^{\left (6 \, x\right )}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^5+(-8*log(3)-6)*exp(x)^4+(-2*log(5)^2*exp(3-x)^2-8*log(3)*log(5)*exp(3-x)+24*log(3)+8)*exp
(x)^3+(6*log(5)^2*exp(3-x)^2+(24*log(3)+4*x-2)*log(5)*exp(3-x)+(8*x-32)*log(3)-2*x-6)*exp(x)^2+(-6*log(5)^2*ex
p(3-x)^2+(-24*log(3)-6*x+4)*log(5)*exp(3-x)+(-8*x+24)*log(3)-2*x^2+4*x+2)*exp(x)+2*log(5)^2*exp(3-x)^2+(8*log(
3)+2*x-2)*log(5)*exp(3-x)-8*log(3)-2*x)/(exp(x)^3-3*exp(x)^2+3*exp(x)-1),x, algorithm="fricas")

[Out]

(e^6*log(5)^2 - 2*(4*log(3) + 1)*e^(5*x) + (2*x + 16*log(3) + 1)*e^(4*x) + 2*(4*e^3*log(5)*log(3) - 4*(x + 1)*
log(3) - x)*e^(3*x) + (e^6*log(5)^2 + x^2 - 2*(x*e^3 + 8*e^3*log(3))*log(5) + 8*x*log(3))*e^(2*x) - 2*(e^6*log
(5)^2 - (x*e^3 + 4*e^3*log(3))*log(5))*e^x + e^(6*x))/(e^(4*x) - 2*e^(3*x) + e^(2*x))

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giac [B]  time = 0.31, size = 179, normalized size = 5.42 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} - 2 \, x e^{\left (2 \, x + 3\right )} \log \relax (5) + 2 \, x e^{\left (x + 3\right )} \log \relax (5) + e^{6} \log \relax (5)^{2} + e^{\left (2 \, x + 6\right )} \log \relax (5)^{2} - 2 \, e^{\left (x + 6\right )} \log \relax (5)^{2} - 8 \, x e^{\left (3 \, x\right )} \log \relax (3) + 8 \, x e^{\left (2 \, x\right )} \log \relax (3) + 8 \, e^{\left (3 \, x + 3\right )} \log \relax (5) \log \relax (3) - 16 \, e^{\left (2 \, x + 3\right )} \log \relax (5) \log \relax (3) + 8 \, e^{\left (x + 3\right )} \log \relax (5) \log \relax (3) + 2 \, x e^{\left (4 \, x\right )} - 2 \, x e^{\left (3 \, x\right )} - 8 \, e^{\left (5 \, x\right )} \log \relax (3) + 16 \, e^{\left (4 \, x\right )} \log \relax (3) - 8 \, e^{\left (3 \, x\right )} \log \relax (3) + e^{\left (6 \, x\right )} - 2 \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^5+(-8*log(3)-6)*exp(x)^4+(-2*log(5)^2*exp(3-x)^2-8*log(3)*log(5)*exp(3-x)+24*log(3)+8)*exp
(x)^3+(6*log(5)^2*exp(3-x)^2+(24*log(3)+4*x-2)*log(5)*exp(3-x)+(8*x-32)*log(3)-2*x-6)*exp(x)^2+(-6*log(5)^2*ex
p(3-x)^2+(-24*log(3)-6*x+4)*log(5)*exp(3-x)+(-8*x+24)*log(3)-2*x^2+4*x+2)*exp(x)+2*log(5)^2*exp(3-x)^2+(8*log(
3)+2*x-2)*log(5)*exp(3-x)-8*log(3)-2*x)/(exp(x)^3-3*exp(x)^2+3*exp(x)-1),x, algorithm="giac")

[Out]

(x^2*e^(2*x) - 2*x*e^(2*x + 3)*log(5) + 2*x*e^(x + 3)*log(5) + e^6*log(5)^2 + e^(2*x + 6)*log(5)^2 - 2*e^(x +
6)*log(5)^2 - 8*x*e^(3*x)*log(3) + 8*x*e^(2*x)*log(3) + 8*e^(3*x + 3)*log(5)*log(3) - 16*e^(2*x + 3)*log(5)*lo
g(3) + 8*e^(x + 3)*log(5)*log(3) + 2*x*e^(4*x) - 2*x*e^(3*x) - 8*e^(5*x)*log(3) + 16*e^(4*x)*log(3) - 8*e^(3*x
)*log(3) + e^(6*x) - 2*e^(5*x) + e^(4*x))/(e^(4*x) - 2*e^(3*x) + e^(2*x))

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maple [B]  time = 0.34, size = 89, normalized size = 2.70

method	result	size
risch	\(-8 \ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{2 x}+2 x +\left (8 \ln \relax (3) \ln \relax (5) {\mathrm e}^{3}+2 x \,{\mathrm e}^{3} \ln \relax (5)\right ) {\mathrm e}^{-x}+\ln \relax (5)^{2} {\mathrm e}^{6-2 x}-\frac {x \left (2 \ln \relax (5) {\mathrm e}^{3+x}+8 \ln \relax (3) {\mathrm e}^{x}-2 \,{\mathrm e}^{3} \ln \relax (5)-8 \ln \relax (3)-x -2 \,{\mathrm e}^{x}+2\right )}{\left ({\mathrm e}^{x}-1\right )^{2}}\)	\(89\)
norman	\(\frac {\left ({\mathrm e}^{6 x}+\left (-2-8 \ln \relax (3)\right ) {\mathrm e}^{5 x}+\left (-2 \,{\mathrm e}^{6} \ln \relax (5)^{2}+8 \ln \relax (3) \ln \relax (5) {\mathrm e}^{3}\right ) {\mathrm e}^{x}+\left (8 \ln \relax (3) \ln \relax (5) {\mathrm e}^{3}+24 \ln \relax (3)+2\right ) {\mathrm e}^{3 x}+\left ({\mathrm e}^{6} \ln \relax (5)^{2}-16 \ln \relax (3) \ln \relax (5) {\mathrm e}^{3}-16 \ln \relax (3)-1\right ) {\mathrm e}^{2 x}+{\mathrm e}^{6} \ln \relax (5)^{2}+{\mathrm e}^{2 x} x^{2}+\left (-2-8 \ln \relax (3)\right ) x \,{\mathrm e}^{3 x}+\left (8 \ln \relax (3)-2 \,{\mathrm e}^{3} \ln \relax (5)\right ) x \,{\mathrm e}^{2 x}+2 x \,{\mathrm e}^{4 x}+2 \,{\mathrm e}^{x} \ln \relax (5) {\mathrm e}^{3} x \right ) {\mathrm e}^{-2 x}}{\left ({\mathrm e}^{x}-1\right )^{2}}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x)^5+(-8*ln(3)-6)*exp(x)^4+(-2*ln(5)^2*exp(3-x)^2-8*ln(3)*ln(5)*exp(3-x)+24*ln(3)+8)*exp(x)^3+(6*ln
(5)^2*exp(3-x)^2+(24*ln(3)+4*x-2)*ln(5)*exp(3-x)+(8*x-32)*ln(3)-2*x-6)*exp(x)^2+(-6*ln(5)^2*exp(3-x)^2+(-24*ln
(3)-6*x+4)*ln(5)*exp(3-x)+(-8*x+24)*ln(3)-2*x^2+4*x+2)*exp(x)+2*ln(5)^2*exp(3-x)^2+(8*ln(3)+2*x-2)*ln(5)*exp(3
-x)-8*ln(3)-2*x)/(exp(x)^3-3*exp(x)^2+3*exp(x)-1),x,method=_RETURNVERBOSE)

[Out]

-8*ln(3)*exp(x)+exp(2*x)+2*x+(8*ln(3)*ln(5)*exp(3)+2*x*exp(3)*ln(5))*exp(-x)+ln(5)^2*exp(6-2*x)-x*(2*ln(5)*exp
(3+x)+8*ln(3)*exp(x)-2*exp(3)*ln(5)-8*ln(3)-x-2*exp(x)+2)/(exp(x)-1)^2

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maxima [B]  time = 0.80, size = 422, normalized size = 12.79 \begin {gather*} -3 \, {\left (2 \, x - \frac {2 \, e^{x} - 3}{e^{\left (2 \, x\right )} - 2 \, e^{x} + 1} - 2 \, \log \left (e^{x} - 1\right )\right )} e^{6} \log \relax (5)^{2} + 12 \, {\left (2 \, x - \frac {2 \, e^{x} - 3}{e^{\left (2 \, x\right )} - 2 \, e^{x} + 1} - 2 \, \log \left (e^{x} - 1\right )\right )} e^{3} \log \relax (5) \log \relax (3) - 2 \, {\left (2 \, x - \frac {2 \, e^{x} - 3}{e^{\left (2 \, x\right )} - 2 \, e^{x} + 1} - 2 \, \log \left (e^{x} - 1\right )\right )} e^{3} \log \relax (5) + 4 \, {\left (2 \, x - \frac {2 \, e^{x} - 3}{e^{\left (2 \, x\right )} - 2 \, e^{x} + 1} - 2 \, \log \left (e^{x} - 1\right )\right )} \log \relax (3) - 2 \, {\left (3 \, e^{6} \log \relax (5)^{2} - 2 \, {\left (6 \, \log \relax (5) \log \relax (3) - \log \relax (5)\right )} e^{3} - 4 \, \log \relax (3)\right )} \log \left (e^{x} - 1\right ) + \frac {e^{6} \log \relax (5)^{2} - 2 \, {\left (4 \, \log \relax (3) + 1\right )} e^{\left (5 \, x\right )} + {\left (2 \, {\left (3 \, e^{6} \log \relax (5)^{2} - 2 \, {\left (6 \, \log \relax (5) \log \relax (3) - \log \relax (5)\right )} e^{3} - 4 \, \log \relax (3) + 1\right )} x + 16 \, \log \relax (3) + 1\right )} e^{\left (4 \, x\right )} - 2 \, {\left (3 \, e^{6} \log \relax (5)^{2} + {\left (6 \, e^{6} \log \relax (5)^{2} - 4 \, {\left (6 \, \log \relax (5) \log \relax (3) - \log \relax (5)\right )} e^{3} - 4 \, \log \relax (3) + 1\right )} x - 2 \, {\left (8 \, \log \relax (5) \log \relax (3) - \log \relax (5)\right )} e^{3}\right )} e^{\left (3 \, x\right )} + {\left (10 \, e^{6} \log \relax (5)^{2} + 2 \, {\left (3 \, e^{6} \log \relax (5)^{2} - {\left (12 \, \log \relax (5) \log \relax (3) - \log \relax (5)\right )} e^{3}\right )} x + x^{2} - 2 \, {\left (26 \, \log \relax (5) \log \relax (3) - 3 \, \log \relax (5)\right )} e^{3} - 12 \, \log \relax (3)\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x e^{3} \log \relax (5) - e^{6} \log \relax (5)^{2} + 4 \, e^{3} \log \relax (5) \log \relax (3)\right )} e^{x} + e^{\left (6 \, x\right )}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)^5+(-8*log(3)-6)*exp(x)^4+(-2*log(5)^2*exp(3-x)^2-8*log(3)*log(5)*exp(3-x)+24*log(3)+8)*exp
(x)^3+(6*log(5)^2*exp(3-x)^2+(24*log(3)+4*x-2)*log(5)*exp(3-x)+(8*x-32)*log(3)-2*x-6)*exp(x)^2+(-6*log(5)^2*ex
p(3-x)^2+(-24*log(3)-6*x+4)*log(5)*exp(3-x)+(-8*x+24)*log(3)-2*x^2+4*x+2)*exp(x)+2*log(5)^2*exp(3-x)^2+(8*log(
3)+2*x-2)*log(5)*exp(3-x)-8*log(3)-2*x)/(exp(x)^3-3*exp(x)^2+3*exp(x)-1),x, algorithm="maxima")

[Out]

-3*(2*x - (2*e^x - 3)/(e^(2*x) - 2*e^x + 1) - 2*log(e^x - 1))*e^6*log(5)^2 + 12*(2*x - (2*e^x - 3)/(e^(2*x) -
2*e^x + 1) - 2*log(e^x - 1))*e^3*log(5)*log(3) - 2*(2*x - (2*e^x - 3)/(e^(2*x) - 2*e^x + 1) - 2*log(e^x - 1))*
e^3*log(5) + 4*(2*x - (2*e^x - 3)/(e^(2*x) - 2*e^x + 1) - 2*log(e^x - 1))*log(3) - 2*(3*e^6*log(5)^2 - 2*(6*lo
g(5)*log(3) - log(5))*e^3 - 4*log(3))*log(e^x - 1) + (e^6*log(5)^2 - 2*(4*log(3) + 1)*e^(5*x) + (2*(3*e^6*log(
5)^2 - 2*(6*log(5)*log(3) - log(5))*e^3 - 4*log(3) + 1)*x + 16*log(3) + 1)*e^(4*x) - 2*(3*e^6*log(5)^2 + (6*e^
6*log(5)^2 - 4*(6*log(5)*log(3) - log(5))*e^3 - 4*log(3) + 1)*x - 2*(8*log(5)*log(3) - log(5))*e^3)*e^(3*x) +
(10*e^6*log(5)^2 + 2*(3*e^6*log(5)^2 - (12*log(5)*log(3) - log(5))*e^3)*x + x^2 - 2*(26*log(5)*log(3) - 3*log(
5))*e^3 - 12*log(3))*e^(2*x) + 2*(x*e^3*log(5) - e^6*log(5)^2 + 4*e^3*log(5)*log(3))*e^x + e^(6*x))/(e^(4*x) -
2*e^(3*x) + e^(2*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {2\,{\mathrm {e}}^{5\,x}-2\,x-8\,\ln \relax (3)+{\mathrm {e}}^{2\,x}\,\left (\ln \relax (3)\,\left (8\,x-32\right )-2\,x+6\,{\mathrm {e}}^{6-2\,x}\,{\ln \relax (5)}^2+{\mathrm {e}}^{3-x}\,\ln \relax (5)\,\left (4\,x+24\,\ln \relax (3)-2\right )-6\right )+{\mathrm {e}}^{3\,x}\,\left (24\,\ln \relax (3)-2\,{\mathrm {e}}^{6-2\,x}\,{\ln \relax (5)}^2-8\,{\mathrm {e}}^{3-x}\,\ln \relax (3)\,\ln \relax (5)+8\right )-{\mathrm {e}}^{4\,x}\,\left (8\,\ln \relax (3)+6\right )+2\,{\mathrm {e}}^{6-2\,x}\,{\ln \relax (5)}^2-{\mathrm {e}}^x\,\left (\ln \relax (3)\,\left (8\,x-24\right )-4\,x+6\,{\mathrm {e}}^{6-2\,x}\,{\ln \relax (5)}^2+2\,x^2+{\mathrm {e}}^{3-x}\,\ln \relax (5)\,\left (6\,x+24\,\ln \relax (3)-4\right )-2\right )+{\mathrm {e}}^{3-x}\,\ln \relax (5)\,\left (2\,x+8\,\ln \relax (3)-2\right )}{3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(5*x) - 2*x - 8*log(3) + exp(2*x)*(log(3)*(8*x - 32) - 2*x + 6*exp(6 - 2*x)*log(5)^2 + exp(3 - x)*l
og(5)*(4*x + 24*log(3) - 2) - 6) + exp(3*x)*(24*log(3) - 2*exp(6 - 2*x)*log(5)^2 - 8*exp(3 - x)*log(3)*log(5)
+ 8) - exp(4*x)*(8*log(3) + 6) + 2*exp(6 - 2*x)*log(5)^2 - exp(x)*(log(3)*(8*x - 24) - 4*x + 6*exp(6 - 2*x)*lo
g(5)^2 + 2*x^2 + exp(3 - x)*log(5)*(6*x + 24*log(3) - 4) - 2) + exp(3 - x)*log(5)*(2*x + 8*log(3) - 2))/(3*exp
(2*x) - exp(3*x) - 3*exp(x) + 1),x)

[Out]

-int((2*exp(5*x) - 2*x - 8*log(3) + exp(2*x)*(log(3)*(8*x - 32) - 2*x + 6*exp(6 - 2*x)*log(5)^2 + exp(3 - x)*l
og(5)*(4*x + 24*log(3) - 2) - 6) + exp(3*x)*(24*log(3) - 2*exp(6 - 2*x)*log(5)^2 - 8*exp(3 - x)*log(3)*log(5)
+ 8) - exp(4*x)*(8*log(3) + 6) + 2*exp(6 - 2*x)*log(5)^2 - exp(x)*(log(3)*(8*x - 24) - 4*x + 6*exp(6 - 2*x)*lo
g(5)^2 + 2*x^2 + exp(3 - x)*log(5)*(6*x + 24*log(3) - 4) - 2) + exp(3 - x)*log(5)*(2*x + 8*log(3) - 2))/(3*exp
(2*x) - exp(3*x) - 3*exp(x) + 1), x)

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sympy [B]  time = 0.40, size = 114, normalized size = 3.45 \begin {gather*} 2 x + \left (2 x e^{3} \log {\relax (5 )} + 8 e^{3} \log {\relax (3 )} \log {\relax (5 )}\right ) e^{- x} + e^{2 x} - 8 e^{x} \log {\relax (3 )} + e^{6} e^{- 2 x} \log {\relax (5 )}^{2} + \frac {x^{2} - 2 x + 8 x \log {\relax (3 )} + 2 x e^{3} \log {\relax (5 )} + \left (- 2 x e^{3} \log {\relax (5 )} - 8 x \log {\relax (3 )} + 2 x\right ) e^{x}}{e^{2 x} - 2 e^{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)**5+(-8*ln(3)-6)*exp(x)**4+(-2*ln(5)**2*exp(3-x)**2-8*ln(3)*ln(5)*exp(3-x)+24*ln(3)+8)*exp(
x)**3+(6*ln(5)**2*exp(3-x)**2+(24*ln(3)+4*x-2)*ln(5)*exp(3-x)+(8*x-32)*ln(3)-2*x-6)*exp(x)**2+(-6*ln(5)**2*exp
(3-x)**2+(-24*ln(3)-6*x+4)*ln(5)*exp(3-x)+(-8*x+24)*ln(3)-2*x**2+4*x+2)*exp(x)+2*ln(5)**2*exp(3-x)**2+(8*ln(3)
+2*x-2)*ln(5)*exp(3-x)-8*ln(3)-2*x)/(exp(x)**3-3*exp(x)**2+3*exp(x)-1),x)

[Out]

2*x + (2*x*exp(3)*log(5) + 8*exp(3)*log(3)*log(5))*exp(-x) + exp(2*x) - 8*exp(x)*log(3) + exp(6)*exp(-2*x)*log
(5)**2 + (x**2 - 2*x + 8*x*log(3) + 2*x*exp(3)*log(5) + (-2*x*exp(3)*log(5) - 8*x*log(3) + 2*x)*exp(x))/(exp(2
*x) - 2*exp(x) + 1)

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