3.21.22 \(\int \frac {16+4 e^{2 x}+(-36-72 x-24 x^2) \log (3)+e^x (-16+(18+18 x-6 x^2-4 x^3) \log (3)-4 \log (4))+(8+(-9-18 x-6 x^2) \log (3)) \log (4)+\log ^2(4)}{16+4 e^{2 x}+e^x (-16-4 \log (4))+8 \log (4)+\log ^2(4)} \, dx\)

Optimal. Leaf size=27 \[ x+\frac {x (3+x) (3+2 x) \log (3)}{-4+2 e^x-\log (4)} \]

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Rubi [C]  time = 3.90, antiderivative size = 936, normalized size of antiderivative = 34.67, number of steps used = 62, number of rules used = 16, integrand size = 102, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.157, Rules used = {6741, 12, 6742, 2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2279, 2391, 36, 29, 31} \begin {gather*} \frac {\log (9) x^4}{8 (2+\log (2))}-\frac {\log (3) (4+\log (4)) x^4}{8 (2+\log (2))^2}-\frac {\log (9) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x^3}{2 (2+\log (2))}+\frac {\log (3) (4+\log (4)) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x^3}{2 (2+\log (2))^2}+\frac {\log (27) x^3}{6 (2+\log (2))}-\frac {\log (3) (6+\log (8)) x^3}{2 (2+\log (2))^2}-\frac {\log (3) (4+\log (4)) x^3}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}+\frac {\log (3) (4+\log (4)) x^3}{2 (2+\log (2))^2}-\frac {\log (27) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x^2}{2 (2+\log (2))}+\frac {3 \log (3) (6+\log (8)) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x^2}{2 (2+\log (2))^2}-\frac {3 \log (3) (4+\log (4)) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x^2}{2 (2+\log (2))^2}-\frac {3 \log (9) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right ) x^2}{2 (2+\log (2))}+\frac {3 \log (3) (4+\log (4)) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right ) x^2}{2 (2+\log (2))^2}-\frac {\log (19683) x^2}{4 (2+\log (2))}-\frac {\log (3) (18+\log (512)) x^2}{4 (2+\log (2))^2}-\frac {3 \log (3) (6+\log (8)) x^2}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}+\frac {3 \log (3) (6+\log (8)) x^2}{2 (2+\log (2))^2}+\frac {\log (19683) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x}{2 (2+\log (2))}+\frac {\log (3) (18+\log (512)) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x}{2 (2+\log (2))^2}-\frac {3 \log (3) (6+\log (8)) \log \left (1-\frac {e^x}{2+\log (2)}\right ) x}{(2+\log (2))^2}-\frac {\log (27) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right ) x}{2+\log (2)}+\frac {3 \log (3) (6+\log (8)) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right ) x}{(2+\log (2))^2}-\frac {3 \log (3) (4+\log (4)) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right ) x}{(2+\log (2))^2}+\frac {3 \log (9) \text {Li}_3\left (\frac {e^x}{2+\log (2)}\right ) x}{2+\log (2)}-\frac {3 \log (3) (4+\log (4)) \text {Li}_3\left (\frac {e^x}{2+\log (2)}\right ) x}{(2+\log (2))^2}-\frac {\log (19683) x}{2 (2+\log (2))}-\frac {\log (3) (18+\log (512)) x}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}+\frac {\log (3) (18+\log (512)) x}{2 (2+\log (2))^2}+x+\frac {\log (19683) \log \left (2-e^x+\log (2)\right )}{2 (2+\log (2))}-\frac {\log (3) (18+\log (512)) \log \left (2-e^x+\log (2)\right )}{2 (2+\log (2))^2}+\frac {\log (19683) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right )}{2 (2+\log (2))}+\frac {\log (3) (18+\log (512)) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right )}{2 (2+\log (2))^2}-\frac {3 \log (3) (6+\log (8)) \text {Li}_2\left (\frac {e^x}{2+\log (2)}\right )}{(2+\log (2))^2}+\frac {\log (27) \text {Li}_3\left (\frac {e^x}{2+\log (2)}\right )}{2+\log (2)}-\frac {3 \log (3) (6+\log (8)) \text {Li}_3\left (\frac {e^x}{2+\log (2)}\right )}{(2+\log (2))^2}+\frac {3 \log (3) (4+\log (4)) \text {Li}_3\left (\frac {e^x}{2+\log (2)}\right )}{(2+\log (2))^2}-\frac {3 \log (9) \text {Li}_4\left (\frac {e^x}{2+\log (2)}\right )}{2+\log (2)}+\frac {3 \log (3) (4+\log (4)) \text {Li}_4\left (\frac {e^x}{2+\log (2)}\right )}{(2+\log (2))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + 4*E^(2*x) + (-36 - 72*x - 24*x^2)*Log[3] + E^x*(-16 + (18 + 18*x - 6*x^2 - 4*x^3)*Log[3] - 4*Log[4])
+ (8 + (-9 - 18*x - 6*x^2)*Log[3])*Log[4] + Log[4]^2)/(16 + 4*E^(2*x) + E^x*(-16 - 4*Log[4]) + 8*Log[4] + Log
[4]^2),x]

[Out]

x + (x^3*Log[3]*(4 + Log[4]))/(2*(2 + Log[2])^2) - (x^4*Log[3]*(4 + Log[4]))/(8*(2 + Log[2])^2) - (x^3*Log[3]*
(4 + Log[4]))/(2*(2 + Log[2])*(2 - E^x + Log[2])) + (3*x^2*Log[3]*(6 + Log[8]))/(2*(2 + Log[2])^2) - (x^3*Log[
3]*(6 + Log[8]))/(2*(2 + Log[2])^2) - (3*x^2*Log[3]*(6 + Log[8]))/(2*(2 + Log[2])*(2 - E^x + Log[2])) + (x^4*L
og[9])/(8*(2 + Log[2])) + (x^3*Log[27])/(6*(2 + Log[2])) + (x*Log[3]*(18 + Log[512]))/(2*(2 + Log[2])^2) - (x^
2*Log[3]*(18 + Log[512]))/(4*(2 + Log[2])^2) - (x*Log[3]*(18 + Log[512]))/(2*(2 + Log[2])*(2 - E^x + Log[2]))
- (x*Log[19683])/(2*(2 + Log[2])) - (x^2*Log[19683])/(4*(2 + Log[2])) - (Log[3]*(18 + Log[512])*Log[2 - E^x +
Log[2]])/(2*(2 + Log[2])^2) + (Log[19683]*Log[2 - E^x + Log[2]])/(2*(2 + Log[2])) - (3*x^2*Log[3]*(4 + Log[4])
*Log[1 - E^x/(2 + Log[2])])/(2*(2 + Log[2])^2) + (x^3*Log[3]*(4 + Log[4])*Log[1 - E^x/(2 + Log[2])])/(2*(2 + L
og[2])^2) - (3*x*Log[3]*(6 + Log[8])*Log[1 - E^x/(2 + Log[2])])/(2 + Log[2])^2 + (3*x^2*Log[3]*(6 + Log[8])*Lo
g[1 - E^x/(2 + Log[2])])/(2*(2 + Log[2])^2) - (x^3*Log[9]*Log[1 - E^x/(2 + Log[2])])/(2*(2 + Log[2])) - (x^2*L
og[27]*Log[1 - E^x/(2 + Log[2])])/(2*(2 + Log[2])) + (x*Log[3]*(18 + Log[512])*Log[1 - E^x/(2 + Log[2])])/(2*(
2 + Log[2])^2) + (x*Log[19683]*Log[1 - E^x/(2 + Log[2])])/(2*(2 + Log[2])) - (3*x*Log[3]*(4 + Log[4])*PolyLog[
2, E^x/(2 + Log[2])])/(2 + Log[2])^2 + (3*x^2*Log[3]*(4 + Log[4])*PolyLog[2, E^x/(2 + Log[2])])/(2*(2 + Log[2]
)^2) - (3*Log[3]*(6 + Log[8])*PolyLog[2, E^x/(2 + Log[2])])/(2 + Log[2])^2 + (3*x*Log[3]*(6 + Log[8])*PolyLog[
2, E^x/(2 + Log[2])])/(2 + Log[2])^2 - (3*x^2*Log[9]*PolyLog[2, E^x/(2 + Log[2])])/(2*(2 + Log[2])) - (x*Log[2
7]*PolyLog[2, E^x/(2 + Log[2])])/(2 + Log[2]) + (Log[3]*(18 + Log[512])*PolyLog[2, E^x/(2 + Log[2])])/(2*(2 +
Log[2])^2) + (Log[19683]*PolyLog[2, E^x/(2 + Log[2])])/(2*(2 + Log[2])) + (3*Log[3]*(4 + Log[4])*PolyLog[3, E^
x/(2 + Log[2])])/(2 + Log[2])^2 - (3*x*Log[3]*(4 + Log[4])*PolyLog[3, E^x/(2 + Log[2])])/(2 + Log[2])^2 - (3*L
og[3]*(6 + Log[8])*PolyLog[3, E^x/(2 + Log[2])])/(2 + Log[2])^2 + (3*x*Log[9]*PolyLog[3, E^x/(2 + Log[2])])/(2
+ Log[2]) + (Log[27]*PolyLog[3, E^x/(2 + Log[2])])/(2 + Log[2]) + (3*Log[3]*(4 + Log[4])*PolyLog[4, E^x/(2 +
Log[2])])/(2 + Log[2])^2 - (3*Log[9]*PolyLog[4, E^x/(2 + Log[2])])/(2 + Log[2])

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 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 {4 e^{2 x}+\left (-36-72 x-24 x^2\right ) \log (3)+e^x \left (-16+\left (18+18 x-6 x^2-4 x^3\right ) \log (3)-4 \log (4)\right )+\left (8+\left (-9-18 x-6 x^2\right ) \log (3)\right ) \log (4)+16 \left (1+\frac {\log ^2(4)}{16}\right )}{\left (2 e^x-4 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx\\ &=\int \frac {4 e^{2 x}+\left (-36-72 x-24 x^2\right ) \log (3)+e^x \left (-16+\left (18+18 x-6 x^2-4 x^3\right ) \log (3)-4 \log (4)\right )+\left (8+\left (-9-18 x-6 x^2\right ) \log (3)\right ) \log (4)+16 \left (1+\frac {\log ^2(4)}{16}\right )}{4 \left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {4 e^{2 x}+\left (-36-72 x-24 x^2\right ) \log (3)+e^x \left (-16+\left (18+18 x-6 x^2-4 x^3\right ) \log (3)-4 \log (4)\right )+\left (8+\left (-9-18 x-6 x^2\right ) \log (3)\right ) \log (4)+16 \left (1+\frac {\log ^2(4)}{16}\right )}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \left (4+\frac {2 x \log (3) \left (-18-x^2 (4+\log (4))-3 x (6+\log (8))-\log (512)\right )}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2}+\frac {2 \left (-x^3 \log (9)-x^2 \log (27)+\log (19683)+x \log (19683)\right )}{e^x-2 \left (1+\frac {\log (2)}{2}\right )}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {-x^3 \log (9)-x^2 \log (27)+\log (19683)+x \log (19683)}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx+\frac {1}{2} \log (3) \int \frac {x \left (-18-x^2 (4+\log (4))-3 x (6+\log (8))-\log (512)\right )}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx\\ &=x+\frac {1}{2} \int \left (\frac {x^3 \log (9)}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )}+\frac {x^2 \log (27)}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )}+\frac {\log (19683)}{e^x-2 \left (1+\frac {\log (2)}{2}\right )}+\frac {x \log (19683)}{e^x-2 \left (1+\frac {\log (2)}{2}\right )}\right ) \, dx+\frac {1}{2} \log (3) \int \left (\frac {x^3 (-4-\log (4))}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2}+\frac {3 x^2 (-6-\log (8))}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2}+\frac {x (-18-\log (512))}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2}\right ) \, dx\\ &=x-\frac {1}{2} (\log (3) (4+\log (4))) \int \frac {x^3}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx-\frac {1}{2} (3 \log (3) (6+\log (8))) \int \frac {x^2}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx+\frac {1}{2} \log (9) \int \frac {x^3}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )} \, dx+\frac {1}{2} \log (27) \int \frac {x^2}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )} \, dx-\frac {1}{2} (\log (3) (18+\log (512))) \int \frac {x}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx+\frac {1}{2} \log (19683) \int \frac {1}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx+\frac {1}{2} \log (19683) \int \frac {x}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx\\ &=x+\frac {x^4 \log (9)}{8 (2+\log (2))}+\frac {x^3 \log (27)}{6 (2+\log (2))}-\frac {x^2 \log (19683)}{4 (2+\log (2))}-\frac {(\log (3) (4+\log (4))) \int \frac {e^x x^3}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx}{2 (2+\log (2))}+\frac {(\log (3) (4+\log (4))) \int \frac {x^3}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}-\frac {(3 \log (3) (6+\log (8))) \int \frac {e^x x^2}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx}{2 (2+\log (2))}+\frac {(3 \log (3) (6+\log (8))) \int \frac {x^2}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}+\frac {\log (9) \int \frac {e^x x^3}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}+\frac {\log (27) \int \frac {e^x x^2}{-e^x+2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}-\frac {(\log (3) (18+\log (512))) \int \frac {e^x x}{\left (e^x-2 \left (1+\frac {\log (2)}{2}\right )\right )^2} \, dx}{2 (2+\log (2))}+\frac {(\log (3) (18+\log (512))) \int \frac {x}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}+\frac {1}{2} \log (19683) \operatorname {Subst}\left (\int \frac {1}{x (-2+x-\log (2))} \, dx,x,e^x\right )+\frac {\log (19683) \int \frac {e^x x}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}\\ &=x-\frac {x^4 \log (3) (4+\log (4))}{8 (2+\log (2))^2}-\frac {x^3 \log (3) (4+\log (4))}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}-\frac {x^3 \log (3) (6+\log (8))}{2 (2+\log (2))^2}-\frac {3 x^2 \log (3) (6+\log (8))}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}+\frac {x^4 \log (9)}{8 (2+\log (2))}+\frac {x^3 \log (27)}{6 (2+\log (2))}-\frac {x^2 \log (3) (18+\log (512))}{4 (2+\log (2))^2}-\frac {x \log (3) (18+\log (512))}{2 (2+\log (2)) \left (2-e^x+\log (2)\right )}-\frac {x^2 \log (19683)}{4 (2+\log (2))}-\frac {x^3 \log (9) \log \left (1-\frac {e^x}{2+\log (2)}\right )}{2 (2+\log (2))}-\frac {x^2 \log (27) \log \left (1-\frac {e^x}{2+\log (2)}\right )}{2 (2+\log (2))}+\frac {x \log (19683) \log \left (1-\frac {e^x}{2+\log (2)}\right )}{2 (2+\log (2))}+\frac {(\log (3) (4+\log (4))) \int \frac {e^x x^3}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))^2}-\frac {(3 \log (3) (4+\log (4))) \int \frac {x^2}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}+\frac {(3 \log (3) (6+\log (8))) \int \frac {e^x x^2}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))^2}-\frac {(3 \log (3) (6+\log (8))) \int \frac {x}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2+\log (2)}+\frac {(3 \log (9)) \int x^2 \log \left (1-\frac {e^x}{2 \left (1+\frac {\log (2)}{2}\right )}\right ) \, dx}{2 (2+\log (2))}+\frac {\log (27) \int x \log \left (1-\frac {e^x}{2 \left (1+\frac {\log (2)}{2}\right )}\right ) \, dx}{2+\log (2)}+\frac {(\log (3) (18+\log (512))) \int \frac {e^x x}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))^2}-\frac {(\log (3) (18+\log (512))) \int \frac {1}{e^x-2 \left (1+\frac {\log (2)}{2}\right )} \, dx}{2 (2+\log (2))}-\frac {\log (19683) \int \log \left (1-\frac {e^x}{2 \left (1+\frac {\log (2)}{2}\right )}\right ) \, dx}{2 (2+\log (2))}-\frac {\log (19683) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{2 (2+\log (2))}+\frac {\log (19683) \operatorname {Subst}\left (\int \frac {1}{-2+x-\log (2)} \, dx,x,e^x\right )}{2 (2+\log (2))}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [C]  time = 1.47, size = 316, normalized size = 11.70 \begin {gather*} x+\frac {x \left (x^2 \log (9)+\log (19683)+x \log (19683)\right )}{-4+2 e^x-\log (4)}-\frac {9 \log (3) \log \left (4-2 e^x+\log (4)\right )}{4+\log (4)}+\frac {\log (4) \log \left (4-2 e^x+\log (4)\right )}{4+\log (4)}+\frac {\log \left (\frac {19683}{4}\right ) \log \left (4-2 e^x+\log (4)\right )}{4+\log (4)}-\frac {2 \log (27) \left (x \log \left (\frac {4-2 e^x+\log (4)}{4+\log (4)}\right )+\text {Li}_2\left (\frac {2 e^x}{4+\log (4)}\right )\right )}{4+\log (4)}+\frac {3 \log (3) \left (x \left (x+2 \log \left (1-\frac {1}{2} e^{-x} (4+\log (4))\right )\right )-2 \text {Li}_2\left (\frac {1}{2} e^{-x} (4+\log (4))\right )\right )}{4+\log (4)}+\frac {6 \log (3) \left (x^2 \log \left (\frac {4-2 e^x+\log (4)}{4+\log (4)}\right )+2 x \text {Li}_2\left (\frac {2 e^x}{4+\log (4)}\right )-2 \text {Li}_3\left (\frac {2 e^x}{4+\log (4)}\right )\right )}{4+\log (4)}-\frac {\log (9) \left (x^2 \left (x+3 \log \left (1-\frac {1}{2} e^{-x} (4+\log (4))\right )\right )-6 x \text {Li}_2\left (\frac {1}{2} e^{-x} (4+\log (4))\right )-6 \text {Li}_3\left (\frac {1}{2} e^{-x} (4+\log (4))\right )\right )}{4+\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 4*E^(2*x) + (-36 - 72*x - 24*x^2)*Log[3] + E^x*(-16 + (18 + 18*x - 6*x^2 - 4*x^3)*Log[3] - 4*L
og[4]) + (8 + (-9 - 18*x - 6*x^2)*Log[3])*Log[4] + Log[4]^2)/(16 + 4*E^(2*x) + E^x*(-16 - 4*Log[4]) + 8*Log[4]
+ Log[4]^2),x]

[Out]

x + (x*(x^2*Log[9] + Log[19683] + x*Log[19683]))/(-4 + 2*E^x - Log[4]) - (9*Log[3]*Log[4 - 2*E^x + Log[4]])/(4
+ Log[4]) + (Log[4]*Log[4 - 2*E^x + Log[4]])/(4 + Log[4]) + (Log[19683/4]*Log[4 - 2*E^x + Log[4]])/(4 + Log[4
]) - (2*Log[27]*(x*Log[(4 - 2*E^x + Log[4])/(4 + Log[4])] + PolyLog[2, (2*E^x)/(4 + Log[4])]))/(4 + Log[4]) +
(3*Log[3]*(x*(x + 2*Log[1 - (4 + Log[4])/(2*E^x)]) - 2*PolyLog[2, (4 + Log[4])/(2*E^x)]))/(4 + Log[4]) + (6*Lo
g[3]*(x^2*Log[(4 - 2*E^x + Log[4])/(4 + Log[4])] + 2*x*PolyLog[2, (2*E^x)/(4 + Log[4])] - 2*PolyLog[3, (2*E^x)
/(4 + Log[4])]))/(4 + Log[4]) - (Log[9]*(x^2*(x + 3*Log[1 - (4 + Log[4])/(2*E^x)]) - 6*x*PolyLog[2, (4 + Log[4
])/(2*E^x)] - 6*PolyLog[3, (4 + Log[4])/(2*E^x)]))/(4 + Log[4])

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fricas [A]  time = 0.91, size = 43, normalized size = 1.59 \begin {gather*} \frac {2 \, x e^{x} + {\left (2 \, x^{3} + 9 \, x^{2} + 9 \, x\right )} \log \relax (3) - 2 \, x \log \relax (2) - 4 \, x}{2 \, {\left (e^{x} - \log \relax (2) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)^2+(-8*log(2)+(-4*x^3-6*x^2+18*x+18)*log(3)-16)*exp(x)+4*log(2)^2+2*((-6*x^2-18*x-9)*log(3)
+8)*log(2)+(-24*x^2-72*x-36)*log(3)+16)/(4*exp(x)^2+(-8*log(2)-16)*exp(x)+4*log(2)^2+16*log(2)+16),x, algorith
m="fricas")

[Out]

1/2*(2*x*e^x + (2*x^3 + 9*x^2 + 9*x)*log(3) - 2*x*log(2) - 4*x)/(e^x - log(2) - 2)

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giac [B]  time = 0.20, size = 119, normalized size = 4.41 \begin {gather*} \frac {2 \, x^{3} \log \relax (3) + 9 \, x^{2} \log \relax (3) + 2 \, x e^{x} + 9 \, x \log \relax (3) - 2 \, x \log \relax (2) + 2 \, e^{x} \log \left (e^{x} - \log \relax (2) - 2\right ) - 2 \, \log \relax (2) \log \left (e^{x} - \log \relax (2) - 2\right ) - 2 \, e^{x} \log \left (-e^{x} + \log \relax (2) + 2\right ) + 2 \, \log \relax (2) \log \left (-e^{x} + \log \relax (2) + 2\right ) - 4 \, x - 4 \, \log \left (e^{x} - \log \relax (2) - 2\right ) + 4 \, \log \left (-e^{x} + \log \relax (2) + 2\right )}{2 \, {\left (e^{x} - \log \relax (2) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)^2+(-8*log(2)+(-4*x^3-6*x^2+18*x+18)*log(3)-16)*exp(x)+4*log(2)^2+2*((-6*x^2-18*x-9)*log(3)
+8)*log(2)+(-24*x^2-72*x-36)*log(3)+16)/(4*exp(x)^2+(-8*log(2)-16)*exp(x)+4*log(2)^2+16*log(2)+16),x, algorith
m="giac")

[Out]

1/2*(2*x^3*log(3) + 9*x^2*log(3) + 2*x*e^x + 9*x*log(3) - 2*x*log(2) + 2*e^x*log(e^x - log(2) - 2) - 2*log(2)*
log(e^x - log(2) - 2) - 2*e^x*log(-e^x + log(2) + 2) + 2*log(2)*log(-e^x + log(2) + 2) - 4*x - 4*log(e^x - log
(2) - 2) + 4*log(-e^x + log(2) + 2))/(e^x - log(2) - 2)

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maple [A]  time = 0.28, size = 28, normalized size = 1.04

method	result	size
risch	\(x -\frac {\left (2 x^{2}+9 x +9\right ) \ln \relax (3) x}{2 \left (\ln \relax (2)-{\mathrm e}^{x}+2\right )}\)	\(28\)
norman	\(\frac {\left (2-\frac {9 \ln \relax (3)}{2}+\ln \relax (2)\right ) x -\frac {9 x^{2} \ln \relax (3)}{2}-x^{3} \ln \relax (3)-{\mathrm e}^{x} x}{\ln \relax (2)-{\mathrm e}^{x}+2}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(x)^2+(-8*ln(2)+(-4*x^3-6*x^2+18*x+18)*ln(3)-16)*exp(x)+4*ln(2)^2+2*((-6*x^2-18*x-9)*ln(3)+8)*ln(2)+
(-24*x^2-72*x-36)*ln(3)+16)/(4*exp(x)^2+(-8*ln(2)-16)*exp(x)+4*ln(2)^2+16*ln(2)+16),x,method=_RETURNVERBOSE)

[Out]

x-1/2*(2*x^2+9*x+9)*ln(3)*x/(ln(2)-exp(x)+2)

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maxima [B]  time = 0.89, size = 434, normalized size = 16.07 \begin {gather*} -\frac {9}{2} \, {\left (\frac {x}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {\log \left (e^{x} - \log \relax (2) - 2\right )}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {1}{{\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4}\right )} \log \relax (3) \log \relax (2) + {\left (\frac {x}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {\log \left (e^{x} - \log \relax (2) - 2\right )}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {1}{{\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4}\right )} \log \relax (2)^{2} - 9 \, {\left (\frac {x}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {\log \left (e^{x} - \log \relax (2) - 2\right )}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {1}{{\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4}\right )} \log \relax (3) + 4 \, {\left (\frac {x}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {\log \left (e^{x} - \log \relax (2) - 2\right )}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {1}{{\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4}\right )} \log \relax (2) - \frac {{\left (9 \, \log \relax (3) - 2 \, \log \relax (2) - 4\right )} \log \left (e^{x} - \log \relax (2) - 2\right )}{2 \, {\left (\log \relax (2) + 2\right )}} + \frac {2 \, {\left (\log \relax (3) \log \relax (2) + 2 \, \log \relax (3)\right )} x^{3} + 9 \, {\left (\log \relax (3) \log \relax (2) + 2 \, \log \relax (3)\right )} x^{2} + 9 \, x e^{x} \log \relax (3) - {\left (9 \, \log \relax (3) - 8\right )} \log \relax (2) + 2 \, \log \relax (2)^{2} - 18 \, \log \relax (3) + 8}{2 \, {\left ({\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4\right )}} + \frac {4 \, x}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {4 \, \log \left (e^{x} - \log \relax (2) - 2\right )}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {4}{{\left (\log \relax (2) + 2\right )} e^{x} - \log \relax (2)^{2} - 4 \, \log \relax (2) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)^2+(-8*log(2)+(-4*x^3-6*x^2+18*x+18)*log(3)-16)*exp(x)+4*log(2)^2+2*((-6*x^2-18*x-9)*log(3)
+8)*log(2)+(-24*x^2-72*x-36)*log(3)+16)/(4*exp(x)^2+(-8*log(2)-16)*exp(x)+4*log(2)^2+16*log(2)+16),x, algorith
m="maxima")

[Out]

-9/2*(x/(log(2)^2 + 4*log(2) + 4) - log(e^x - log(2) - 2)/(log(2)^2 + 4*log(2) + 4) - 1/((log(2) + 2)*e^x - lo
g(2)^2 - 4*log(2) - 4))*log(3)*log(2) + (x/(log(2)^2 + 4*log(2) + 4) - log(e^x - log(2) - 2)/(log(2)^2 + 4*log
(2) + 4) - 1/((log(2) + 2)*e^x - log(2)^2 - 4*log(2) - 4))*log(2)^2 - 9*(x/(log(2)^2 + 4*log(2) + 4) - log(e^x
- log(2) - 2)/(log(2)^2 + 4*log(2) + 4) - 1/((log(2) + 2)*e^x - log(2)^2 - 4*log(2) - 4))*log(3) + 4*(x/(log(
2)^2 + 4*log(2) + 4) - log(e^x - log(2) - 2)/(log(2)^2 + 4*log(2) + 4) - 1/((log(2) + 2)*e^x - log(2)^2 - 4*lo
g(2) - 4))*log(2) - 1/2*(9*log(3) - 2*log(2) - 4)*log(e^x - log(2) - 2)/(log(2) + 2) + 1/2*(2*(log(3)*log(2) +
2*log(3))*x^3 + 9*(log(3)*log(2) + 2*log(3))*x^2 + 9*x*e^x*log(3) - (9*log(3) - 8)*log(2) + 2*log(2)^2 - 18*l
og(3) + 8)/((log(2) + 2)*e^x - log(2)^2 - 4*log(2) - 4) + 4*x/(log(2)^2 + 4*log(2) + 4) - 4*log(e^x - log(2) -
2)/(log(2)^2 + 4*log(2) + 4) - 4/((log(2) + 2)*e^x - log(2)^2 - 4*log(2) - 4)

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mupad [B]  time = 1.27, size = 35, normalized size = 1.30 \begin {gather*} x-\frac {x\,\left (\ln \relax (9)\,x^2+9\,\ln \relax (3)\,x+2\,\ln \relax (2)+\ln \left (\frac {19683}{4}\right )\right )}{2\,\left (\ln \relax (2)-{\mathrm {e}}^x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(2*x) - log(3)*(72*x + 24*x^2 + 36) - 2*log(2)*(log(3)*(18*x + 6*x^2 + 9) - 8) + 4*log(2)^2 - exp(x)
*(8*log(2) - log(3)*(18*x - 6*x^2 - 4*x^3 + 18) + 16) + 16)/(4*exp(2*x) + 16*log(2) - exp(x)*(8*log(2) + 16) +
4*log(2)^2 + 16),x)

[Out]

x - (x*(2*log(2) + log(19683/4) + 9*x*log(3) + x^2*log(9)))/(2*(log(2) - exp(x) + 2))

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sympy [A]  time = 0.17, size = 36, normalized size = 1.33 \begin {gather*} x + \frac {2 x^{3} \log {\relax (3 )} + 9 x^{2} \log {\relax (3 )} + 9 x \log {\relax (3 )}}{2 e^{x} - 4 - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)**2+(-8*ln(2)+(-4*x**3-6*x**2+18*x+18)*ln(3)-16)*exp(x)+4*ln(2)**2+2*((-6*x**2-18*x-9)*ln(3
)+8)*ln(2)+(-24*x**2-72*x-36)*ln(3)+16)/(4*exp(x)**2+(-8*ln(2)-16)*exp(x)+4*ln(2)**2+16*ln(2)+16),x)

[Out]

x + (2*x**3*log(3) + 9*x**2*log(3) + 9*x*log(3))/(2*exp(x) - 4 - 2*log(2))

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