3.5.20 \(\int \frac {392+112 x+6 x^2+e^{e^{e^3}} (16 x+24 e^x x+12 e^{2 x} x+2 e^{3 x} x)+(112+16 x) \log (5)+8 \log ^2(5)+e^{3 x} (49+28 x+3 x^2+(14+4 x) \log (5)+\log ^2(5))+e^{2 x} (294+140 x+26 x^2+2 x^3+(84+20 x+2 x^2) \log (5)+6 \log ^2(5))+e^x (588+224 x+43 x^2+2 x^3+(168+32 x+4 x^2) \log (5)+12 \log ^2(5))}{8+12 e^x+6 e^{2 x}+e^{3 x}} \, dx\)

Optimal. Leaf size=32 \[ x \left (e^{e^{e^3}} x+\left (-7-x+\frac {x}{2+e^x}-\log (5)\right )^2\right ) \]

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Rubi [C]  time = 3.74, antiderivative size = 381, normalized size of antiderivative = 11.91, number of steps used = 72, number of rules used = 12, integrand size = 186, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6741, 6742, 2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2279, 2391} \begin {gather*} -\frac {9}{2} x \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {3}{2} \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {9}{2} \text {Li}_3\left (-\frac {e^x}{2}\right )+\frac {1}{2} x (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (7+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_3\left (-\frac {e^x}{2}\right )+2 (4+\log (5)) \text {Li}_3\left (-\frac {e^x}{2}\right )-\frac {2 x^3}{e^x+2}+\frac {x^3}{\left (e^x+2\right )^2}+\frac {7 x^3}{4}-\frac {1}{12} x^3 (25+\log (625))+\frac {1}{3} x^3 (4+\log (5))-\frac {3 x^2}{2 \left (e^x+2\right )}+\frac {3 x^2}{4}+\frac {1}{4} x^2 (25+\log (625)) \log \left (\frac {e^x}{2}+1\right )-x^2 (4+\log (5)) \log \left (\frac {e^x}{2}+1\right )-\frac {9}{4} x^2 \log \left (\frac {e^x}{2}+1\right )-\frac {x^2 (25+\log (625))}{2 \left (e^x+2\right )}+\frac {1}{4} x^2 (25+\log (625))+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )-x^2 (7+\log (5))-\frac {1}{2} x (25+\log (625)) \log \left (\frac {e^x}{2}+1\right )+2 x (7+\log (5)) \log \left (\frac {e^x}{2}+1\right )-\frac {3}{2} x \log \left (\frac {e^x}{2}+1\right )+x (7+\log (5))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(392 + 112*x + 6*x^2 + E^E^E^3*(16*x + 24*E^x*x + 12*E^(2*x)*x + 2*E^(3*x)*x) + (112 + 16*x)*Log[5] + 8*Lo
g[5]^2 + E^(3*x)*(49 + 28*x + 3*x^2 + (14 + 4*x)*Log[5] + Log[5]^2) + E^(2*x)*(294 + 140*x + 26*x^2 + 2*x^3 +
(84 + 20*x + 2*x^2)*Log[5] + 6*Log[5]^2) + E^x*(588 + 224*x + 43*x^2 + 2*x^3 + (168 + 32*x + 4*x^2)*Log[5] + 1
2*Log[5]^2))/(8 + 12*E^x + 6*E^(2*x) + E^(3*x)),x]

[Out]

(3*x^2)/4 - (3*x^2)/(2*(2 + E^x)) + (7*x^3)/4 + x^3/(2 + E^x)^2 - (2*x^3)/(2 + E^x) + (x^3*(4 + Log[5]))/3 - x
^2*(7 + Log[5]) + x*(7 + Log[5])^2 + x^2*(14 + E^E^E^3 + Log[25]) + (x^2*(25 + Log[625]))/4 - (x^2*(25 + Log[6
25]))/(2*(2 + E^x)) - (x^3*(25 + Log[625]))/12 - (3*x*Log[1 + E^x/2])/2 - (9*x^2*Log[1 + E^x/2])/4 - x^2*(4 +
Log[5])*Log[1 + E^x/2] + 2*x*(7 + Log[5])*Log[1 + E^x/2] - (x*(25 + Log[625])*Log[1 + E^x/2])/2 + (x^2*(25 + L
og[625])*Log[1 + E^x/2])/4 - (3*PolyLog[2, -1/2*E^x])/2 - (9*x*PolyLog[2, -1/2*E^x])/2 - 2*x*(4 + Log[5])*Poly
Log[2, -1/2*E^x] + 2*(7 + Log[5])*PolyLog[2, -1/2*E^x] - ((25 + Log[625])*PolyLog[2, -1/2*E^x])/2 + (x*(25 + L
og[625])*PolyLog[2, -1/2*E^x])/2 + (9*PolyLog[3, -1/2*E^x])/2 + 2*(4 + Log[5])*PolyLog[3, -1/2*E^x] - ((25 + L
og[625])*PolyLog[3, -1/2*E^x])/2

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 {112 x+6 x^2+e^{e^{e^3}} \left (16 x+24 e^x x+12 e^{2 x} x+2 e^{3 x} x\right )+(112+16 x) \log (5)+392 \left (1+\frac {\log ^2(5)}{49}\right )+e^{3 x} \left (49+28 x+3 x^2+(14+4 x) \log (5)+\log ^2(5)\right )+e^{2 x} \left (294+140 x+26 x^2+2 x^3+\left (84+20 x+2 x^2\right ) \log (5)+6 \log ^2(5)\right )+e^x \left (588+224 x+43 x^2+2 x^3+\left (168+32 x+4 x^2\right ) \log (5)+12 \log ^2(5)\right )}{\left (2+e^x\right )^3} \, dx\\ &=\int \left (3 x^2+\frac {4 x^3}{\left (2+e^x\right )^3}+\frac {2 x \left (x^2+x (4+\log (5))-2 (7+\log (5))\right )}{2+e^x}+49 \left (1+\frac {1}{49} \log (5) (14+\log (5))\right )+28 x \left (1+\frac {1}{14} \left (e^{e^{e^3}}+\log (25)\right )\right )-\frac {x^2 (25+6 x+\log (625))}{\left (2+e^x\right )^2}\right ) \, dx\\ &=x^3+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+2 \int \frac {x \left (x^2+x (4+\log (5))-2 (7+\log (5))\right )}{2+e^x} \, dx+4 \int \frac {x^3}{\left (2+e^x\right )^3} \, dx-\int \frac {x^2 (25+6 x+\log (625))}{\left (2+e^x\right )^2} \, dx\\ &=x^3+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )-2 \int \frac {e^x x^3}{\left (2+e^x\right )^3} \, dx+2 \int \frac {x^3}{\left (2+e^x\right )^2} \, dx+2 \int \left (\frac {x^3}{2+e^x}+\frac {x^2 (4+\log (5))}{2+e^x}-\frac {2 x (7+\log (5))}{2+e^x}\right ) \, dx-\int \left (\frac {6 x^3}{\left (2+e^x\right )^2}+\frac {x^2 (25+\log (625))}{\left (2+e^x\right )^2}\right ) \, dx\\ &=x^3+\frac {x^3}{\left (2+e^x\right )^2}+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+2 \int \frac {x^3}{2+e^x} \, dx-3 \int \frac {x^2}{\left (2+e^x\right )^2} \, dx-6 \int \frac {x^3}{\left (2+e^x\right )^2} \, dx+(2 (4+\log (5))) \int \frac {x^2}{2+e^x} \, dx-(4 (7+\log (5))) \int \frac {x}{2+e^x} \, dx-(25+\log (625)) \int \frac {x^2}{\left (2+e^x\right )^2} \, dx-\int \frac {e^x x^3}{\left (2+e^x\right )^2} \, dx+\int \frac {x^3}{2+e^x} \, dx\\ &=x^3+\frac {x^3}{\left (2+e^x\right )^2}+\frac {x^3}{2+e^x}+\frac {3 x^4}{8}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )-\frac {1}{2} \int \frac {e^x x^3}{2+e^x} \, dx+\frac {3}{2} \int \frac {e^x x^2}{\left (2+e^x\right )^2} \, dx-\frac {3}{2} \int \frac {x^2}{2+e^x} \, dx-3 \int \frac {x^2}{2+e^x} \, dx+3 \int \frac {e^x x^3}{\left (2+e^x\right )^2} \, dx-3 \int \frac {x^3}{2+e^x} \, dx+(-4-\log (5)) \int \frac {e^x x^2}{2+e^x} \, dx+(2 (7+\log (5))) \int \frac {e^x x}{2+e^x} \, dx-\frac {1}{2} (-25-\log (625)) \int \frac {e^x x^2}{\left (2+e^x\right )^2} \, dx-\frac {1}{2} (25+\log (625)) \int \frac {x^2}{2+e^x} \, dx-\int \frac {e^x x^3}{2+e^x} \, dx\\ &=-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))-\frac {3}{2} x^3 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )+\frac {3}{4} \int \frac {e^x x^2}{2+e^x} \, dx+\frac {3}{2} \int \frac {e^x x^2}{2+e^x} \, dx+\frac {3}{2} \int \frac {e^x x^3}{2+e^x} \, dx+\frac {3}{2} \int x^2 \log \left (1+\frac {e^x}{2}\right ) \, dx+3 \int \frac {x}{2+e^x} \, dx+3 \int x^2 \log \left (1+\frac {e^x}{2}\right ) \, dx+9 \int \frac {x^2}{2+e^x} \, dx+(2 (4+\log (5))) \int x \log \left (1+\frac {e^x}{2}\right ) \, dx-(2 (7+\log (5))) \int \log \left (1+\frac {e^x}{2}\right ) \, dx-\frac {1}{4} (-25-\log (625)) \int \frac {e^x x^2}{2+e^x} \, dx-(-25-\log (625)) \int \frac {x}{2+e^x} \, dx\\ &=\frac {3 x^2}{4}-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {7 x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+\frac {1}{4} x^2 (25+\log (625))-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))+\frac {9}{4} x^2 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )+\frac {1}{4} x^2 (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )-\frac {9}{2} x^2 \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {3}{2} \int \frac {e^x x}{2+e^x} \, dx-\frac {3}{2} \int x \log \left (1+\frac {e^x}{2}\right ) \, dx-3 \int x \log \left (1+\frac {e^x}{2}\right ) \, dx+3 \int x \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx-\frac {9}{2} \int \frac {e^x x^2}{2+e^x} \, dx-\frac {9}{2} \int x^2 \log \left (1+\frac {e^x}{2}\right ) \, dx+6 \int x \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx+(2 (4+\log (5))) \int \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx-(2 (7+\log (5))) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-\frac {1}{2} (25+\log (625)) \int \frac {e^x x}{2+e^x} \, dx-\frac {1}{2} (25+\log (625)) \int x \log \left (1+\frac {e^x}{2}\right ) \, dx\\ &=\frac {3 x^2}{4}-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {7 x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+\frac {1}{4} x^2 (25+\log (625))-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))-\frac {3}{2} x \log \left (1+\frac {e^x}{2}\right )-\frac {9}{4} x^2 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )-\frac {1}{2} x (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )+\frac {1}{4} x^2 (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )+\frac {9}{2} x \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (7+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {1}{2} x (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+9 x \text {Li}_3\left (-\frac {e^x}{2}\right )+\frac {3}{2} \int \log \left (1+\frac {e^x}{2}\right ) \, dx-\frac {3}{2} \int \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx-3 \int \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx-3 \int \text {Li}_3\left (-\frac {e^x}{2}\right ) \, dx-6 \int \text {Li}_3\left (-\frac {e^x}{2}\right ) \, dx+9 \int x \log \left (1+\frac {e^x}{2}\right ) \, dx-9 \int x \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx+(2 (4+\log (5))) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-\frac {1}{2} (-25-\log (625)) \int \log \left (1+\frac {e^x}{2}\right ) \, dx-\frac {1}{2} (25+\log (625)) \int \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx\\ &=\frac {3 x^2}{4}-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {7 x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+\frac {1}{4} x^2 (25+\log (625))-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))-\frac {3}{2} x \log \left (1+\frac {e^x}{2}\right )-\frac {9}{4} x^2 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )-\frac {1}{2} x (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )+\frac {1}{4} x^2 (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )-\frac {9}{2} x \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (7+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {1}{2} x (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (4+\log (5)) \text {Li}_3\left (-\frac {e^x}{2}\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-\frac {3}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-3 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-3 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )+9 \int \text {Li}_2\left (-\frac {e^x}{2}\right ) \, dx+9 \int \text {Li}_3\left (-\frac {e^x}{2}\right ) \, dx-\frac {1}{2} (-25-\log (625)) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx,x,e^x\right )-\frac {1}{2} (25+\log (625)) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {3 x^2}{4}-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {7 x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+\frac {1}{4} x^2 (25+\log (625))-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))-\frac {3}{2} x \log \left (1+\frac {e^x}{2}\right )-\frac {9}{4} x^2 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )-\frac {1}{2} x (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )+\frac {1}{4} x^2 (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )-\frac {3}{2} \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {9}{2} x \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (7+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {1}{2} x (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {9}{2} \text {Li}_3\left (-\frac {e^x}{2}\right )+2 (4+\log (5)) \text {Li}_3\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_3\left (-\frac {e^x}{2}\right )-9 \text {Li}_4\left (-\frac {e^x}{2}\right )+9 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )+9 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{2}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {3 x^2}{4}-\frac {3 x^2}{2 \left (2+e^x\right )}+\frac {7 x^3}{4}+\frac {x^3}{\left (2+e^x\right )^2}-\frac {2 x^3}{2+e^x}+\frac {1}{3} x^3 (4+\log (5))-x^2 (7+\log (5))+x (7+\log (5))^2+x^2 \left (14+e^{e^{e^3}}+\log (25)\right )+\frac {1}{4} x^2 (25+\log (625))-\frac {x^2 (25+\log (625))}{2 \left (2+e^x\right )}-\frac {1}{12} x^3 (25+\log (625))-\frac {3}{2} x \log \left (1+\frac {e^x}{2}\right )-\frac {9}{4} x^2 \log \left (1+\frac {e^x}{2}\right )-x^2 (4+\log (5)) \log \left (1+\frac {e^x}{2}\right )+2 x (7+\log (5)) \log \left (1+\frac {e^x}{2}\right )-\frac {1}{2} x (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )+\frac {1}{4} x^2 (25+\log (625)) \log \left (1+\frac {e^x}{2}\right )-\frac {3}{2} \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {9}{2} x \text {Li}_2\left (-\frac {e^x}{2}\right )-2 x (4+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )+2 (7+\log (5)) \text {Li}_2\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {1}{2} x (25+\log (625)) \text {Li}_2\left (-\frac {e^x}{2}\right )+\frac {9}{2} \text {Li}_3\left (-\frac {e^x}{2}\right )+2 (4+\log (5)) \text {Li}_3\left (-\frac {e^x}{2}\right )-\frac {1}{2} (25+\log (625)) \text {Li}_3\left (-\frac {e^x}{2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 51, normalized size = 1.59 \begin {gather*} x \left (x^2+\frac {x^2}{\left (2+e^x\right )^2}+(7+\log (5))^2-\frac {2 x (7+x+\log (5))}{2+e^x}+x \left (14+e^{e^{e^3}}+\log (25)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(392 + 112*x + 6*x^2 + E^E^E^3*(16*x + 24*E^x*x + 12*E^(2*x)*x + 2*E^(3*x)*x) + (112 + 16*x)*Log[5]
+ 8*Log[5]^2 + E^(3*x)*(49 + 28*x + 3*x^2 + (14 + 4*x)*Log[5] + Log[5]^2) + E^(2*x)*(294 + 140*x + 26*x^2 + 2*
x^3 + (84 + 20*x + 2*x^2)*Log[5] + 6*Log[5]^2) + E^x*(588 + 224*x + 43*x^2 + 2*x^3 + (168 + 32*x + 4*x^2)*Log[
5] + 12*Log[5]^2))/(8 + 12*E^x + 6*E^(2*x) + E^(3*x)),x]

[Out]

x*(x^2 + x^2/(2 + E^x)^2 + (7 + Log[5])^2 - (2*x*(7 + x + Log[5]))/(2 + E^x) + x*(14 + E^E^E^3 + Log[25]))

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fricas [B]  time = 0.67, size = 138, normalized size = 4.31 \begin {gather*} \frac {x^{3} + 4 \, x \log \relax (5)^{2} + 28 \, x^{2} + {\left (x^{3} + x \log \relax (5)^{2} + 14 \, x^{2} + 2 \, {\left (x^{2} + 7 \, x\right )} \log \relax (5) + 49 \, x\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{3} + 2 \, x \log \relax (5)^{2} + 21 \, x^{2} + {\left (3 \, x^{2} + 28 \, x\right )} \log \relax (5) + 98 \, x\right )} e^{x} + {\left (x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} e^{x} + 4 \, x^{2}\right )} e^{\left (e^{\left (e^{3}\right )}\right )} + 4 \, {\left (x^{2} + 14 \, x\right )} \log \relax (5) + 196 \, x}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^3+12*x*exp(x)^2+24*exp(x)*x+16*x)*exp(exp(exp(3)))+(log(5)^2+(4*x+14)*log(5)+3*x^2+28*x
+49)*exp(x)^3+(6*log(5)^2+(2*x^2+20*x+84)*log(5)+2*x^3+26*x^2+140*x+294)*exp(x)^2+(12*log(5)^2+(4*x^2+32*x+168
)*log(5)+2*x^3+43*x^2+224*x+588)*exp(x)+8*log(5)^2+(16*x+112)*log(5)+6*x^2+112*x+392)/(exp(x)^3+6*exp(x)^2+12*
exp(x)+8),x, algorithm="fricas")

[Out]

(x^3 + 4*x*log(5)^2 + 28*x^2 + (x^3 + x*log(5)^2 + 14*x^2 + 2*(x^2 + 7*x)*log(5) + 49*x)*e^(2*x) + 2*(x^3 + 2*
x*log(5)^2 + 21*x^2 + (3*x^2 + 28*x)*log(5) + 98*x)*e^x + (x^2*e^(2*x) + 4*x^2*e^x + 4*x^2)*e^(e^(e^3)) + 4*(x
^2 + 14*x)*log(5) + 196*x)/(e^(2*x) + 4*e^x + 4)

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giac [B]  time = 0.51, size = 174, normalized size = 5.44 \begin {gather*} \frac {x^{3} e^{\left (2 \, x\right )} + 2 \, x^{3} e^{x} + 2 \, x^{2} e^{\left (2 \, x\right )} \log \relax (5) + 6 \, x^{2} e^{x} \log \relax (5) + x e^{\left (2 \, x\right )} \log \relax (5)^{2} + 4 \, x e^{x} \log \relax (5)^{2} + x^{3} + 14 \, x^{2} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x + e^{\left (e^{3}\right )}\right )} + 4 \, x^{2} e^{\left (x + e^{\left (e^{3}\right )}\right )} + 42 \, x^{2} e^{x} + 4 \, x^{2} e^{\left (e^{\left (e^{3}\right )}\right )} + 4 \, x^{2} \log \relax (5) + 14 \, x e^{\left (2 \, x\right )} \log \relax (5) + 56 \, x e^{x} \log \relax (5) + 4 \, x \log \relax (5)^{2} + 28 \, x^{2} + 49 \, x e^{\left (2 \, x\right )} + 196 \, x e^{x} + 56 \, x \log \relax (5) + 196 \, x}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^3+12*x*exp(x)^2+24*exp(x)*x+16*x)*exp(exp(exp(3)))+(log(5)^2+(4*x+14)*log(5)+3*x^2+28*x
+49)*exp(x)^3+(6*log(5)^2+(2*x^2+20*x+84)*log(5)+2*x^3+26*x^2+140*x+294)*exp(x)^2+(12*log(5)^2+(4*x^2+32*x+168
)*log(5)+2*x^3+43*x^2+224*x+588)*exp(x)+8*log(5)^2+(16*x+112)*log(5)+6*x^2+112*x+392)/(exp(x)^3+6*exp(x)^2+12*
exp(x)+8),x, algorithm="giac")

[Out]

(x^3*e^(2*x) + 2*x^3*e^x + 2*x^2*e^(2*x)*log(5) + 6*x^2*e^x*log(5) + x*e^(2*x)*log(5)^2 + 4*x*e^x*log(5)^2 + x
^3 + 14*x^2*e^(2*x) + x^2*e^(2*x + e^(e^3)) + 4*x^2*e^(x + e^(e^3)) + 42*x^2*e^x + 4*x^2*e^(e^(e^3)) + 4*x^2*l
og(5) + 14*x*e^(2*x)*log(5) + 56*x*e^x*log(5) + 4*x*log(5)^2 + 28*x^2 + 49*x*e^(2*x) + 196*x*e^x + 56*x*log(5)
+ 196*x)/(e^(2*x) + 4*e^x + 4)

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maple [B]  time = 0.38, size = 74, normalized size = 2.31

method	result	size
risch	\(x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+x \ln \relax (5)^{2}+2 x^{2} \ln \relax (5)+x^{3}+14 x \ln \relax (5)+14 x^{2}+49 x -\frac {x^{2} \left (2 \,{\mathrm e}^{x} \ln \relax (5)+2 \,{\mathrm e}^{x} x +4 \ln \relax (5)+3 x +14 \,{\mathrm e}^{x}+28\right )}{\left ({\mathrm e}^{x}+2\right )^{2}}\)	\(74\)
norman	\(\frac {x^{3}+\left (28+4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+4 \ln \relax (5)\right ) x^{2}+\left (4 \ln \relax (5)^{2}+56 \ln \relax (5)+196\right ) x +{\mathrm e}^{2 x} x^{3}+\left (14+{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+2 \ln \relax (5)\right ) x^{2} {\mathrm e}^{2 x}+\left (42+4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+6 \ln \relax (5)\right ) x^{2} {\mathrm e}^{x}+\left (49+\ln \relax (5)^{2}+14 \ln \relax (5)\right ) x \,{\mathrm e}^{2 x}+\left (4 \ln \relax (5)^{2}+56 \ln \relax (5)+196\right ) x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{3}}{\left ({\mathrm e}^{x}+2\right )^{2}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x)^3+12*x*exp(x)^2+24*exp(x)*x+16*x)*exp(exp(exp(3)))+(ln(5)^2+(4*x+14)*ln(5)+3*x^2+28*x+49)*exp
(x)^3+(6*ln(5)^2+(2*x^2+20*x+84)*ln(5)+2*x^3+26*x^2+140*x+294)*exp(x)^2+(12*ln(5)^2+(4*x^2+32*x+168)*ln(5)+2*x
^3+43*x^2+224*x+588)*exp(x)+8*ln(5)^2+(16*x+112)*ln(5)+6*x^2+112*x+392)/(exp(x)^3+6*exp(x)^2+12*exp(x)+8),x,me
thod=_RETURNVERBOSE)

[Out]

x^2*exp(exp(exp(3)))+x*ln(5)^2+2*x^2*ln(5)+x^3+14*x*ln(5)+14*x^2+49*x-x^2*(2*exp(x)*ln(5)+2*exp(x)*x+4*ln(5)+3
*x+14*exp(x)+28)/(exp(x)+2)^2

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maxima [B]  time = 0.81, size = 207, normalized size = 6.47 \begin {gather*} {\left (x + \frac {2 \, {\left (e^{x} + 3\right )}}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} - \log \left (e^{x} + 2\right )\right )} \log \relax (5)^{2} + 14 \, {\left (x + \frac {2 \, {\left (e^{x} + 3\right )}}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} - \log \left (e^{x} + 2\right )\right )} \log \relax (5) + {\left (\log \relax (5)^{2} + 14 \, \log \relax (5) + 49\right )} \log \left (e^{x} + 2\right ) + 49 \, x + \frac {x^{3} + 4 \, x^{2} {\left (e^{\left (e^{\left (e^{3}\right )}\right )} + \log \relax (5) + 7\right )} + {\left (x^{3} + x^{2} {\left (e^{\left (e^{\left (e^{3}\right )}\right )} + 2 \, \log \relax (5) + 14\right )}\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{3} + x^{2} {\left (2 \, e^{\left (e^{\left (e^{3}\right )}\right )} + 3 \, \log \relax (5) + 21\right )} - \log \relax (5)^{2} - 14 \, \log \relax (5) - 49\right )} e^{x} - 6 \, \log \relax (5)^{2} - 84 \, \log \relax (5) - 294}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} + \frac {98 \, {\left (e^{x} + 3\right )}}{e^{\left (2 \, x\right )} + 4 \, e^{x} + 4} - 49 \, \log \left (e^{x} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)^3+12*x*exp(x)^2+24*exp(x)*x+16*x)*exp(exp(exp(3)))+(log(5)^2+(4*x+14)*log(5)+3*x^2+28*x
+49)*exp(x)^3+(6*log(5)^2+(2*x^2+20*x+84)*log(5)+2*x^3+26*x^2+140*x+294)*exp(x)^2+(12*log(5)^2+(4*x^2+32*x+168
)*log(5)+2*x^3+43*x^2+224*x+588)*exp(x)+8*log(5)^2+(16*x+112)*log(5)+6*x^2+112*x+392)/(exp(x)^3+6*exp(x)^2+12*
exp(x)+8),x, algorithm="maxima")

[Out]

(x + 2*(e^x + 3)/(e^(2*x) + 4*e^x + 4) - log(e^x + 2))*log(5)^2 + 14*(x + 2*(e^x + 3)/(e^(2*x) + 4*e^x + 4) -
log(e^x + 2))*log(5) + (log(5)^2 + 14*log(5) + 49)*log(e^x + 2) + 49*x + (x^3 + 4*x^2*(e^(e^(e^3)) + log(5) +
7) + (x^3 + x^2*(e^(e^(e^3)) + 2*log(5) + 14))*e^(2*x) + 2*(x^3 + x^2*(2*e^(e^(e^3)) + 3*log(5) + 21) - log(5)
^2 - 14*log(5) - 49)*e^x - 6*log(5)^2 - 84*log(5) - 294)/(e^(2*x) + 4*e^x + 4) + 98*(e^x + 3)/(e^(2*x) + 4*e^x
+ 4) - 49*log(e^x + 2)

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mupad [B]  time = 0.57, size = 126, normalized size = 3.94 \begin {gather*} \frac {2\,x^3\,{\mathrm {e}}^x+x\,\left (56\,\ln \relax (5)+4\,{\ln \relax (5)}^2+196\right )+x^3\,{\mathrm {e}}^{2\,x}+x^2\,\left (4\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}}+\ln \left (625\right )+28\right )+x^3+x\,{\mathrm {e}}^{2\,x}\,\left (14\,\ln \relax (5)+{\ln \relax (5)}^2+49\right )+x\,{\mathrm {e}}^x\,\left (56\,\ln \relax (5)+4\,{\ln \relax (5)}^2+196\right )+x^2\,{\mathrm {e}}^{2\,x}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}}+\ln \left (25\right )+14\right )+x^2\,{\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}}+6\,\ln \relax (5)+42\right )}{{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((112*x + log(5)*(16*x + 112) + exp(x)*(224*x + log(5)*(32*x + 4*x^2 + 168) + 12*log(5)^2 + 43*x^2 + 2*x^3
+ 588) + exp(exp(exp(3)))*(16*x + 12*x*exp(2*x) + 2*x*exp(3*x) + 24*x*exp(x)) + exp(2*x)*(140*x + log(5)*(20*x
+ 2*x^2 + 84) + 6*log(5)^2 + 26*x^2 + 2*x^3 + 294) + 8*log(5)^2 + 6*x^2 + exp(3*x)*(28*x + log(5)*(4*x + 14)
+ log(5)^2 + 3*x^2 + 49) + 392)/(6*exp(2*x) + exp(3*x) + 12*exp(x) + 8),x)

[Out]

(2*x^3*exp(x) + x*(56*log(5) + 4*log(5)^2 + 196) + x^3*exp(2*x) + x^2*(4*exp(exp(exp(3))) + log(625) + 28) + x
^3 + x*exp(2*x)*(14*log(5) + log(5)^2 + 49) + x*exp(x)*(56*log(5) + 4*log(5)^2 + 196) + x^2*exp(2*x)*(exp(exp(
exp(3))) + log(25) + 14) + x^2*exp(x)*(4*exp(exp(exp(3))) + 6*log(5) + 42))/(exp(2*x) + 4*exp(x) + 4)

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sympy [B]  time = 0.22, size = 87, normalized size = 2.72 \begin {gather*} x^{3} + x^{2} \left (2 \log {\relax (5 )} + 14 + e^{e^{e^{3}}}\right ) + x \left (\log {\relax (5 )}^{2} + 14 \log {\relax (5 )} + 49\right ) + \frac {- 3 x^{3} - 28 x^{2} - 4 x^{2} \log {\relax (5 )} + \left (- 2 x^{3} - 14 x^{2} - 2 x^{2} \log {\relax (5 )}\right ) e^{x}}{e^{2 x} + 4 e^{x} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)**3+12*x*exp(x)**2+24*exp(x)*x+16*x)*exp(exp(exp(3)))+(ln(5)**2+(4*x+14)*ln(5)+3*x**2+28
*x+49)*exp(x)**3+(6*ln(5)**2+(2*x**2+20*x+84)*ln(5)+2*x**3+26*x**2+140*x+294)*exp(x)**2+(12*ln(5)**2+(4*x**2+3
2*x+168)*ln(5)+2*x**3+43*x**2+224*x+588)*exp(x)+8*ln(5)**2+(16*x+112)*ln(5)+6*x**2+112*x+392)/(exp(x)**3+6*exp
(x)**2+12*exp(x)+8),x)

[Out]

x**3 + x**2*(2*log(5) + 14 + exp(exp(exp(3)))) + x*(log(5)**2 + 14*log(5) + 49) + (-3*x**3 - 28*x**2 - 4*x**2*
log(5) + (-2*x**3 - 14*x**2 - 2*x**2*log(5))*exp(x))/(exp(2*x) + 4*exp(x) + 4)

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