3.78.22 \(\int \frac {e^{e^2} (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} (12-4 x-3 x^2+x^3))+e^{e^2} (12 x-8 x^2+3 x^3+e^{-1+x} (-8 x-2 x^2+4 x^3-x^4)) \log (\frac {5}{x})}{16 x-8 x^3+x^5} \, dx\)

Optimal. Leaf size=35 \[ \frac {e^{e^2} (3-x) \left (-e^{-1+x}+x\right ) \log \left (\frac {5}{x}\right )}{4-x^2} \]

________________________________________________________________________________________

Rubi [B]  time = 6.01, antiderivative size = 97, normalized size of antiderivative = 2.77, number of steps used = 81, number of rules used = 22, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.198, Rules used = {1594, 28, 6742, 199, 207, 261, 288, 266, 43, 2323, 2324, 12, 5912, 2335, 260, 2351, 6688, 2177, 2178, 2269, 6725, 2554} \begin {gather*} -\frac {e^{e^2} x^2 \log \left (\frac {5}{x}\right )}{4-x^2}+\frac {3 e^{e^2} x \log \left (\frac {5}{x}\right )}{4-x^2}-\frac {e^{x+e^2-1} \log \left (\frac {5}{x}\right )}{4 (2-x)}-\frac {5 e^{x+e^2-1} \log \left (\frac {5}{x}\right )}{4 (x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^2*(-12*x + 4*x^2 + 3*x^3 - x^4 + E^(-1 + x)*(12 - 4*x - 3*x^2 + x^3)) + E^E^2*(12*x - 8*x^2 + 3*x^3 +
E^(-1 + x)*(-8*x - 2*x^2 + 4*x^3 - x^4))*Log[5/x])/(16*x - 8*x^3 + x^5),x]

[Out]

-1/4*(E^(-1 + E^2 + x)*Log[5/x])/(2 - x) - (5*E^(-1 + E^2 + x)*Log[5/x])/(4*(2 + x)) + (3*E^E^2*x*Log[5/x])/(4
- x^2) - (E^E^2*x^2*Log[5/x])/(4 - x^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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2269

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d +
e*x)^n), 1/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x]

Rule 2323

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(q + 1
)*(a + b*Log[c*x^n]))/(2*d*(q + 1)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[c*
x^n]), x], x] + Dist[(b*n)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &&
LtQ[q, -1]

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rule 6688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

________________________________________________________________________________________

Mathematica [A]  time = 2.23, size = 44, normalized size = 1.26 \begin {gather*} e^{-1+e^2} \left (-\frac {\left (e^x (-3+x)+e (-4+3 x)\right ) \log \left (\frac {5}{x}\right )}{-4+x^2}-e \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^2*(-12*x + 4*x^2 + 3*x^3 - x^4 + E^(-1 + x)*(12 - 4*x - 3*x^2 + x^3)) + E^E^2*(12*x - 8*x^2 + 3
*x^3 + E^(-1 + x)*(-8*x - 2*x^2 + 4*x^3 - x^4))*Log[5/x])/(16*x - 8*x^3 + x^5),x]

[Out]

E^(-1 + E^2)*(-(((E^x*(-3 + x) + E*(-4 + 3*x))*Log[5/x])/(-4 + x^2)) - E*Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 33, normalized size = 0.94 \begin {gather*} \frac {{\left (x^{2} - {\left (x - 3\right )} e^{\left (x - 1\right )} - 3 \, x\right )} e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right )}{x^{2} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(x-1)+3*x^3-8*x^2+12*x)*exp(exp(2))*log(5/x)+((x^3-3*x^2-4*x+12)*exp(x-1
)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2)))/(x^5-8*x^3+16*x),x, algorithm="fricas")

[Out]

(x^2 - (x - 3)*e^(x - 1) - 3*x)*e^(e^2)*log(5/x)/(x^2 - 4)

________________________________________________________________________________________

giac [B]  time = 0.21, size = 77, normalized size = 2.20 \begin {gather*} -\frac {x^{2} e^{\left (e^{2}\right )} \log \relax (x) + x e^{\left (x + e^{2} - 1\right )} \log \left (\frac {5}{x}\right ) + 3 \, x e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right ) - 4 \, e^{\left (e^{2}\right )} \log \relax (x) - 3 \, e^{\left (x + e^{2} - 1\right )} \log \left (\frac {5}{x}\right ) - 4 \, e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right )}{x^{2} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(x-1)+3*x^3-8*x^2+12*x)*exp(exp(2))*log(5/x)+((x^3-3*x^2-4*x+12)*exp(x-1
)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2)))/(x^5-8*x^3+16*x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.39, size = 82, normalized size = 2.34

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{2}} \left (x \,{\mathrm e}^{x -1}+3 x -3 \,{\mathrm e}^{x -1}-4\right ) \ln \relax (x )}{x^{2}-4}-\frac {{\mathrm e}^{{\mathrm e}^{2}} \left (2 x^{2} \ln \relax (x )+2 \ln \relax (5) x \,{\mathrm e}^{x -1}+6 x \ln \relax (5)-6 \ln \relax (5) {\mathrm e}^{x -1}-8 \ln \relax (x )-8 \ln \relax (5)\right )}{2 \left (x^{2}-4\right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^4+4*x^3-2*x^2-8*x)*exp(x-1)+3*x^3-8*x^2+12*x)*exp(exp(2))*ln(5/x)+((x^3-3*x^2-4*x+12)*exp(x-1)-x^4+3
*x^3+4*x^2-12*x)*exp(exp(2)))/(x^5-8*x^3+16*x),x,method=_RETURNVERBOSE)

[Out]

exp(exp(2))*(x*exp(x-1)+3*x-3*exp(x-1)-4)/(x^2-4)*ln(x)-1/2*exp(exp(2))*(2*x^2*ln(x)+2*ln(5)*x*exp(x-1)+6*x*ln
(5)-6*ln(5)*exp(x-1)-8*ln(x)-8*ln(5))/(x^2-4)

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 140, normalized size = 4.00 \begin {gather*} \frac {3}{8} \, {\left (\frac {4 \, x}{x^{2} - 4} - \log \left (x + 2\right ) + \log \left (x - 2\right )\right )} e^{\left (e^{2}\right )} + \frac {3}{8} \, e^{\left (e^{2}\right )} \log \left (x + 2\right ) - \frac {3}{8} \, e^{\left (e^{2}\right )} \log \left (x - 2\right ) - \frac {3 \, x {\left (2 \, \log \relax (5) + 1\right )} e^{\left (e^{2} + 1\right )} + 2 \, {\left (x e^{\left (e^{2}\right )} \log \relax (5) - 3 \, e^{\left (e^{2}\right )} \log \relax (5) - {\left (x e^{\left (e^{2}\right )} - 3 \, e^{\left (e^{2}\right )}\right )} \log \relax (x)\right )} e^{x} - 8 \, e^{\left (e^{2} + 1\right )} \log \relax (5) + 2 \, {\left (x^{2} e^{\left (e^{2} + 1\right )} - 3 \, x e^{\left (e^{2} + 1\right )}\right )} \log \relax (x)}{2 \, {\left (x^{2} e - 4 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(x-1)+3*x^3-8*x^2+12*x)*exp(exp(2))*log(5/x)+((x^3-3*x^2-4*x+12)*exp(x-1
)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2)))/(x^5-8*x^3+16*x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (12\,x+{\mathrm {e}}^{x-1}\,\left (-x^3+3\,x^2+4\,x-12\right )-4\,x^2-3\,x^3+x^4\right )-{\mathrm {e}}^{{\mathrm {e}}^2}\,\ln \left (\frac {5}{x}\right )\,\left (12\,x-{\mathrm {e}}^{x-1}\,\left (x^4-4\,x^3+2\,x^2+8\,x\right )-8\,x^2+3\,x^3\right )}{x^5-8\,x^3+16\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(2))*(12*x + exp(x - 1)*(4*x + 3*x^2 - x^3 - 12) - 4*x^2 - 3*x^3 + x^4) - exp(exp(2))*log(5/x)*(1
2*x - exp(x - 1)*(8*x + 2*x^2 - 4*x^3 + x^4) - 8*x^2 + 3*x^3))/(16*x - 8*x^3 + x^5),x)

[Out]

int(-(exp(exp(2))*(12*x + exp(x - 1)*(4*x + 3*x^2 - x^3 - 12) - 4*x^2 - 3*x^3 + x^4) - exp(exp(2))*log(5/x)*(1
2*x - exp(x - 1)*(8*x + 2*x^2 - 4*x^3 + x^4) - 8*x^2 + 3*x^3))/(16*x - 8*x^3 + x^5), x)

________________________________________________________________________________________

sympy [B]  time = 0.56, size = 66, normalized size = 1.89 \begin {gather*} - e^{e^{2}} \log {\relax (x )} + \frac {\left (- 3 x e^{e^{2}} + 4 e^{e^{2}}\right ) \log {\left (\frac {5}{x} \right )}}{x^{2} - 4} + \frac {\left (- x e^{e^{2}} \log {\left (\frac {5}{x} \right )} + 3 e^{e^{2}} \log {\left (\frac {5}{x} \right )}\right ) e^{x - 1}}{x^{2} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**4+4*x**3-2*x**2-8*x)*exp(x-1)+3*x**3-8*x**2+12*x)*exp(exp(2))*ln(5/x)+((x**3-3*x**2-4*x+12)*e
xp(x-1)-x**4+3*x**3+4*x**2-12*x)*exp(exp(2)))/(x**5-8*x**3+16*x),x)

[Out]

-exp(exp(2))*log(x) + (-3*x*exp(exp(2)) + 4*exp(exp(2)))*log(5/x)/(x**2 - 4) + (-x*exp(exp(2))*log(5/x) + 3*ex
p(exp(2))*log(5/x))*exp(x - 1)/(x**2 - 4)

________________________________________________________________________________________