3.95.72 \(\int \frac {32+32 e^3-32 e^{\frac {e^3}{15}}-32 x^2+(2+2 e^3-2 e^{\frac {e^3}{15}}-2 x^2) \log (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x})}{-x-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx\)

Optimal. Leaf size=28 \[ \left (16+\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )\right )^2 \]

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Rubi [C]  time = 3.02, antiderivative size = 902, normalized size of antiderivative = 32.21, number of steps used = 46, number of rules used = 16, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6, 1594, 6728, 1628, 628, 2528, 2524, 2357, 2301, 2317, 2391, 2418, 2394, 2315, 2390, 2393} \begin {gather*} -\log ^2\left (-2 \left (-x-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2\right )\right )-\log ^2\left (-2 \left (-x+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2\right )\right )-\log ^2(x)-32 \log (x)-2 \log (x) \log \left (x-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}\right )-2 \log \left (-\frac {i \left (-x+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2\right )}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 x-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )+2 \log \left (\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 x-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )+2 \log \left (x-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}\right ) \log \left (2 x-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )-2 \log \left (\frac {i \left (-x-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2\right )}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 x-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )+2 \log \left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 x-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )+2 \log \left (x-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}\right ) \log \left (2 x-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )\right )+2 \log (x) \log \left (1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \log (x) \log \left (1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+32 \log \left (x^2-4 x-e^{\frac {e^3}{15}}+e^3+1\right )-2 \text {Li}_2\left (-\frac {-i x-\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 i}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )-2 \text {Li}_2\left (\frac {-i x+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 i}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 + 32*E^3 - 32*E^(E^3/15) - 32*x^2 + (2 + 2*E^3 - 2*E^(E^3/15) - 2*x^2)*Log[(1 + E^3 - E^(E^3/15) - 4*x
+ x^2)/x])/(-x - E^3*x + E^(E^3/15)*x + 4*x^2 - x^3),x]

[Out]

-Log[-2*(2 - I*Sqrt[-3 + E^3 - E^(E^3/15)] - x)]^2 - Log[-2*(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)] - x)]^2 - 32*Lo
g[x] - Log[x]^2 - 2*Log[x]*Log[-4 + (1 + E^3 - E^(E^3/15))/x + x] - 2*Log[((-1/2*I)*(2 + I*Sqrt[-3 + E^3 - E^(
E^3/15)] - x))/Sqrt[-3 + E^3 - E^(E^3/15)]]*Log[-2*(2 - I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] + 2*Log[x/(2 - I
*Sqrt[-3 + E^3 - E^(E^3/15)])]*Log[-2*(2 - I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] + 2*Log[-4 + (1 + E^3 - E^(E^
3/15))/x + x]*Log[-2*(2 - I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] - 2*Log[((I/2)*(2 - I*Sqrt[-3 + E^3 - E^(E^3/1
5)] - x))/Sqrt[-3 + E^3 - E^(E^3/15)]]*Log[-2*(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] + 2*Log[x/(2 + I*Sqrt
[-3 + E^3 - E^(E^3/15)])]*Log[-2*(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] + 2*Log[-4 + (1 + E^3 - E^(E^3/15)
)/x + x]*Log[-2*(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)]) + 2*x] + 2*Log[x]*Log[1 - x/(2 - I*Sqrt[-3 + E^3 - E^(E^3/
15)])] + 2*Log[x]*Log[1 - x/(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)])] + 32*Log[1 + E^3 - E^(E^3/15) - 4*x + x^2] -
2*PolyLog[2, -1/2*(2*I - Sqrt[-3 + E^3 - E^(E^3/15)] - I*x)/Sqrt[-3 + E^3 - E^(E^3/15)]] - 2*PolyLog[2, (2*I +
Sqrt[-3 + E^3 - E^(E^3/15)] - I*x)/(2*Sqrt[-3 + E^3 - E^(E^3/15)])] + 2*PolyLog[2, x/(2 - I*Sqrt[-3 + E^3 - E
^(E^3/15)])] + 2*PolyLog[2, x/(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)])] + 2*PolyLog[2, 1 - x/(2 - I*Sqrt[-3 + E^3 -
E^(E^3/15)])] + 2*PolyLog[2, 1 - x/(2 + I*Sqrt[-3 + E^3 - E^(E^3/15)])]

Rule 6

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2301

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

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 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

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

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32+32 e^3-32 e^{\frac {e^3}{15}}-32 x^2+\left (2+2 e^3-2 e^{\frac {e^3}{15}}-2 x^2\right ) \log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )}{e^{\frac {e^3}{15}} x+\left (-1-e^3\right ) x+4 x^2-x^3} \, dx\\ &=\int \frac {32+32 e^3-32 e^{\frac {e^3}{15}}-32 x^2+\left (2+2 e^3-2 e^{\frac {e^3}{15}}-2 x^2\right ) \log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )}{\left (-1-e^3+e^{\frac {e^3}{15}}\right ) x+4 x^2-x^3} \, dx\\ &=\int \frac {32+32 e^3-32 e^{\frac {e^3}{15}}-32 x^2+\left (2+2 e^3-2 e^{\frac {e^3}{15}}-2 x^2\right ) \log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )}{x \left (-1-e^3+e^{\frac {e^3}{15}}+4 x-x^2\right )} \, dx\\ &=\int \left (\frac {32 \left (-1-e^3+e^{\frac {e^3}{15}}+x^2\right )}{x \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )}+\frac {2 \left (-1-e^3+e^{\frac {e^3}{15}}+x^2\right ) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{x \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )}\right ) \, dx\\ &=2 \int \frac {\left (-1-e^3+e^{\frac {e^3}{15}}+x^2\right ) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{x \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )} \, dx+32 \int \frac {-1-e^3+e^{\frac {e^3}{15}}+x^2}{x \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )} \, dx\\ &=2 \int \left (-\frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{x}+\frac {2 (-2+x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}\right ) \, dx+32 \int \left (-\frac {1}{x}+\frac {2 (-2+x)}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}\right ) \, dx\\ &=-32 \log (x)-2 \int \frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{x} \, dx+4 \int \frac {(-2+x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2} \, dx+64 \int \frac {-2+x}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2} \, dx\\ &=-32 \log (x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )+2 \int \frac {\left (1-\frac {1+e^3-e^{\frac {e^3}{15}}}{x^2}\right ) \log (x)}{-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x} \, dx+4 \int \left (\frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}+\frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}\right ) \, dx\\ &=-32 \log (x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )+2 \int \left (-\frac {\log (x)}{x}+\frac {2 (-2+x) \log (x)}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}\right ) \, dx+4 \int \frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx+4 \int \frac {\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx\\ &=-32 \log (x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )-2 \int \frac {\log (x)}{x} \, dx-2 \int \frac {\left (1-\frac {1+e^3-e^{\frac {e^3}{15}}}{x^2}\right ) \log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x} \, dx-2 \int \frac {\left (1-\frac {1+e^3-e^{\frac {e^3}{15}}}{x^2}\right ) \log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x} \, dx+4 \int \frac {(-2+x) \log (x)}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2} \, dx\\ &=-32 \log (x)-\log ^2(x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )-2 \int \left (-\frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{x}+\frac {2 (-2+x) \log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}\right ) \, dx-2 \int \left (-\frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{x}+\frac {2 (-2+x) \log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}\right ) \, dx+4 \int \left (\frac {\log (x)}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}+\frac {\log (x)}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}\right ) \, dx\\ &=-32 \log (x)-\log ^2(x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )+2 \int \frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{x} \, dx+2 \int \frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{x} \, dx+4 \int \frac {\log (x)}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx+4 \int \frac {\log (x)}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \frac {(-2+x) \log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2} \, dx-4 \int \frac {(-2+x) \log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{1+e^3-e^{\frac {e^3}{15}}-4 x+x^2} \, dx\\ &=-32 \log (x)-\log ^2(x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+2 \log \left (\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log (x) \log \left (1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \log (x) \log \left (1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )-2 \int \frac {\log \left (1+\frac {2 x}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )}{x} \, dx-2 \int \frac {\log \left (1+\frac {2 x}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )}{x} \, dx-4 \int \frac {\log \left (\frac {2 x}{4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \frac {\log \left (\frac {2 x}{4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \left (\frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}+\frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}\right ) \, dx-4 \int \left (\frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}+\frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x}\right ) \, dx\\ &=-32 \log (x)-\log ^2(x)-2 \log (x) \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )+2 \log \left (\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right ) \log \left (-2 \left (2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}\right )+2 x\right )+2 \log (x) \log \left (1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \log (x) \log \left (1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+32 \log \left (1+e^3-e^{\frac {e^3}{15}}-4 x+x^2\right )+2 \text {Li}_2\left (\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \text {Li}_2\left (1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )-4 \int \frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \frac {\log \left (-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4-2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx-4 \int \frac {\log \left (-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x\right )}{-4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+2 x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.37, size = 949, normalized size = 33.89 \begin {gather*} 2 \left (\log \left (\frac {2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}-i x}{2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log (x)+\log \left (\frac {-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+i x}{-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log (x)-\frac {\log ^2(x)}{2}-\log \left (\frac {2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}-i x}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 \left (-2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )+\log \left (\frac {2 x}{4+2 i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 \left (-2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )-\frac {1}{2} \log ^2\left (2 \left (-2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )-\log \left (\frac {-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+i x}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 \left (-2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )+\log \left (\frac {i x}{2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right ) \log \left (2 \left (-2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )-\frac {1}{2} \log ^2\left (2 \left (-2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right )-\log (x) \left (16+\log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )\right )+\log \left (2 \left (-2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right ) \left (16+\log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )\right )+\log \left (2 \left (-2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}+x\right )\right ) \left (16+\log \left (\frac {1+e^3-e^{\frac {e^3}{15}}-4 x+x^2}{x}\right )\right )-\text {Li}_2\left (\frac {2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}-i x}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+\text {Li}_2\left (\frac {2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}-i x}{2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )-\text {Li}_2\left (\frac {-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+i x}{2 \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+\text {Li}_2\left (\frac {-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}+i x}{-2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+\text {Li}_2\left (\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+\text {Li}_2\left (\frac {i x}{2 i+\sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 + 32*E^3 - 32*E^(E^3/15) - 32*x^2 + (2 + 2*E^3 - 2*E^(E^3/15) - 2*x^2)*Log[(1 + E^3 - E^(E^3/15)
- 4*x + x^2)/x])/(-x - E^3*x + E^(E^3/15)*x + 4*x^2 - x^3),x]

[Out]

2*(Log[(2*I + Sqrt[-3 + E^3 - E^(E^3/15)] - I*x)/(2*I + Sqrt[-3 + E^3 - E^(E^3/15)])]*Log[x] + Log[(-2*I + Sqr
t[-3 + E^3 - E^(E^3/15)] + I*x)/(-2*I + Sqrt[-3 + E^3 - E^(E^3/15)])]*Log[x] - Log[x]^2/2 - Log[(2*I + Sqrt[-3
+ E^3 - E^(E^3/15)] - I*x)/(2*Sqrt[-3 + E^3 - E^(E^3/15)])]*Log[2*(-2 - I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)] +
Log[(2*x)/(4 + (2*I)*Sqrt[-3 + E^3 - E^(E^3/15)])]*Log[2*(-2 - I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)] - Log[2*(-
2 - I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)]^2/2 - Log[(-2*I + Sqrt[-3 + E^3 - E^(E^3/15)] + I*x)/(2*Sqrt[-3 + E^3
- E^(E^3/15)])]*Log[2*(-2 + I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)] + Log[(I*x)/(2*I + Sqrt[-3 + E^3 - E^(E^3/15)]
)]*Log[2*(-2 + I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)] - Log[2*(-2 + I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)]^2/2 - Log
[x]*(16 + Log[(1 + E^3 - E^(E^3/15) - 4*x + x^2)/x]) + Log[2*(-2 - I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)]*(16 + L
og[(1 + E^3 - E^(E^3/15) - 4*x + x^2)/x]) + Log[2*(-2 + I*Sqrt[-3 + E^3 - E^(E^3/15)] + x)]*(16 + Log[(1 + E^3
- E^(E^3/15) - 4*x + x^2)/x]) - PolyLog[2, (2*I + Sqrt[-3 + E^3 - E^(E^3/15)] - I*x)/(2*Sqrt[-3 + E^3 - E^(E^
3/15)])] + PolyLog[2, (2*I + Sqrt[-3 + E^3 - E^(E^3/15)] - I*x)/(2*I + Sqrt[-3 + E^3 - E^(E^3/15)])] - PolyLog
[2, (-2*I + Sqrt[-3 + E^3 - E^(E^3/15)] + I*x)/(2*Sqrt[-3 + E^3 - E^(E^3/15)])] + PolyLog[2, (-2*I + Sqrt[-3 +
E^3 - E^(E^3/15)] + I*x)/(-2*I + Sqrt[-3 + E^3 - E^(E^3/15)])] + PolyLog[2, x/(2 + I*Sqrt[-3 + E^3 - E^(E^3/1
5)])] + PolyLog[2, (I*x)/(2*I + Sqrt[-3 + E^3 - E^(E^3/15)])])

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fricas [B]  time = 0.60, size = 49, normalized size = 1.75 \begin {gather*} \log \left (\frac {x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1}{x}\right )^{2} + 32 \, \log \left (\frac {x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(
3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*exp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm="fricas")

[Out]

log((x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/x)^2 + 32*log((x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (16 \, x^{2} + {\left (x^{2} - e^{3} + e^{\left (\frac {1}{15} \, e^{3}\right )} - 1\right )} \log \left (\frac {x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1}{x}\right ) - 16 \, e^{3} + 16 \, e^{\left (\frac {1}{15} \, e^{3}\right )} - 16\right )}}{x^{3} - 4 \, x^{2} + x e^{3} - x e^{\left (\frac {1}{15} \, e^{3}\right )} + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(
3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*exp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm="giac")

[Out]

integrate(2*(16*x^2 + (x^2 - e^3 + e^(1/15*e^3) - 1)*log((x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/x) - 16*e^3 + 16
*e^(1/15*e^3) - 16)/(x^3 - 4*x^2 + x*e^3 - x*e^(1/15*e^3) + x), x)

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maple [B]  time = 0.53, size = 50, normalized size = 1.79

method	result	size
norman	\(\ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )^{2}+32 \ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )\)	\(50\)
default	error in gcdex: invalid arguments\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*ln((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(3))+32*
exp(3)-32*x^2+32)/(x*exp(1/15*exp(3))-x*exp(3)-x^3+4*x^2-x),x,method=_RETURNVERBOSE)

[Out]

ln((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)^2+32*ln((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)

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maxima [B]  time = 3.55, size = 399, normalized size = 14.25 \begin {gather*} 16 \, {\left (\frac {\log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {2 \, \log \relax (x)}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {4 \, \arctan \left (\frac {x - 2}{\sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )}{{\left (e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )} \sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )} e^{3} - 16 \, {\left (\frac {\log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {2 \, \log \relax (x)}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {4 \, \arctan \left (\frac {x - 2}{\sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )}{{\left (e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )} \sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )} e^{\left (\frac {1}{15} \, e^{3}\right )} + \log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )^{2} - 2 \, \log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right ) \log \relax (x) + \log \relax (x)^{2} + \frac {64 \, \arctan \left (\frac {x - 2}{\sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )}{\sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}} + \frac {16 \, \log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {32 \, \log \relax (x)}{e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1} - \frac {64 \, \arctan \left (\frac {x - 2}{\sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}}\right )}{{\left (e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right )} \sqrt {e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} - 3}} + 16 \, \log \left (x^{2} - 4 \, x + e^{3} - e^{\left (\frac {1}{15} \, e^{3}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+exp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(
3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*exp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm="maxima")

[Out]

16*(log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/(e^3 - e^(1/15*e^3) + 1) - 2*log(x)/(e^3 - e^(1/15*e^3) + 1) - 4*a
rctan((x - 2)/sqrt(e^3 - e^(1/15*e^3) - 3))/((e^3 - e^(1/15*e^3) + 1)*sqrt(e^3 - e^(1/15*e^3) - 3)))*e^3 - 16*
(log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/(e^3 - e^(1/15*e^3) + 1) - 2*log(x)/(e^3 - e^(1/15*e^3) + 1) - 4*arct
an((x - 2)/sqrt(e^3 - e^(1/15*e^3) - 3))/((e^3 - e^(1/15*e^3) + 1)*sqrt(e^3 - e^(1/15*e^3) - 3)))*e^(1/15*e^3)
+ log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)^2 - 2*log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)*log(x) + log(x)^2 + 6
4*arctan((x - 2)/sqrt(e^3 - e^(1/15*e^3) - 3))/sqrt(e^3 - e^(1/15*e^3) - 3) + 16*log(x^2 - 4*x + e^3 - e^(1/15
*e^3) + 1)/(e^3 - e^(1/15*e^3) + 1) - 32*log(x)/(e^3 - e^(1/15*e^3) + 1) - 64*arctan((x - 2)/sqrt(e^3 - e^(1/1
5*e^3) - 3))/((e^3 - e^(1/15*e^3) + 1)*sqrt(e^3 - e^(1/15*e^3) - 3)) + 16*log(x^2 - 4*x + e^3 - e^(1/15*e^3) +
1)

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mupad [B]  time = 33.35, size = 49, normalized size = 1.75 \begin {gather*} {\ln \left (\frac {x^2-4\,x+{\mathrm {e}}^3-{\left ({\mathrm {e}}^{{\mathrm {e}}^3}\right )}^{1/15}+1}{x}\right )}^2+32\,\ln \left (x^2-4\,x-{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{15}}+{\mathrm {e}}^3+1\right )-32\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*exp(exp(3)/15) - 32*exp(3) + 32*x^2 + log((exp(3) - exp(exp(3)/15) - 4*x + x^2 + 1)/x)*(2*exp(exp(3)/1
5) - 2*exp(3) + 2*x^2 - 2) - 32)/(x - x*exp(exp(3)/15) + x*exp(3) - 4*x^2 + x^3),x)

[Out]

32*log(exp(3) - exp(exp(3)/15) - 4*x + x^2 + 1) - 32*log(x) + log((exp(3) - 4*x - exp(exp(3))^(1/15) + x^2 + 1
)/x)^2

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sympy [B]  time = 49.07, size = 49, normalized size = 1.75 \begin {gather*} - 32 \log {\relax (x )} + \log {\left (\frac {x^{2} - 4 x - e^{\frac {e^{3}}{15}} + 1 + e^{3}}{x} \right )}^{2} + 32 \log {\left (x^{2} - 4 x - e^{\frac {e^{3}}{15}} + 1 + e^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x**2+2)*ln((-exp(1/15*exp(3))+exp(3)+x**2-4*x+1)/x)-32*exp(1/15*exp
(3))+32*exp(3)-32*x**2+32)/(x*exp(1/15*exp(3))-x*exp(3)-x**3+4*x**2-x),x)

[Out]

-32*log(x) + log((x**2 - 4*x - exp(exp(3)/15) + 1 + exp(3))/x)**2 + 32*log(x**2 - 4*x - exp(exp(3)/15) + 1 + e
xp(3))

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