Integrand size = 108, antiderivative size = 28 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx=\left (16+\log \left (-4+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}+x\right )\right )^2 \end {dmath*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.43 (sec) , antiderivative size = 949, normalized size of antiderivative = 33.89 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx =\text {Too large to display} \end {dmath*}
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]
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 + Sqrt[-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*S qrt[-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 + Log[(1 + E^3 - E^(E^3/15) - 4*x + x^2)/x]) + Log[2*(-2 + I*Sqrt[-3 + E^3 - E^(E^3/1 5)] + 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^3/15)])] + PolyLog[2, (I*x)/(2*I + Sqrt[-3 + E^3 -...
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.96 (sec) , antiderivative size = 902, normalized size of antiderivative = 32.21, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6, 6, 2026, 7279, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {-32 x^2+\left (-2 x^2-2 e^{\frac {e^3}{15}}+2 e^3+2\right ) \log \left (\frac {x^2-4 x-e^{\frac {e^3}{15}}+e^3+1}{x}\right )-32 e^{\frac {e^3}{15}}+32 e^3+32}{-x^3+4 x^2+e^{\frac {e^3}{15}} x-e^3 x-x} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {-32 x^2+\left (-2 x^2-2 e^{\frac {e^3}{15}}+2 e^3+2\right ) \log \left (\frac {x^2-4 x-e^{\frac {e^3}{15}}+e^3+1}{x}\right )-32 e^{\frac {e^3}{15}}+32 e^3+32}{-x^3+4 x^2+\left (-1-e^3\right ) x+e^{\frac {e^3}{15}} x}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {-32 x^2+\left (-2 x^2-2 e^{\frac {e^3}{15}}+2 e^3+2\right ) \log \left (\frac {x^2-4 x-e^{\frac {e^3}{15}}+e^3+1}{x}\right )-32 e^{\frac {e^3}{15}}+32 e^3+32}{-x^3+4 x^2+\left (-1-e^3+e^{\frac {e^3}{15}}\right ) x}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-32 x^2+\left (-2 x^2-2 e^{\frac {e^3}{15}}+2 e^3+2\right ) \log \left (\frac {x^2-4 x-e^{\frac {e^3}{15}}+e^3+1}{x}\right )-32 e^{\frac {e^3}{15}}+32 e^3+32}{x \left (-x^2+4 x+e^{\frac {e^3}{15}}-e^3-1\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {32 \left (x^2+e^{\frac {e^3}{15}}-e^3-1\right )}{x \left (x^2-4 x-e^{\frac {e^3}{15}}+e^3+1\right )}+\frac {2 \left (x^2+e^{\frac {e^3}{15}}-e^3-1\right ) \log \left (x+\frac {1+e^3-e^{\frac {e^3}{15}}}{x}-4\right )}{x \left (x^2-4 x-e^{\frac {e^3}{15}}+e^3+1\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,\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 \operatorname {PolyLog}\left (2,\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \operatorname {PolyLog}\left (2,\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \operatorname {PolyLog}\left (2,1-\frac {x}{2-i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )+2 \operatorname {PolyLog}\left (2,1-\frac {x}{2+i \sqrt {-3+e^3-e^{\frac {e^3}{15}}}}\right )\) |
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]
-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*Log[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^3/15)] - 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[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)])]
3.5.72.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(23)=46\).
Time = 0.49 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79
| method | result | size |
| default | \(-32 \ln \left (x \right )+32 \ln \left (-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1\right )+\ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )^{2}\) | \(50\) |
| norman | \(\ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )^{2}+32 \ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )\) | \(50\) |
| risch | \(-32 \ln \left (x \right )+32 \ln \left (-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1\right )+\ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )^{2}\) | \(50\) |
| parts | \(-32 \ln \left (x \right )+32 \ln \left (-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1\right )+\ln \left (\frac {-{\mathrm e}^{\frac {{\mathrm e}^{3}}{15}}+{\mathrm e}^{3}+x^{2}-4 x +1}{x}\right )^{2}\) | \(50\) |
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)
-32*ln(x)+32*ln(-exp(exp(3))^(1/15)+exp(3)+x^2-4*x+1)+ln((-exp(exp(3))^(1/ 15)+exp(3)+x^2-4*x+1)/x)^2
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx=\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 {dmath*}
integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+e xp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*e xp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm=\
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)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 3.88 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx=- 32 \log {\left (x \right )} + \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 {dmath*}
integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x**2+2)*ln((-exp(1/15*exp(3))+e xp(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)
-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 + exp(3))
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (23) = 46\).
Time = 2.91 (sec) , antiderivative size = 399, normalized size of antiderivative = 14.25 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx=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 \left (x\right )}{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 \left (x\right )}{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 \left (x\right ) + \log \left (x\right )^{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 \left (x\right )}{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 {dmath*}
integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+e xp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*e xp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm=\
16*(log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)/(e^3 - e^(1/15*e^3) + 1) - 2*l og(x)/(e^3 - e^(1/15*e^3) + 1) - 4*arctan((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*(l og(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*arctan((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) + l og(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 + 64*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*ar ctan((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)) + 16*log(x^2 - 4*x + e^3 - e^(1/15*e^3) + 1)
\begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx=\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 {dmath*}
integrate(((-2*exp(1/15*exp(3))+2*exp(3)-2*x^2+2)*log((-exp(1/15*exp(3))+e xp(3)+x^2-4*x+1)/x)-32*exp(1/15*exp(3))+32*exp(3)-32*x^2+32)/(x*exp(1/15*e xp(3))-x*exp(3)-x^3+4*x^2-x),x, algorithm=\
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)
Time = 37.89 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \begin {dmath*} \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-e^3 x+e^{\frac {e^3}{15}} x+4 x^2-x^3} \, dx={\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 \left (x\right ) \end {dmath*}
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)/15) - 2*exp(3) + 2*x^2 - 2) - 32)/(x - x*exp(exp(3)/15) + x*exp(3) - 4*x^2 + x^3),x)