Integrand size = 75, antiderivative size = 31 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{(1-x) x}\right )} \]
Time = 1.56 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{x-x^2}\right )} \]
Integrate[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/ (-x + x^2) + 48*Log[5]^2)*(-1 + 2*x))/(x^2 - 2*x^3 + x^4),x]
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.
\(\displaystyle \int \frac {(2 x-1) \exp \left (\frac {\left (e^{e^4} \left (2 x^2-2 x\right )-1\right ) e^{48 \log ^2(5)-e^4}}{x^2-x}-e^4+48 \log ^2(5)\right )}{x^4-2 x^3+x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(2 x-1) \exp \left (\frac {\left (e^{e^4} \left (2 x^2-2 x\right )-1\right ) e^{48 \log ^2(5)-e^4}}{x^2-x}-e^4+48 \log ^2(5)\right )}{x^2 \left (x^2-2 x+1\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}+48 \log ^2(5)-e^4\right ) (1-2 x)}{4 (1-x)^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}+48 \log ^2(5)-e^4\right ) (1-2 x)}{(1-x)^2 x^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right ) (1-2 x)}{(1-x)^2 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2}-\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(x-1)^2}dx-\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2}dx\) |
Int[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + x^2) + 48*Log[5]^2)*(-1 + 2*x))/(x^2 - 2*x^3 + x^4),x]
3.18.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_.))^(p_.), x_Symbol] :> Simp[1/(4^p*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] && !AlgebraicFu nctionQ[u, x]
Time = 0.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\) | \(38\) |
gosper | \({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\) | \(39\) |
risch | \({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\) | \(39\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}-x \,{\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}}{x \left (-1+x \right )}\) | \(94\) |
int((-1+2*x)*exp(48*ln(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*ln(5)^ 2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x,method=_RETURNVERBOS E)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.32 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\left (-48 \, \log \left (5\right )^{2} + \frac {48 \, {\left (x^{2} - x\right )} \log \left (5\right )^{2} - {\left (x^{2} - x\right )} e^{4} + {\left (2 \, {\left (x^{2} - x\right )} e^{\left (e^{4}\right )} - 1\right )} e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x^{2} - x} + e^{4}\right )} \]
integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 *log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=\
e^(-48*log(5)^2 + (48*(x^2 - x)*log(5)^2 - (x^2 - x)*e^4 + (2*(x^2 - x)*e^ (e^4) - 1)*e^(48*log(5)^2 - e^4))/(x^2 - x) + e^4)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\frac {\left (\left (2 x^{2} - 2 x\right ) e^{e^{4}} - 1\right ) e^{48 \log {\left (5 \right )}^{2}}}{\left (x^{2} - x\right ) e^{e^{4}}}} \]
integrate((-1+2*x)*exp(48*ln(5)**2)*exp(((2*x**2-2*x)*exp(exp(4))-1)*exp(4 8*ln(5)**2)/(x**2-x)/exp(exp(4)))/(x**4-2*x**3+x**2)/exp(exp(4)),x)
Time = 0.53 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\left (-\frac {e^{\left (48 \, \log \left (5\right )^{2}\right )}}{x e^{\left (e^{4}\right )} - e^{\left (e^{4}\right )}} + \frac {e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x} + 2 \, e^{\left (48 \, \log \left (5\right )^{2}\right )}\right )} \]
integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 *log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (26) = 52\).
Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 4.39 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\left (\frac {48 \, x^{2} \log \left (5\right )^{2}}{x^{2} - x} - \frac {x^{2} e^{4}}{x^{2} - x} + \frac {2 \, x^{2} e^{\left (48 \, \log \left (5\right )^{2}\right )}}{x^{2} - x} - \frac {48 \, x \log \left (5\right )^{2}}{x^{2} - x} - 48 \, \log \left (5\right )^{2} + \frac {x e^{4}}{x^{2} - x} - \frac {2 \, x e^{\left (48 \, \log \left (5\right )^{2}\right )}}{x^{2} - x} - \frac {e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x^{2} - x} + e^{4}\right )} \]
integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 *log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=\
e^(48*x^2*log(5)^2/(x^2 - x) - x^2*e^4/(x^2 - x) + 2*x^2*e^(48*log(5)^2)/( x^2 - x) - 48*x*log(5)^2/(x^2 - x) - 48*log(5)^2 + x*e^4/(x^2 - x) - 2*x*e ^(48*log(5)^2)/(x^2 - x) - e^(48*log(5)^2 - e^4)/(x^2 - x) + e^4)
Time = 12.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx={\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}}\,{\mathrm {e}}^{-\frac {2\,x^2\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}} \]
int((exp(-exp(4))*exp(48*log(5)^2)*exp((exp(-exp(4))*exp(48*log(5)^2)*(exp (exp(4))*(2*x - 2*x^2) + 1))/(x - x^2))*(2*x - 1))/(x^2 - 2*x^3 + x^4),x)