Integrand size = 193, antiderivative size = 32 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^x \left (x+\log \left (1+x-x^2\right )\right )+\log \left (\frac {1}{2}-e^3+x-\log (x)\right ) \] Output:
ln(1/2-exp(3)+x-ln(x))-(ln(-x^2+x+1)+x)*exp(x)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^x x-e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right ) \] Input:
Integrate[(-2 + 4*x^2 - 2*x^3 + E^x*(-2*x - 4*x^2 + x^4 + 2*x^5 + E^3*(4*x - 2*x^4)) + E^x*(-x - 3*x^2 - x^3 + 2*x^4 + E^3*(2*x + 2*x^2 - 2*x^3))*Lo g[1 + x - x^2] + Log[x]*(E^x*(4*x - 2*x^4) + E^x*(2*x + 2*x^2 - 2*x^3)*Log [1 + x - x^2]))/(x + 3*x^2 + x^3 - 2*x^4 + E^3*(-2*x - 2*x^2 + 2*x^3) + (- 2*x - 2*x^2 + 2*x^3)*Log[x]),x]
Output:
-(E^x*x) - E^x*Log[1 + x - x^2] + Log[1 - 2*E^3 + 2*x - 2*Log[x]]
Time = 2.69 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {7292, 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.
\(\displaystyle \int \frac {-2 x^3+4 x^2+e^x \left (2 x^5+x^4+e^3 \left (4 x-2 x^4\right )-4 x^2-2 x\right )+e^x \left (2 x^4-x^3-3 x^2+e^3 \left (-2 x^3+2 x^2+2 x\right )-x\right ) \log \left (-x^2+x+1\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (-2 x^3+2 x^2+2 x\right ) \log \left (-x^2+x+1\right )\right )-2}{-2 x^4+x^3+3 x^2+e^3 \left (2 x^3-2 x^2-2 x\right )+\left (2 x^3-2 x^2-2 x\right ) \log (x)+x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-2 x^3+4 x^2+e^x \left (2 x^5+x^4+e^3 \left (4 x-2 x^4\right )-4 x^2-2 x\right )+e^x \left (2 x^4-x^3-3 x^2+e^3 \left (-2 x^3+2 x^2+2 x\right )-x\right ) \log \left (-x^2+x+1\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (-2 x^3+2 x^2+2 x\right ) \log \left (-x^2+x+1\right )\right )-2}{x \left (-x^2+x+1\right ) \left (2 x-2 \log (x)-2 e^3+1\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2 (x-1)}{x \left (2 x-2 \log (x)-2 e^3+1\right )}-\frac {e^x \left (x^3+x^2 \log \left (-x^2+x+1\right )-x \log \left (-x^2+x+1\right )-\log \left (-x^2+x+1\right )-2\right )}{x^2-x-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^x \log \left (-x^2+x+1\right )-e^x x+\log \left (2 x-2 \log (x)-2 e^3+1\right )\) |
Input:
Int[(-2 + 4*x^2 - 2*x^3 + E^x*(-2*x - 4*x^2 + x^4 + 2*x^5 + E^3*(4*x - 2*x ^4)) + E^x*(-x - 3*x^2 - x^3 + 2*x^4 + E^3*(2*x + 2*x^2 - 2*x^3))*Log[1 + x - x^2] + Log[x]*(E^x*(4*x - 2*x^4) + E^x*(2*x + 2*x^2 - 2*x^3)*Log[1 + x - x^2]))/(x + 3*x^2 + x^3 - 2*x^4 + E^3*(-2*x - 2*x^2 + 2*x^3) + (-2*x - 2*x^2 + 2*x^3)*Log[x]),x]
Output:
-(E^x*x) - E^x*Log[1 + x - x^2] + Log[1 - 2*E^3 + 2*x - 2*Log[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]
Time = 1.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
\[-{\mathrm e}^{x} \ln \left (-x^{2}+x +1\right )-{\mathrm e}^{x} x +\ln \left ({\mathrm e}^{3}-x +\ln \left (x \right )-\frac {1}{2}\right )\]
Input:
int((((-2*x^3+2*x^2+2*x)*exp(x)*ln(-x^2+x+1)+(-2*x^4+4*x)*exp(x))*ln(x)+(( -2*x^3+2*x^2+2*x)*exp(3)+2*x^4-x^3-3*x^2-x)*exp(x)*ln(-x^2+x+1)+((-2*x^4+4 *x)*exp(3)+2*x^5+x^4-4*x^2-2*x)*exp(x)-2*x^3+4*x^2-2)/((2*x^3-2*x^2-2*x)*l n(x)+(2*x^3-2*x^2-2*x)*exp(3)-2*x^4+x^3+3*x^2+x),x)
Output:
-exp(x)*ln(-x^2+x+1)-exp(x)*x+ln(exp(3)-x+ln(x)-1/2)
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-2 \, x + 2 \, e^{3} + 2 \, \log \left (x\right ) - 1\right ) \] Input:
integrate((((-2*x^3+2*x^2+2*x)*exp(x)*log(-x^2+x+1)+(-2*x^4+4*x)*exp(x))*l og(x)+((-2*x^3+2*x^2+2*x)*exp(3)+2*x^4-x^3-3*x^2-x)*exp(x)*log(-x^2+x+1)+( (-2*x^4+4*x)*exp(3)+2*x^5+x^4-4*x^2-2*x)*exp(x)-2*x^3+4*x^2-2)/((2*x^3-2*x ^2-2*x)*log(x)+(2*x^3-2*x^2-2*x)*exp(3)-2*x^4+x^3+3*x^2+x),x, algorithm="f ricas")
Output:
-x*e^x - e^x*log(-x^2 + x + 1) + log(-2*x + 2*e^3 + 2*log(x) - 1)
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=\left (- x - \log {\left (- x^{2} + x + 1 \right )}\right ) e^{x} + \log {\left (- x + \log {\left (x \right )} - \frac {1}{2} + e^{3} \right )} \] Input:
integrate((((-2*x**3+2*x**2+2*x)*exp(x)*ln(-x**2+x+1)+(-2*x**4+4*x)*exp(x) )*ln(x)+((-2*x**3+2*x**2+2*x)*exp(3)+2*x**4-x**3-3*x**2-x)*exp(x)*ln(-x**2 +x+1)+((-2*x**4+4*x)*exp(3)+2*x**5+x**4-4*x**2-2*x)*exp(x)-2*x**3+4*x**2-2 )/((2*x**3-2*x**2-2*x)*ln(x)+(2*x**3-2*x**2-2*x)*exp(3)-2*x**4+x**3+3*x**2 +x),x)
Output:
(-x - log(-x**2 + x + 1))*exp(x) + log(-x + log(x) - 1/2 + exp(3))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-x + e^{3} + \log \left (x\right ) - \frac {1}{2}\right ) \] Input:
integrate((((-2*x^3+2*x^2+2*x)*exp(x)*log(-x^2+x+1)+(-2*x^4+4*x)*exp(x))*l og(x)+((-2*x^3+2*x^2+2*x)*exp(3)+2*x^4-x^3-3*x^2-x)*exp(x)*log(-x^2+x+1)+( (-2*x^4+4*x)*exp(3)+2*x^5+x^4-4*x^2-2*x)*exp(x)-2*x^3+4*x^2-2)/((2*x^3-2*x ^2-2*x)*log(x)+(2*x^3-2*x^2-2*x)*exp(3)-2*x^4+x^3+3*x^2+x),x, algorithm="m axima")
Output:
-x*e^x - e^x*log(-x^2 + x + 1) + log(-x + e^3 + log(x) - 1/2)
Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-2 \, x + 2 \, e^{3} + 2 \, \log \left (x\right ) - 1\right ) \] Input:
integrate((((-2*x^3+2*x^2+2*x)*exp(x)*log(-x^2+x+1)+(-2*x^4+4*x)*exp(x))*l og(x)+((-2*x^3+2*x^2+2*x)*exp(3)+2*x^4-x^3-3*x^2-x)*exp(x)*log(-x^2+x+1)+( (-2*x^4+4*x)*exp(3)+2*x^5+x^4-4*x^2-2*x)*exp(x)-2*x^3+4*x^2-2)/((2*x^3-2*x ^2-2*x)*log(x)+(2*x^3-2*x^2-2*x)*exp(3)-2*x^4+x^3+3*x^2+x),x, algorithm="g iac")
Output:
-x*e^x - e^x*log(-x^2 + x + 1) + log(-2*x + 2*e^3 + 2*log(x) - 1)
Time = 3.83 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=\ln \left ({\mathrm {e}}^3-x+\ln \left (x\right )-\frac {1}{2}\right )-x\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (-x^2+x+1\right ) \] Input:
int((log(x)*(exp(x)*(4*x - 2*x^4) + exp(x)*log(x - x^2 + 1)*(2*x + 2*x^2 - 2*x^3)) + 4*x^2 - 2*x^3 + exp(x)*(exp(3)*(4*x - 2*x^4) - 2*x - 4*x^2 + x^ 4 + 2*x^5) - exp(x)*log(x - x^2 + 1)*(x - exp(3)*(2*x + 2*x^2 - 2*x^3) + 3 *x^2 + x^3 - 2*x^4) - 2)/(x - exp(3)*(2*x + 2*x^2 - 2*x^3) + 3*x^2 + x^3 - 2*x^4 - log(x)*(2*x + 2*x^2 - 2*x^3)),x)
Output:
log(exp(3) - x + log(x) - 1/2) - x*exp(x) - exp(x)*log(x - x^2 + 1)
Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^{x} \mathrm {log}\left (-x^{2}+x +1\right )-e^{x} x +\mathrm {log}\left (2 \,\mathrm {log}\left (x \right )+2 e^{3}-2 x -1\right ) \] Input:
int((((-2*x^3+2*x^2+2*x)*exp(x)*log(-x^2+x+1)+(-2*x^4+4*x)*exp(x))*log(x)+ ((-2*x^3+2*x^2+2*x)*exp(3)+2*x^4-x^3-3*x^2-x)*exp(x)*log(-x^2+x+1)+((-2*x^ 4+4*x)*exp(3)+2*x^5+x^4-4*x^2-2*x)*exp(x)-2*x^3+4*x^2-2)/((2*x^3-2*x^2-2*x )*log(x)+(2*x^3-2*x^2-2*x)*exp(3)-2*x^4+x^3+3*x^2+x),x)
Output:
- e**x*log( - x**2 + x + 1) - e**x*x + log(2*log(x) + 2*e**3 - 2*x - 1)