Integrand size = 170, antiderivative size = 33 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=5-\frac {\left (x-\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right )\right ) (x+2 \log (x+\log (9)))}{x} \end {dmath*}
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-x+\log \left (2-e^x+x\right )-2 \log (x+\log (9))+\frac {2 \log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \log (x+\log (9))}{x} \end {dmath*}
Integrate[(4*x^2 - 2*E^x*x^2 + 3*x^3 + x^4 + (x^2 + x^3)*Log[9] + (-2*x^2 - 2*x*Log[9] + E^x*(2*x^2 + 2*x*Log[9]))*Log[x + Log[9]] + Log[(-2 + E^x - x)/5]*(-4*x + 2*E^x*x - 2*x^2 + (4*x + 2*x^2 + E^x*(-2*x - 2*Log[9]) + (4 + 2*x)*Log[9])*Log[x + Log[9]]))/(-2*x^3 - x^4 + (-2*x^2 - x^3)*Log[9] + E^x*(x^3 + x^2*Log[9])),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.
| \(\displaystyle \int \frac {x^4+3 x^3-2 e^x x^2+4 x^2+\left (-2 x^2+e^x \left (2 x^2+2 x \log (9)\right )-2 x \log (9)\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-x+e^x-2\right )\right ) \left (-2 x^2+\left (2 x^2+4 x+e^x (-2 x-2 \log (9))+(2 x+4) \log (9)\right ) \log (x+\log (9))+2 e^x x-4 x\right )+\left (x^3+x^2\right ) \log (9)}{-x^4-2 x^3+e^x \left (x^3+x^2 \log (9)\right )+\left (-x^3-2 x^2\right ) \log (9)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (-x+e^x-2\right ) \log \left (\frac {1}{5} \left (-x+e^x-2\right )\right ) ((x+\log (9)) \log (x+\log (9))-x)-x \left (x \left (x^2-2 e^x+x (3+\log (9))+4+\log (9)\right )+2 \left (e^x-1\right ) (x+\log (9)) \log (x+\log (9))\right )}{x^2 \left (x-e^x+2\right ) (x+\log (9))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 \left (x-\log \left (-x+e^x-2\right )+\log (5)\right ) (-x+x \log (x+\log (9))+\log (9) \log (x+\log (9)))}{x^2 (x+\log (9))}+\frac {(x+1) (x+2 \log (x+\log (9)))}{\left (-x+e^x-2\right ) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {1}{-x+e^x-2}dx+\int \frac {x}{-x+e^x-2}dx+\frac {2 \log \left (-x+e^x-2\right ) \log (x+\log (9))}{x}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{x}\) |
Int[(4*x^2 - 2*E^x*x^2 + 3*x^3 + x^4 + (x^2 + x^3)*Log[9] + (-2*x^2 - 2*x* Log[9] + E^x*(2*x^2 + 2*x*Log[9]))*Log[x + Log[9]] + Log[(-2 + E^x - x)/5] *(-4*x + 2*E^x*x - 2*x^2 + (4*x + 2*x^2 + E^x*(-2*x - 2*Log[9]) + (4 + 2*x )*Log[9])*Log[x + Log[9]]))/(-2*x^3 - x^4 + (-2*x^2 - x^3)*Log[9] + E^x*(x ^3 + x^2*Log[9])),x]
3.7.4.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33
| method | result | size |
| risch | \(\frac {2 \ln \left (2 \ln \left (3\right )+x \right ) \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right )}{x}-2 \ln \left (2 \ln \left (3\right )+x \right )-x +\ln \left ({\mathrm e}^{x}-2-x \right )\) | \(44\) |
| parallelrisch | \(\frac {4 x \ln \left (3\right )-x^{2}-2 \ln \left (2 \ln \left (3\right )+x \right ) x +\ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) x +2 \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) \ln \left (2 \ln \left (3\right )+x \right )}{x}\) | \(57\) |
int(((((-4*ln(3)-2*x)*exp(x)+2*(4+2*x)*ln(3)+2*x^2+4*x)*ln(2*ln(3)+x)+2*ex p(x)*x-2*x^2-4*x)*ln(1/5*exp(x)-2/5-1/5*x)+((4*x*ln(3)+2*x^2)*exp(x)-4*x*l n(3)-2*x^2)*ln(2*ln(3)+x)-2*exp(x)*x^2+2*(x^3+x^2)*ln(3)+x^4+3*x^3+4*x^2)/ ((2*x^2*ln(3)+x^3)*exp(x)+2*(-x^3-2*x^2)*ln(3)-x^4-2*x^3),x,method=_RETURN VERBOSE)
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{x} - \frac {2}{5}\right )}{x} \end {dmath*}
integrate(((((-4*log(3)-2*x)*exp(x)+2*(4+2*x)*log(3)+2*x^2+4*x)*log(2*log( 3)+x)+2*exp(x)*x-2*x^2-4*x)*log(1/5*exp(x)-2/5-1/5*x)+((4*x*log(3)+2*x^2)* exp(x)-4*x*log(3)-2*x^2)*log(2*log(3)+x)-2*exp(x)*x^2+2*(x^3+x^2)*log(3)+x ^4+3*x^3+4*x^2)/((2*x^2*log(3)+x^3)*exp(x)+2*(-x^3-2*x^2)*log(3)-x^4-2*x^3 ),x, algorithm=\
Time = 0.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=- x - 2 \log {\left (x + 2 \log {\left (3 \right )} \right )} + \log {\left (- x + e^{x} - 2 \right )} + \frac {2 \log {\left (x + 2 \log {\left (3 \right )} \right )} \log {\left (- \frac {x}{5} + \frac {e^{x}}{5} - \frac {2}{5} \right )}}{x} \end {dmath*}
integrate(((((-4*ln(3)-2*x)*exp(x)+2*(4+2*x)*ln(3)+2*x**2+4*x)*ln(2*ln(3)+ x)+2*exp(x)*x-2*x**2-4*x)*ln(1/5*exp(x)-2/5-1/5*x)+((4*x*ln(3)+2*x**2)*exp (x)-4*x*ln(3)-2*x**2)*ln(2*ln(3)+x)-2*exp(x)*x**2+2*(x**3+x**2)*ln(3)+x**4 +3*x**3+4*x**2)/((2*x**2*ln(3)+x**3)*exp(x)+2*(-x**3-2*x**2)*ln(3)-x**4-2* x**3),x)
-x - 2*log(x + 2*log(3)) + log(-x + exp(x) - 2) + 2*log(x + 2*log(3))*log( -x/5 + exp(x)/5 - 2/5)/x
Time = 0.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, {\left (x + \log \left (5\right )\right )} \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-x + e^{x} - 2\right )}{x} \end {dmath*}
integrate(((((-4*log(3)-2*x)*exp(x)+2*(4+2*x)*log(3)+2*x^2+4*x)*log(2*log( 3)+x)+2*exp(x)*x-2*x^2-4*x)*log(1/5*exp(x)-2/5-1/5*x)+((4*x*log(3)+2*x^2)* exp(x)-4*x*log(3)-2*x^2)*log(2*log(3)+x)-2*exp(x)*x^2+2*(x^3+x^2)*log(3)+x ^4+3*x^3+4*x^2)/((2*x^2*log(3)+x^3)*exp(x)+2*(-x^3-2*x^2)*log(3)-x^4-2*x^3 ),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} - x \log \left (x - e^{x} + 2\right ) + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) + 2 \, \log \left (5\right ) \log \left (x + 2 \, \log \left (3\right )\right ) - 2 \, \log \left (x + 2 \, \log \left (3\right )\right ) \log \left (-x + e^{x} - 2\right )}{x} \end {dmath*}
integrate(((((-4*log(3)-2*x)*exp(x)+2*(4+2*x)*log(3)+2*x^2+4*x)*log(2*log( 3)+x)+2*exp(x)*x-2*x^2-4*x)*log(1/5*exp(x)-2/5-1/5*x)+((4*x*log(3)+2*x^2)* exp(x)-4*x*log(3)-2*x^2)*log(2*log(3)+x)-2*exp(x)*x^2+2*(x^3+x^2)*log(3)+x ^4+3*x^3+4*x^2)/((2*x^2*log(3)+x^3)*exp(x)+2*(-x^3-2*x^2)*log(3)-x^4-2*x^3 ),x, algorithm=\
-(x^2 - x*log(x - e^x + 2) + 2*x*log(x + 2*log(3)) + 2*log(5)*log(x + 2*lo g(3)) - 2*log(x + 2*log(3))*log(-x + e^x - 2))/x
Timed out. \begin {dmath*} \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=\int -\frac {4\,x^2-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x\,\ln \left (3\right )-{\mathrm {e}}^x\,\left (2\,x^2+4\,\ln \left (3\right )\,x\right )+2\,x^2\right )-\ln \left (\frac {{\mathrm {e}}^x}{5}-\frac {x}{5}-\frac {2}{5}\right )\,\left (4\,x-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x+2\,\ln \left (3\right )\,\left (2\,x+4\right )-{\mathrm {e}}^x\,\left (2\,x+4\,\ln \left (3\right )\right )+2\,x^2\right )-2\,x\,{\mathrm {e}}^x+2\,x^2\right )-2\,x^2\,{\mathrm {e}}^x+3\,x^3+x^4+2\,\ln \left (3\right )\,\left (x^3+x^2\right )}{2\,\ln \left (3\right )\,\left (x^3+2\,x^2\right )-{\mathrm {e}}^x\,\left (x^3+2\,\ln \left (3\right )\,x^2\right )+2\,x^3+x^4} \,d x \end {dmath*}
int(-(4*x^2 - log(x + 2*log(3))*(4*x*log(3) - exp(x)*(4*x*log(3) + 2*x^2) + 2*x^2) - log(exp(x)/5 - x/5 - 2/5)*(4*x - log(x + 2*log(3))*(4*x + 2*log (3)*(2*x + 4) - exp(x)*(2*x + 4*log(3)) + 2*x^2) - 2*x*exp(x) + 2*x^2) - 2 *x^2*exp(x) + 3*x^3 + x^4 + 2*log(3)*(x^2 + x^3))/(2*log(3)*(2*x^2 + x^3) - exp(x)*(2*x^2*log(3) + x^3) + 2*x^3 + x^4),x)
int(-(4*x^2 - log(x + 2*log(3))*(4*x*log(3) - exp(x)*(4*x*log(3) + 2*x^2) + 2*x^2) - log(exp(x)/5 - x/5 - 2/5)*(4*x - log(x + 2*log(3))*(4*x + 2*log (3)*(2*x + 4) - exp(x)*(2*x + 4*log(3)) + 2*x^2) - 2*x*exp(x) + 2*x^2) - 2 *x^2*exp(x) + 3*x^3 + x^4 + 2*log(3)*(x^2 + x^3))/(2*log(3)*(2*x^2 + x^3) - exp(x)*(2*x^2*log(3) + x^3) + 2*x^3 + x^4), x)