Integrand size = 247, antiderivative size = 33 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=5-x+\frac {3+x}{x-e^{-1+x} \log (4)}-\log (\log (3+x-\log (x))) \]
Time = 0.55 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=-x+\frac {-3 e-e x}{-e x+e^x \log (4)}-\log (\log (3+x-\log (x))) \]
Integrate[(-x^2 + x^3 + E^(-1 + x)*(2*x - 2*x^2)*Log[4] + E^(-2 + 2*x)*(-1 + x)*Log[4]^2 + (9*x + 3*x^2 + 3*x^3 + x^4 + E^(-1 + x)*(-6*x - 11*x^2 - 3*x^3)*Log[4] + E^(-2 + 2*x)*(3*x + x^2)*Log[4]^2 + (-3*x - x^3 + E^(-1 + x)*(2*x + 3*x^2)*Log[4] - E^(-2 + 2*x)*x*Log[4]^2)*Log[x])*Log[3 + x - Log [x]])/((-3*x^3 - x^4 + E^(-1 + x)*(6*x^2 + 2*x^3)*Log[4] + E^(-2 + 2*x)*(- 3*x - x^2)*Log[4]^2 + (x^3 - 2*E^(-1 + x)*x^2*Log[4] + E^(-2 + 2*x)*x*Log[ 4]^2)*Log[x])*Log[3 + x - Log[x]]),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.
| \(\displaystyle \int \frac {x^3-x^2+e^{x-1} \left (2 x-2 x^2\right ) \log (4)+\left (x^4+3 x^3+3 x^2+e^{2 x-2} \left (x^2+3 x\right ) \log ^2(4)+\left (-x^3+e^{x-1} \left (3 x^2+2 x\right ) \log (4)-3 x-e^{2 x-2} x \log ^2(4)\right ) \log (x)+e^{x-1} \left (-3 x^3-11 x^2-6 x\right ) \log (4)+9 x\right ) \log (x-\log (x)+3)+e^{2 x-2} (x-1) \log ^2(4)}{\left (-x^4-3 x^3+e^{2 x-2} \left (-x^2-3 x\right ) \log ^2(4)+\left (x^3-2 e^{x-1} x^2 \log (4)+e^{2 x-2} x \log ^2(4)\right ) \log (x)+e^{x-1} \left (2 x^3+6 x^2\right ) \log (4)\right ) \log (x-\log (x)+3)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {-e^2 \left (x^2+3\right )-e^{2 x} \log ^2(4)+e^{x+1} (3 x+2) \log (4)}{\left (e x-e^x \log (4)\right )^2}+\frac {1-x}{x (x-\log (x)+3) \log (x-\log (x)+3)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^2 \int \frac {x^2}{\left (e x-e^x \log (4)\right )^2}dx-3 e^2 \int \frac {1}{\left (e x-e^x \log (4)\right )^2}dx+2 e^2 \int \frac {x}{\left (e x-e^x \log (4)\right )^2}dx-2 e \int \frac {1}{e x-e^x \log (4)}dx-e \int \frac {x}{e x-e^x \log (4)}dx-x-\log (\log (x-\log (x)+3))\) |
Int[(-x^2 + x^3 + E^(-1 + x)*(2*x - 2*x^2)*Log[4] + E^(-2 + 2*x)*(-1 + x)* Log[4]^2 + (9*x + 3*x^2 + 3*x^3 + x^4 + E^(-1 + x)*(-6*x - 11*x^2 - 3*x^3) *Log[4] + E^(-2 + 2*x)*(3*x + x^2)*Log[4]^2 + (-3*x - x^3 + E^(-1 + x)*(2* x + 3*x^2)*Log[4] - E^(-2 + 2*x)*x*Log[4]^2)*Log[x])*Log[3 + x - Log[x]])/ ((-3*x^3 - x^4 + E^(-1 + x)*(6*x^2 + 2*x^3)*Log[4] + E^(-2 + 2*x)*(-3*x - x^2)*Log[4]^2 + (x^3 - 2*E^(-1 + x)*x^2*Log[4] + E^(-2 + 2*x)*x*Log[4]^2)* Log[x])*Log[3 + x - Log[x]]),x]
3.9.62.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 9.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39
| method | result | size |
| risch | \(-\frac {2 x \ln \left (2\right ) {\mathrm e}^{-1+x}-x^{2}+x +3}{2 \ln \left (2\right ) {\mathrm e}^{-1+x}-x}-\ln \left (\ln \left (-\ln \left (x \right )+3+x \right )\right )\) | \(46\) |
| parallelrisch | \(\frac {-2 \ln \left (2\right ) \ln \left (\ln \left (-\ln \left (x \right )+3+x \right )\right ) {\mathrm e}^{-1+x}-2 x \ln \left (2\right ) {\mathrm e}^{-1+x}-3-2 \ln \left (2\right ) {\mathrm e}^{-1+x}+\ln \left (\ln \left (-\ln \left (x \right )+3+x \right )\right ) x +x^{2}}{2 \ln \left (2\right ) {\mathrm e}^{-1+x}-x}\) | \(66\) |
int((((-4*x*ln(2)^2*exp(-1+x)^2+2*(3*x^2+2*x)*ln(2)*exp(-1+x)-x^3-3*x)*ln( x)+4*(x^2+3*x)*ln(2)^2*exp(-1+x)^2+2*(-3*x^3-11*x^2-6*x)*ln(2)*exp(-1+x)+x ^4+3*x^3+3*x^2+9*x)*ln(-ln(x)+3+x)+4*(-1+x)*ln(2)^2*exp(-1+x)^2+2*(-2*x^2+ 2*x)*ln(2)*exp(-1+x)+x^3-x^2)/((4*x*ln(2)^2*exp(-1+x)^2-4*x^2*ln(2)*exp(-1 +x)+x^3)*ln(x)+4*(-x^2-3*x)*ln(2)^2*exp(-1+x)^2+2*(2*x^3+6*x^2)*ln(2)*exp( -1+x)-x^4-3*x^3)/ln(-ln(x)+3+x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=-\frac {2 \, x e^{\left (x - 1\right )} \log \left (2\right ) - x^{2} + {\left (2 \, e^{\left (x - 1\right )} \log \left (2\right ) - x\right )} \log \left (\log \left (x - \log \left (x\right ) + 3\right )\right ) + x + 3}{2 \, e^{\left (x - 1\right )} \log \left (2\right ) - x} \]
integrate((((-4*x*log(2)^2*exp(-1+x)^2+2*(3*x^2+2*x)*log(2)*exp(-1+x)-x^3- 3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(-1+x)^2+2*(-3*x^3-11*x^2-6*x)*log(2)* exp(-1+x)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(-1+x)*log(2)^2*exp(-1+x )^2+2*(-2*x^2+2*x)*log(2)*exp(-1+x)+x^3-x^2)/((4*x*log(2)^2*exp(-1+x)^2-4* x^2*log(2)*exp(-1+x)+x^3)*log(x)+4*(-x^2-3*x)*log(2)^2*exp(-1+x)^2+2*(2*x^ 3+6*x^2)*log(2)*exp(-1+x)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm=\
-(2*x*e^(x - 1)*log(2) - x^2 + (2*e^(x - 1)*log(2) - x)*log(log(x - log(x) + 3)) + x + 3)/(2*e^(x - 1)*log(2) - x)
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=- x + \frac {- x - 3}{- x + 2 e^{x - 1} \log {\left (2 \right )}} - \log {\left (\log {\left (x - \log {\left (x \right )} + 3 \right )} \right )} \]
integrate((((-4*x*ln(2)**2*exp(-1+x)**2+2*(3*x**2+2*x)*ln(2)*exp(-1+x)-x** 3-3*x)*ln(x)+4*(x**2+3*x)*ln(2)**2*exp(-1+x)**2+2*(-3*x**3-11*x**2-6*x)*ln (2)*exp(-1+x)+x**4+3*x**3+3*x**2+9*x)*ln(-ln(x)+3+x)+4*(-1+x)*ln(2)**2*exp (-1+x)**2+2*(-2*x**2+2*x)*ln(2)*exp(-1+x)+x**3-x**2)/((4*x*ln(2)**2*exp(-1 +x)**2-4*x**2*ln(2)*exp(-1+x)+x**3)*ln(x)+4*(-x**2-3*x)*ln(2)**2*exp(-1+x) **2+2*(2*x**3+6*x**2)*ln(2)*exp(-1+x)-x**4-3*x**3)/ln(-ln(x)+3+x),x)
Time = 0.37 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=-\frac {x^{2} e - 2 \, x e^{x} \log \left (2\right ) - x e - 3 \, e}{x e - 2 \, e^{x} \log \left (2\right )} - \log \left (\log \left (x - \log \left (x\right ) + 3\right )\right ) \]
integrate((((-4*x*log(2)^2*exp(-1+x)^2+2*(3*x^2+2*x)*log(2)*exp(-1+x)-x^3- 3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(-1+x)^2+2*(-3*x^3-11*x^2-6*x)*log(2)* exp(-1+x)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(-1+x)*log(2)^2*exp(-1+x )^2+2*(-2*x^2+2*x)*log(2)*exp(-1+x)+x^3-x^2)/((4*x*log(2)^2*exp(-1+x)^2-4* x^2*log(2)*exp(-1+x)+x^3)*log(x)+4*(-x^2-3*x)*log(2)^2*exp(-1+x)^2+2*(2*x^ 3+6*x^2)*log(2)*exp(-1+x)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=-\frac {x^{2} e - 2 \, x e^{x} \log \left (2\right ) + x e \log \left (\log \left (x - \log \left (x\right ) + 3\right )\right ) - 2 \, e^{x} \log \left (2\right ) \log \left (\log \left (x - \log \left (x\right ) + 3\right )\right ) - x e - 3 \, e}{x e - 2 \, e^{x} \log \left (2\right )} \]
integrate((((-4*x*log(2)^2*exp(-1+x)^2+2*(3*x^2+2*x)*log(2)*exp(-1+x)-x^3- 3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(-1+x)^2+2*(-3*x^3-11*x^2-6*x)*log(2)* exp(-1+x)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(-1+x)*log(2)^2*exp(-1+x )^2+2*(-2*x^2+2*x)*log(2)*exp(-1+x)+x^3-x^2)/((4*x*log(2)^2*exp(-1+x)^2-4* x^2*log(2)*exp(-1+x)+x^3)*log(x)+4*(-x^2-3*x)*log(2)^2*exp(-1+x)^2+2*(2*x^ 3+6*x^2)*log(2)*exp(-1+x)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm=\
-(x^2*e - 2*x*e^x*log(2) + x*e*log(log(x - log(x) + 3)) - 2*e^x*log(2)*log (log(x - log(x) + 3)) - x*e - 3*e)/(x*e - 2*e^x*log(2))
Time = 12.64 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.09 \[ \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx=-x-\ln \left (\ln \left (x-\ln \left (x\right )+3\right )\right )-\frac {6\,\ln \left (2\right )-x\,\ln \left (16\right )+4\,x^2\,\ln \left (2\right )-x^2\,\ln \left (64\right )}{4\,\ln \left (2\right )\,\left (\ln \left (2\right )-x\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{x-1}-\frac {x}{2\,\ln \left (2\right )}\right )} \]
int(-(x^3 - x^2 + log(x - log(x) + 3)*(9*x - log(x)*(3*x + x^3 + 4*x*exp(2 *x - 2)*log(2)^2 - 2*exp(x - 1)*log(2)*(2*x + 3*x^2)) + 3*x^2 + 3*x^3 + x^ 4 - 2*exp(x - 1)*log(2)*(6*x + 11*x^2 + 3*x^3) + 4*exp(2*x - 2)*log(2)^2*( 3*x + x^2)) + 2*exp(x - 1)*log(2)*(2*x - 2*x^2) + 4*exp(2*x - 2)*log(2)^2* (x - 1))/(log(x - log(x) + 3)*(3*x^3 - log(x)*(x^3 + 4*x*exp(2*x - 2)*log( 2)^2 - 4*x^2*exp(x - 1)*log(2)) + x^4 + 4*exp(2*x - 2)*log(2)^2*(3*x + x^2 ) - 2*exp(x - 1)*log(2)*(6*x^2 + 2*x^3))),x)