Integrand size = 96, antiderivative size = 28 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \left (-x+x^2\right ) \] Output:
(exp(5/(-1+x)+x/ln(x))+2*x)*(x^2-x)
Time = 1.49 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=(-1+x) x \left (e^{\frac {5}{-1+x}+\frac {x}{\log (x)}}+2 x\right ) \] Input:
Integrate[((4*x - 10*x^2 + 6*x^3)*Log[x]^2 + E^((-x + x^2 + 5*Log[x])/((-1 + x)*Log[x]))*(-x + 2*x^2 - x^3 + (x - 2*x^2 + x^3)*Log[x] + (1 - 8*x + 2 *x^2)*Log[x]^2))/((-1 + x)*Log[x]^2),x]
Output:
(-1 + x)*x*(E^(5/(-1 + x) + x/Log[x]) + 2*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.
\(\displaystyle \int \frac {\left (6 x^3-10 x^2+4 x\right ) \log ^2(x)+e^{\frac {x^2-x+5 \log (x)}{(x-1) \log (x)}} \left (-x^3+2 x^2+\left (2 x^2-8 x+1\right ) \log ^2(x)+\left (x^3-2 x^2+x\right ) \log (x)-x\right )}{(x-1) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {5}{x-1}+\frac {x}{\log (x)}} \left (-x^3+x^3 \log (x)+2 x^2+2 x^2 \log ^2(x)-2 x^2 \log (x)-x-8 x \log ^2(x)+\log ^2(x)+x \log (x)\right )}{(x-1) \log ^2(x)}+2 x (3 x-2)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{\frac {x}{\log (x)}+\frac {5}{x-1}} x^2}{\log ^2(x)}dx+\int \frac {e^{\frac {x}{\log (x)}+\frac {5}{x-1}} x^2}{\log (x)}dx+\int \frac {e^{\frac {x}{\log (x)}+\frac {5}{x-1}} x}{\log ^2(x)}dx-6 \int e^{\frac {x}{\log (x)}+\frac {5}{x-1}}dx-5 \int \frac {e^{\frac {x}{\log (x)}+\frac {5}{x-1}}}{x-1}dx+2 \int e^{\frac {x}{\log (x)}+\frac {5}{x-1}} xdx-\int \frac {e^{\frac {x}{\log (x)}+\frac {5}{x-1}} x}{\log (x)}dx+2 x^3-2 x^2\) |
Input:
Int[((4*x - 10*x^2 + 6*x^3)*Log[x]^2 + E^((-x + x^2 + 5*Log[x])/((-1 + x)* Log[x]))*(-x + 2*x^2 - x^3 + (x - 2*x^2 + x^3)*Log[x] + (1 - 8*x + 2*x^2)* Log[x]^2))/((-1 + x)*Log[x]^2),x]
Output:
$Aborted
Time = 1.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50
method | result | size |
risch | \(2 x^{3}-2 x^{2}+\left (x^{2}-x \right ) {\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}}\) | \(42\) |
parallelrisch | \(2 x^{3}+{\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}} x^{2}-2 x^{2}-x \,{\mathrm e}^{\frac {5 \ln \left (x \right )+x^{2}-x}{\left (-1+x \right ) \ln \left (x \right )}}+\frac {2}{3}\) | \(64\) |
Input:
int((((2*x^2-8*x+1)*ln(x)^2+(x^3-2*x^2+x)*ln(x)-x^3+2*x^2-x)*exp((5*ln(x)+ x^2-x)/(-1+x)/ln(x))+(6*x^3-10*x^2+4*x)*ln(x)^2)/(-1+x)/ln(x)^2,x,method=_ RETURNVERBOSE)
Output:
2*x^3-2*x^2+(x^2-x)*exp((5*ln(x)+x^2-x)/(-1+x)/ln(x))
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x^{2} - x + 5 \, \log \left (x\right )}{{\left (x - 1\right )} \log \left (x\right )}\right )} \] Input:
integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp(( 5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^ 2,x, algorithm="fricas")
Output:
2*x^3 - 2*x^2 + (x^2 - x)*e^((x^2 - x + 5*log(x))/((x - 1)*log(x)))
Time = 7.62 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 x^{3} - 2 x^{2} + \left (x^{2} - x\right ) e^{\frac {x^{2} - x + 5 \log {\left (x \right )}}{\left (x - 1\right ) \log {\left (x \right )}}} \] Input:
integrate((((2*x**2-8*x+1)*ln(x)**2+(x**3-2*x**2+x)*ln(x)-x**3+2*x**2-x)*e xp((5*ln(x)+x**2-x)/(-1+x)/ln(x))+(6*x**3-10*x**2+4*x)*ln(x)**2)/(-1+x)/ln (x)**2,x)
Output:
2*x**3 - 2*x**2 + (x**2 - x)*exp((x**2 - x + 5*log(x))/((x - 1)*log(x)))
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=2 \, x^{3} - 2 \, x^{2} + {\left (x^{2} - x\right )} e^{\left (\frac {x}{\log \left (x\right )} + \frac {5}{x - 1}\right )} \] Input:
integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp(( 5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^ 2,x, algorithm="maxima")
Output:
2*x^3 - 2*x^2 + (x^2 - x)*e^(x/log(x) + 5/(x - 1))
Exception generated. \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp(( 5*log(x)+x^2-x)/(-1+x)/log(x))+(6*x^3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^ 2,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{4,[1,28]%%%}+%%%{-74,[1,27]%%%}+%%%{630,[1,26]%%%}+%%%{-32 90,[1,25]
Time = 2.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=x\,\left (2\,x+\frac {{\mathrm {e}}^{\frac {x}{\ln \left (x\right )-x\,\ln \left (x\right )}-\frac {x^2}{\ln \left (x\right )-x\,\ln \left (x\right )}}}{x^{\frac {5}{\ln \left (x\right )-x\,\ln \left (x\right )}}}\right )\,\left (x-1\right ) \] Input:
int((log(x)^2*(4*x - 10*x^2 + 6*x^3) + exp((5*log(x) - x + x^2)/(log(x)*(x - 1)))*(log(x)^2*(2*x^2 - 8*x + 1) - x + log(x)*(x - 2*x^2 + x^3) + 2*x^2 - x^3))/(log(x)^2*(x - 1)),x)
Output:
x*(2*x + exp(x/(log(x) - x*log(x)) - x^2/(log(x) - x*log(x)))/x^(5/(log(x) - x*log(x))))*(x - 1)
Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.79 \[ \int \frac {\left (4 x-10 x^2+6 x^3\right ) \log ^2(x)+e^{\frac {-x+x^2+5 \log (x)}{(-1+x) \log (x)}} \left (-x+2 x^2-x^3+\left (x-2 x^2+x^3\right ) \log (x)+\left (1-8 x+2 x^2\right ) \log ^2(x)\right )}{(-1+x) \log ^2(x)} \, dx=\frac {x \left (e^{\frac {5 \,\mathrm {log}\left (x \right )+x^{2}}{\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}} x -e^{\frac {5 \,\mathrm {log}\left (x \right )+x^{2}}{\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}}+2 e^{\frac {x}{\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}} x^{2}-2 e^{\frac {x}{\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}} x \right )}{e^{\frac {x}{\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )}}} \] Input:
int((((2*x^2-8*x+1)*log(x)^2+(x^3-2*x^2+x)*log(x)-x^3+2*x^2-x)*exp((5*log( x)+x^2-x)/(-1+x)/log(x))+(6*x^3-10*x^2+4*x)*log(x)^2)/(-1+x)/log(x)^2,x)
Output:
(x*(e**((5*log(x) + x**2)/(log(x)*x - log(x)))*x - e**((5*log(x) + x**2)/( log(x)*x - log(x))) + 2*e**(x/(log(x)*x - log(x)))*x**2 - 2*e**(x/(log(x)* x - log(x)))*x))/e**(x/(log(x)*x - log(x)))