Integrand size = 82, antiderivative size = 30 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=5 e^{-x} \left (\frac {1+x}{5}+\log ^2(3)\right ) \left (2+x+\frac {x^2}{\log (x)}\right ) \]
Time = 5.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {e^{-x} \left (1+x+5 \log ^2(3)\right ) \left (x^2+(2+x) \log (x)\right )}{\log (x)} \]
Integrate[(-x - x^2 - 5*x*Log[3]^2 + (2*x + 2*x^2 - x^3 + (10*x - 5*x^2)*L og[3]^2)*Log[x] + (1 - x - x^2 + (-5 - 5*x)*Log[3]^2)*Log[x]^2)/(E^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.
\(\displaystyle \int \frac {e^{-x} \left (-x^2+\left (-x^2-x+(-5 x-5) \log ^2(3)+1\right ) \log ^2(x)+\left (-x^3+2 x^2+\left (10 x-5 x^2\right ) \log ^2(3)+2 x\right ) \log (x)-x-5 x \log ^2(3)\right )}{\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{-x} \left (-x^2+\left (-x^2-x+(-5 x-5) \log ^2(3)+1\right ) \log ^2(x)+\left (-x^3+2 x^2+\left (10 x-5 x^2\right ) \log ^2(3)+2 x\right ) \log (x)+x \left (-1-5 \log ^2(3)\right )\right )}{\log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-e^{-x} x^2+\frac {e^{-x} x \left (-x^2+x \left (2-5 \log ^2(3)\right )+2 \left (1+5 \log ^2(3)\right )\right )}{\log (x)}-\frac {e^{-x} x \left (x+1+5 \log ^2(3)\right )}{\log ^2(x)}-e^{-x} x \left (1+5 \log ^2(3)\right )+e^{-x} \left (1-5 \log ^2(3)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{-x} x^3}{\log (x)}dx-\int \frac {e^{-x} x^2}{\log ^2(x)}dx+\left (2-5 \log ^2(3)\right ) \int \frac {e^{-x} x^2}{\log (x)}dx-\left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log ^2(x)}dx+2 \left (1+5 \log ^2(3)\right ) \int \frac {e^{-x} x}{\log (x)}dx+e^{-x} x^2+2 e^{-x} x+2 e^{-x}+e^{-x} x \left (1+5 \log ^2(3)\right )+e^{-x} \left (1+5 \log ^2(3)\right )-e^{-x} \left (1-5 \log ^2(3)\right )\) |
Int[(-x - x^2 - 5*x*Log[3]^2 + (2*x + 2*x^2 - x^3 + (10*x - 5*x^2)*Log[3]^ 2)*Log[x] + (1 - x - x^2 + (-5 - 5*x)*Log[3]^2)*Log[x]^2)/(E^x*Log[x]^2),x ]
3.3.41.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63
method | result | size |
risch | \(\left (5 x \ln \left (3\right )^{2}+10 \ln \left (3\right )^{2}+x^{2}+3 x +2\right ) {\mathrm e}^{-x}+\frac {x^{2} {\mathrm e}^{-x} \left (5 \ln \left (3\right )^{2}+x +1\right )}{\ln \left (x \right )}\) | \(49\) |
parallelrisch | \(\frac {{\mathrm e}^{-x} \left (5 \ln \left (x \right ) \ln \left (3\right )^{2} x +5 x^{2} \ln \left (3\right )^{2}+10 \ln \left (3\right )^{2} \ln \left (x \right )+x^{2} \ln \left (x \right )+x^{3}+3 x \ln \left (x \right )+x^{2}+2 \ln \left (x \right )\right )}{\ln \left (x \right )}\) | \(58\) |
int((((-5*x-5)*ln(3)^2-x^2-x+1)*ln(x)^2+((-5*x^2+10*x)*ln(3)^2-x^3+2*x^2+2 *x)*ln(x)-5*x*ln(3)^2-x^2-x)/exp(x)/ln(x)^2,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {{\left (5 \, {\left (x + 2\right )} \log \left (3\right )^{2} + x^{2} + 3 \, x + 2\right )} e^{\left (-x\right )} \log \left (x\right ) + {\left (5 \, x^{2} \log \left (3\right )^{2} + x^{3} + x^{2}\right )} e^{\left (-x\right )}}{\log \left (x\right )} \]
integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^ 3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2-x)/exp(x)/log(x)^2,x, algorithm=\
((5*(x + 2)*log(3)^2 + x^2 + 3*x + 2)*e^(-x)*log(x) + (5*x^2*log(3)^2 + x^ 3 + x^2)*e^(-x))/log(x)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {\left (x^{3} + x^{2} \log {\left (x \right )} + x^{2} + 5 x^{2} \log {\left (3 \right )}^{2} + 3 x \log {\left (x \right )} + 5 x \log {\left (3 \right )}^{2} \log {\left (x \right )} + 2 \log {\left (x \right )} + 10 \log {\left (3 \right )}^{2} \log {\left (x \right )}\right ) e^{- x}}{\log {\left (x \right )}} \]
integrate((((-5*x-5)*ln(3)**2-x**2-x+1)*ln(x)**2+((-5*x**2+10*x)*ln(3)**2- x**3+2*x**2+2*x)*ln(x)-5*x*ln(3)**2-x**2-x)/exp(x)/ln(x)**2,x)
(x**3 + x**2*log(x) + x**2 + 5*x**2*log(3)**2 + 3*x*log(x) + 5*x*log(3)**2 *log(x) + 2*log(x) + 10*log(3)**2*log(x))*exp(-x)/log(x)
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (25) = 50\).
Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.53 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=5 \, {\left (x + 1\right )} e^{\left (-x\right )} \log \left (3\right )^{2} + 5 \, e^{\left (-x\right )} \log \left (3\right )^{2} + {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + {\left (x + 1\right )} e^{\left (-x\right )} + \frac {{\left ({\left (5 \, \log \left (3\right )^{2} + 1\right )} x^{2} + x^{3}\right )} e^{\left (-x\right )}}{\log \left (x\right )} - e^{\left (-x\right )} \]
integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^ 3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2-x)/exp(x)/log(x)^2,x, algorithm=\
5*(x + 1)*e^(-x)*log(3)^2 + 5*e^(-x)*log(3)^2 + (x^2 + 2*x + 2)*e^(-x) + ( x + 1)*e^(-x) + ((5*log(3)^2 + 1)*x^2 + x^3)*e^(-x)/log(x) - e^(-x)
\[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\int { -\frac {{\left (5 \, x \log \left (3\right )^{2} + {\left (5 \, {\left (x + 1\right )} \log \left (3\right )^{2} + x^{2} + x - 1\right )} \log \left (x\right )^{2} + x^{2} + {\left (x^{3} + 5 \, {\left (x^{2} - 2 \, x\right )} \log \left (3\right )^{2} - 2 \, x^{2} - 2 \, x\right )} \log \left (x\right ) + x\right )} e^{\left (-x\right )}}{\log \left (x\right )^{2}} \,d x } \]
integrate((((-5*x-5)*log(3)^2-x^2-x+1)*log(x)^2+((-5*x^2+10*x)*log(3)^2-x^ 3+2*x^2+2*x)*log(x)-5*x*log(3)^2-x^2-x)/exp(x)/log(x)^2,x, algorithm=\
integrate(-(5*x*log(3)^2 + (5*(x + 1)*log(3)^2 + x^2 + x - 1)*log(x)^2 + x ^2 + (x^3 + 5*(x^2 - 2*x)*log(3)^2 - 2*x^2 - 2*x)*log(x) + x)*e^(-x)/log(x )^2, x)
Time = 8.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-x} \left (-x-x^2-5 x \log ^2(3)+\left (2 x+2 x^2-x^3+\left (10 x-5 x^2\right ) \log ^2(3)\right ) \log (x)+\left (1-x-x^2+(-5-5 x) \log ^2(3)\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx={\mathrm {e}}^{-x}\,\left (x+2\right )\,\left (x+5\,{\ln \left (3\right )}^2+1\right )+\frac {x^2\,{\mathrm {e}}^{-x}\,\left (x+5\,{\ln \left (3\right )}^2+1\right )}{\ln \left (x\right )} \]