Integrand size = 73, antiderivative size = 32 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\left (-3+e^{-x} \left (3+\frac {1}{4} \left (-x+x^2+\log (3)\right )\right )\right ) \left (-e^3+\log (x)\right ) \] Output:
(ln(x)-exp(3))*((1/4*x^2+1/4*ln(3)-1/4*x+3)/exp(x)-3)
Time = 5.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\frac {1}{4} e^{-x} \left (-e^3 \left (12-x+x^2+\log (3)\right )+\left (12-12 e^x-x+x^2+\log (3)\right ) \log (x)\right ) \] Input:
Integrate[(12 - 12*E^x - x + x^2 + E^3*(13*x - 3*x^2 + x^3) + (1 + E^3*x)* Log[3] + (-13*x + 3*x^2 - x^3 - x*Log[3])*Log[x])/(4*E^x*x),x]
Output:
(-(E^3*(12 - x + x^2 + Log[3])) + (12 - 12*E^x - x + x^2 + Log[3])*Log[x]) /(4*E^x)
Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(32)=64\).
Time = 1.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 4.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {27, 7293, 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.
\(\displaystyle \int \frac {e^{-x} \left (x^2+e^3 \left (x^3-3 x^2+13 x\right )+\left (-x^3+3 x^2-13 x-x \log (3)\right ) \log (x)-x-12 e^x+\left (e^3 x+1\right ) \log (3)+12\right )}{4 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{-x} \left (x^2-x-12 e^x+e^3 \left (x^3-3 x^2+13 x\right )-\left (x^3-3 x^2+\log (3) x+13 x\right ) \log (x)+\left (e^3 x+1\right ) \log (3)+12\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {e^{-x} \left (-\log (x) x^3+e^3 x^3+3 \log (x) x^2+\left (1-3 e^3\right ) x^2-13 \left (1+\frac {\log (3)}{13}\right ) \log (x) x-\left (1-e^3 (13+\log (3))\right ) x+12 \left (1+\frac {\log (3)}{12}\right )\right )}{x}-\frac {12}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-e^{3-x} x^2+e^{-x} x^2 \log (x)-2 e^{3-x} x+e^{-x} x-\left (1-3 e^3\right ) e^{-x} x-2 e^{3-x}-\left (1-3 e^3\right ) e^{-x}-e^{-x} x \log (x)-e^{-x} \log (x)+e^{-x} (13+\log (3)) \log (x)-12 \log (x)+e^{-x} \left (1-e^3 (13+\log (3))\right )\right )\) |
Input:
Int[(12 - 12*E^x - x + x^2 + E^3*(13*x - 3*x^2 + x^3) + (1 + E^3*x)*Log[3] + (-13*x + 3*x^2 - x^3 - x*Log[3])*Log[x])/(4*E^x*x),x]
Output:
(-2*E^(3 - x) - (1 - 3*E^3)/E^x - 2*E^(3 - x)*x + x/E^x - ((1 - 3*E^3)*x)/ E^x - E^(3 - x)*x^2 + (1 - E^3*(13 + Log[3]))/E^x - 12*Log[x] - Log[x]/E^x - (x*Log[x])/E^x + (x^2*Log[x])/E^x + ((13 + Log[3])*Log[x])/E^x)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.81 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62
method | result | size |
default | \(\frac {\left (x \,{\mathrm e}^{3}+x^{2} \ln \left (x \right )+\left (\ln \left (3\right )+12\right ) \ln \left (x \right )-x \ln \left (x \right )-x^{2} {\mathrm e}^{3}-12 \,{\mathrm e}^{3}-{\mathrm e}^{3} \ln \left (3\right )\right ) {\mathrm e}^{-x}}{4}-3 \ln \left (x \right )\) | \(52\) |
risch | \(\frac {\left (x^{2}+\ln \left (3\right )-x +12\right ) {\mathrm e}^{-x} \ln \left (x \right )}{4}-\frac {\left (x^{2} {\mathrm e}^{3}+{\mathrm e}^{3} \ln \left (3\right )+12 \,{\mathrm e}^{x} \ln \left (x \right )-x \,{\mathrm e}^{3}+12 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-x}}{4}\) | \(53\) |
parallelrisch | \(-\frac {\left (x^{2} {\mathrm e}^{3}-x^{2} \ln \left (x \right )+{\mathrm e}^{3} \ln \left (3\right )-\ln \left (x \right ) \ln \left (3\right )-x \,{\mathrm e}^{3}+x \ln \left (x \right )+12 \,{\mathrm e}^{x} \ln \left (x \right )+12 \,{\mathrm e}^{3}-12 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{4}\) | \(55\) |
parts | \(\left (\left (\frac {\ln \left (3\right )}{4}+3\right ) \ln \left (x \right )+\frac {x \,{\mathrm e}^{3}}{4}-\frac {x \ln \left (x \right )}{4}-\frac {x^{2} {\mathrm e}^{3}}{4}+\frac {x^{2} \ln \left (x \right )}{4}-3 \,{\mathrm e}^{3}-\frac {{\mathrm e}^{3} \ln \left (3\right )}{4}\right ) {\mathrm e}^{-x}-3 \ln \left (x \right )\) | \(55\) |
norman | \(\left (\left (\frac {\ln \left (3\right )}{4}+3\right ) \ln \left (x \right )-3 \,{\mathrm e}^{x} \ln \left (x \right )+\frac {x \,{\mathrm e}^{3}}{4}-\frac {x \ln \left (x \right )}{4}-\frac {x^{2} {\mathrm e}^{3}}{4}+\frac {x^{2} \ln \left (x \right )}{4}-3 \,{\mathrm e}^{3}-\frac {{\mathrm e}^{3} \ln \left (3\right )}{4}\right ) {\mathrm e}^{-x}\) | \(56\) |
Input:
int(1/4*((-x*ln(3)-x^3+3*x^2-13*x)*ln(x)-12*exp(x)+(x*exp(3)+1)*ln(3)+(x^3 -3*x^2+13*x)*exp(3)+x^2-x+12)/exp(x)/x,x,method=_RETURNVERBOSE)
Output:
1/4*(x*exp(3)+x^2*ln(x)+(ln(3)+12)*ln(x)-x*ln(x)-x^2*exp(3)-12*exp(3)-exp( 3)*ln(3))/exp(x)-3*ln(x)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=-\frac {1}{4} \, {\left ({\left (x^{2} - x + 12\right )} e^{3} + e^{3} \log \left (3\right ) - {\left (x^{2} - x - 12 \, e^{x} + \log \left (3\right ) + 12\right )} \log \left (x\right )\right )} e^{\left (-x\right )} \] Input:
integrate(1/4*((-x*log(3)-x^3+3*x^2-13*x)*log(x)-12*exp(x)+(x*exp(3)+1)*lo g(3)+(x^3-3*x^2+13*x)*exp(3)+x^2-x+12)/exp(x)/x,x, algorithm="fricas")
Output:
-1/4*((x^2 - x + 12)*e^3 + e^3*log(3) - (x^2 - x - 12*e^x + log(3) + 12)*l og(x))*e^(-x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\frac {\left (x^{2} \log {\left (x \right )} - x^{2} e^{3} - x \log {\left (x \right )} + x e^{3} + \log {\left (3 \right )} \log {\left (x \right )} + 12 \log {\left (x \right )} - 12 e^{3} - e^{3} \log {\left (3 \right )}\right ) e^{- x}}{4} - 3 \log {\left (x \right )} \] Input:
integrate(1/4*((-x*ln(3)-x**3+3*x**2-13*x)*ln(x)-12*exp(x)+(x*exp(3)+1)*ln (3)+(x**3-3*x**2+13*x)*exp(3)+x**2-x+12)/exp(x)/x,x)
Output:
(x**2*log(x) - x**2*exp(3) - x*log(x) + x*exp(3) + log(3)*log(x) + 12*log( x) - 12*exp(3) - exp(3)*log(3))*exp(-x)/4 - 3*log(x)
\[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\int { \frac {{\left (x^{2} + {\left (x^{3} - 3 \, x^{2} + 13 \, x\right )} e^{3} + {\left (x e^{3} + 1\right )} \log \left (3\right ) - {\left (x^{3} - 3 \, x^{2} + x \log \left (3\right ) + 13 \, x\right )} \log \left (x\right ) - x - 12 \, e^{x} + 12\right )} e^{\left (-x\right )}}{4 \, x} \,d x } \] Input:
integrate(1/4*((-x*log(3)-x^3+3*x^2-13*x)*log(x)-12*exp(x)+(x*exp(3)+1)*lo g(3)+(x^3-3*x^2+13*x)*exp(3)+x^2-x+12)/exp(x)/x,x, algorithm="maxima")
Output:
1/4*(x^2 - x - 1)*e^(-x)*log(x) + 1/4*e^(-x)*log(3)*log(x) - 1/4*(x^2*e^3 + 2*x*e^3 + 2*e^3)*e^(-x) + 3/4*(x*e^3 + e^3)*e^(-x) - 1/4*(x + 1)*e^(-x) - 1/4*e^(-x + 3)*log(3) + 13/4*e^(-x)*log(x) - 1/4*Ei(-x) + 1/4*e^(-x) - 1 3/4*e^(-x + 3) - 1/4*integrate((x^2 - x - 1)*e^(-x)/x, x) - 3*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (26) = 52\).
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.53 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\frac {1}{4} \, x^{2} e^{\left (-x\right )} \log \left (x\right ) - \frac {1}{4} \, x^{2} e^{\left (-x + 3\right )} - \frac {1}{4} \, x e^{\left (-x\right )} \log \left (x\right ) + \frac {1}{4} \, e^{\left (-x\right )} \log \left (3\right ) \log \left (x\right ) + \frac {1}{4} \, x e^{\left (-x + 3\right )} - \frac {1}{4} \, e^{\left (-x + 3\right )} \log \left (3\right ) + 3 \, e^{\left (-x\right )} \log \left (x\right ) - 3 \, e^{\left (-x + 3\right )} - 3 \, \log \left (x\right ) \] Input:
integrate(1/4*((-x*log(3)-x^3+3*x^2-13*x)*log(x)-12*exp(x)+(x*exp(3)+1)*lo g(3)+(x^3-3*x^2+13*x)*exp(3)+x^2-x+12)/exp(x)/x,x, algorithm="giac")
Output:
1/4*x^2*e^(-x)*log(x) - 1/4*x^2*e^(-x + 3) - 1/4*x*e^(-x)*log(x) + 1/4*e^( -x)*log(3)*log(x) + 1/4*x*e^(-x + 3) - 1/4*e^(-x + 3)*log(3) + 3*e^(-x)*lo g(x) - 3*e^(-x + 3) - 3*log(x)
Time = 2.54 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\frac {x\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^3-\ln \left (x\right )\right )}{4}-\frac {{\mathrm {e}}^{-x}\,\left (12\,{\mathrm {e}}^x\,\ln \left (x\right )-\ln \left (x\right )\,\left (\ln \left (3\right )+12\right )+{\mathrm {e}}^3\,\left (\ln \left (3\right )+12\right )\right )}{4}-\frac {x^2\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^3-\ln \left (x\right )\right )}{4} \] Input:
int((exp(-x)*((exp(3)*(13*x - 3*x^2 + x^3))/4 - 3*exp(x) - x/4 + (log(3)*( x*exp(3) + 1))/4 - (log(x)*(13*x + x*log(3) - 3*x^2 + x^3))/4 + x^2/4 + 3) )/x,x)
Output:
(x*exp(-x)*(exp(3) - log(x)))/4 - (exp(-x)*(12*exp(x)*log(x) - log(x)*(log (3) + 12) + exp(3)*(log(3) + 12)))/4 - (x^2*exp(-x)*(exp(3) - log(x)))/4
Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-x} \left (12-12 e^x-x+x^2+e^3 \left (13 x-3 x^2+x^3\right )+\left (1+e^3 x\right ) \log (3)+\left (-13 x+3 x^2-x^3-x \log (3)\right ) \log (x)\right )}{4 x} \, dx=\frac {-12 e^{x} \mathrm {log}\left (x \right )+\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )+\mathrm {log}\left (x \right ) x^{2}-\mathrm {log}\left (x \right ) x +12 \,\mathrm {log}\left (x \right )-\mathrm {log}\left (3\right ) e^{3}-e^{3} x^{2}+e^{3} x -12 e^{3}}{4 e^{x}} \] Input:
int(1/4*((-x*log(3)-x^3+3*x^2-13*x)*log(x)-12*exp(x)+(x*exp(3)+1)*log(3)+( x^3-3*x^2+13*x)*exp(3)+x^2-x+12)/exp(x)/x,x)
Output:
( - 12*e**x*log(x) + log(x)*log(3) + log(x)*x**2 - log(x)*x + 12*log(x) - log(3)*e**3 - e**3*x**2 + e**3*x - 12*e**3)/(4*e**x)