Integrand size = 43, antiderivative size = 26 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 e^{-x} \left (4+x+\frac {1}{4} e^5 (\log (3)+\log (4))-\log (x)\right ) \] Output:
3*(4+1/4*exp(5)*(ln(3)+2*ln(2))+x-ln(x))/exp(x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3}{4} e^{-x} \left (16+4 x+e^5 \log (12)-4 \log (x)\right ) \] Input:
Integrate[(-12 - 36*x - 12*x^2 - 3*E^5*x*Log[3] - 3*E^5*x*Log[4] + 12*x*Lo g[x])/(4*E^x*x),x]
Output:
(3*(16 + 4*x + E^5*Log[12] - 4*Log[x]))/(4*E^x)
Time = 0.58 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6, 6, 27, 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.
\(\displaystyle \int \frac {e^{-x} \left (-12 x^2-36 x+12 x \log (x)-3 e^5 x \log (4)-3 e^5 x \log (3)-12\right )}{4 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{-x} \left (-12 x^2+12 x \log (x)-3 e^5 x \log (4)+x \left (-36-3 e^5 \log (3)\right )-12\right )}{4 x}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{-x} \left (-12 x^2+12 x \log (x)+x \left (-36-3 e^5 \log (3)-3 e^5 \log (4)\right )-12\right )}{4 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {3 e^{-x} \left (4 x^2-4 \log (x) x+\left (12+e^5 \log (12)\right ) x+4\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{4} \int \frac {e^{-x} \left (4 x^2-4 \log (x) x+\left (12+e^5 \log (12)\right ) x+4\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3}{4} \int \left (\frac {e^{-x} \left (4 x^2+\left (12+e^5 \log (12)\right ) x+4\right )}{x}-4 e^{-x} \log (x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{4} \left (-4 e^{-x} x-4 e^{-x}+4 e^{-x} \log (x)-e^{-x} \left (12+e^5 \log (12)\right )\right )\) |
Input:
Int[(-12 - 36*x - 12*x^2 - 3*E^5*x*Log[3] - 3*E^5*x*Log[4] + 12*x*Log[x])/ (4*E^x*x),x]
Output:
(-3*(-4/E^x - (4*x)/E^x - (12 + E^5*Log[12])/E^x + (4*Log[x])/E^x))/4
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]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\left (3 x +12-3 \ln \left (x \right )+\frac {3 \,{\mathrm e}^{5} \ln \left (2\right )}{2}+\frac {3 \ln \left (3\right ) {\mathrm e}^{5}}{4}\right ) {\mathrm e}^{-x}\) | \(27\) |
parallelrisch | \(\frac {\left (48+3 \ln \left (3\right ) {\mathrm e}^{5}+6 \,{\mathrm e}^{5} \ln \left (2\right )+12 x -12 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{4}\) | \(28\) |
risch | \(-3 \ln \left (x \right ) {\mathrm e}^{-x}+\frac {3 \left (\ln \left (3\right ) {\mathrm e}^{5}+2 \,{\mathrm e}^{5} \ln \left (2\right )+4 x +16\right ) {\mathrm e}^{-x}}{4}\) | \(32\) |
orering | \(-\frac {\left (2 x^{2}-2 x -1\right ) \left (12 x \ln \left (x \right )-6 x \,{\mathrm e}^{5} \ln \left (2\right )-3 x \,{\mathrm e}^{5} \ln \left (3\right )-12 x^{2}-36 x -12\right ) {\mathrm e}^{-x}}{4 \left (x^{2}-x -1\right ) x}-\frac {\left (-1+x \right ) x \left (\frac {\left (12 \ln \left (x \right )-24-6 \,{\mathrm e}^{5} \ln \left (2\right )-3 \ln \left (3\right ) {\mathrm e}^{5}-24 x \right ) {\mathrm e}^{-x}}{4 x}-\frac {\left (12 x \ln \left (x \right )-6 x \,{\mathrm e}^{5} \ln \left (2\right )-3 x \,{\mathrm e}^{5} \ln \left (3\right )-12 x^{2}-36 x -12\right ) {\mathrm e}^{-x}}{4 x}-\frac {\left (12 x \ln \left (x \right )-6 x \,{\mathrm e}^{5} \ln \left (2\right )-3 x \,{\mathrm e}^{5} \ln \left (3\right )-12 x^{2}-36 x -12\right ) {\mathrm e}^{-x}}{4 x^{2}}\right )}{x^{2}-x -1}\) | \(183\) |
Input:
int(1/4*(12*x*ln(x)-6*x*exp(5)*ln(2)-3*x*exp(5)*ln(3)-12*x^2-36*x-12)/exp( x)/x,x,method=_RETURNVERBOSE)
Output:
(3*x+12-3*ln(x)+3/2*exp(5)*ln(2)+3/4*ln(3)*exp(5))/exp(x)
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3}{4} \, {\left (e^{5} \log \left (3\right ) + 2 \, e^{5} \log \left (2\right ) + 4 \, x + 16\right )} e^{\left (-x\right )} - 3 \, e^{\left (-x\right )} \log \left (x\right ) \] Input:
integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x -12)/exp(x)/x,x, algorithm="fricas")
Output:
3/4*(e^5*log(3) + 2*e^5*log(2) + 4*x + 16)*e^(-x) - 3*e^(-x)*log(x)
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {\left (12 x - 12 \log {\left (x \right )} + 48 + 3 e^{5} \log {\left (3 \right )} + 6 e^{5} \log {\left (2 \right )}\right ) e^{- x}}{4} \] Input:
integrate(1/4*(12*x*ln(x)-6*x*exp(5)*ln(2)-3*x*exp(5)*ln(3)-12*x**2-36*x-1 2)/exp(x)/x,x)
Output:
(12*x - 12*log(x) + 48 + 3*exp(5)*log(3) + 6*exp(5)*log(2))*exp(-x)/4
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \left (3\right ) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \left (2\right ) - 3 \, e^{\left (-x\right )} \log \left (x\right ) + 9 \, e^{\left (-x\right )} \] Input:
integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x -12)/exp(x)/x,x, algorithm="maxima")
Output:
3*(x + 1)*e^(-x) + 3/4*e^(-x + 5)*log(3) + 3/2*e^(-x + 5)*log(2) - 3*e^(-x )*log(x) + 9*e^(-x)
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 \, x e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \left (3\right ) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \left (2\right ) - 3 \, e^{\left (-x\right )} \log \left (x\right ) + 12 \, e^{\left (-x\right )} \] Input:
integrate(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x -12)/exp(x)/x,x, algorithm="giac")
Output:
3*x*e^(-x) + 3/4*e^(-x + 5)*log(3) + 3/2*e^(-x + 5)*log(2) - 3*e^(-x)*log( x) + 12*e^(-x)
Time = 3.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3\,{\mathrm {e}}^{-x}\,\left (4\,x-4\,\ln \left (x\right )+2\,{\mathrm {e}}^5\,\ln \left (2\right )+{\mathrm {e}}^5\,\ln \left (3\right )+16\right )}{4} \] Input:
int(-(exp(-x)*(9*x - 3*x*log(x) + 3*x^2 + (3*x*exp(5)*log(2))/2 + (3*x*exp (5)*log(3))/4 + 3))/x,x)
Output:
(3*exp(-x)*(4*x - 4*log(x) + 2*exp(5)*log(2) + exp(5)*log(3) + 16))/4
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {-3 \,\mathrm {log}\left (x \right )+\frac {3 \,\mathrm {log}\left (3\right ) e^{5}}{4}+\frac {3 \,\mathrm {log}\left (2\right ) e^{5}}{2}+3 x +12}{e^{x}} \] Input:
int(1/4*(12*x*log(x)-6*x*exp(5)*log(2)-3*x*exp(5)*log(3)-12*x^2-36*x-12)/e xp(x)/x,x)
Output:
(3*( - 4*log(x) + log(3)*e**5 + 2*log(2)*e**5 + 4*x + 16))/(4*e**x)