Integrand size = 39, antiderivative size = 26 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=-e^5+\left (-\frac {1}{2 x^2}+x-6 e^{-1+x} x\right ) \log (4) \] Output:
2*ln(2)*(x-1/2/x^2-6*x/exp(1-x))-exp(5)
Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=\left (-\frac {1}{2 x^2}+x-6 e^{-1+x} x\right ) \log (4) \] Input:
Integrate[(E^(-1 + x)*(E^(1 - x)*(1 + x^3)*Log[4] + (-6*x^3 - 6*x^4)*Log[4 ]))/x^3,x]
Output:
(-1/2*1/x^2 + x - 6*E^(-1 + x)*x)*Log[4]
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {e^{x-1} \left (e^{1-x} \left (x^3+1\right ) \log (4)+\left (-6 x^4-6 x^3\right ) \log (4)\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {1}{x^3}-6 e^{x-1} (x+1)+1\right ) \log (4)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (4) \int \left (-6 e^{x-1} (x+1)+\frac {1}{x^3}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (-\frac {1}{2 x^2}+x+6 e^{x-1}-6 e^{x-1} (x+1)\right ) \log (4)\) |
Input:
Int[(E^(-1 + x)*(E^(1 - x)*(1 + x^3)*Log[4] + (-6*x^3 - 6*x^4)*Log[4]))/x^ 3,x]
Output:
(6*E^(-1 + x) - 1/(2*x^2) + x - 6*E^(-1 + x)*(1 + x))*Log[4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \(2 x \ln \left (2\right )-\frac {\ln \left (2\right )}{x^{2}}-12 x \ln \left (2\right ) {\mathrm e}^{-1+x}\) | \(23\) |
parts | \(-12 x \ln \left (2\right ) {\mathrm e}^{-1+x}+2 \ln \left (2\right ) \left (x -\frac {1}{2 x^{2}}\right )\) | \(26\) |
norman | \(\frac {\left (-12 x^{3} \ln \left (2\right )-\ln \left (2\right ) {\mathrm e}^{1-x}+2 x^{3} \ln \left (2\right ) {\mathrm e}^{1-x}\right ) {\mathrm e}^{-1+x}}{x^{2}}\) | \(44\) |
parallelrisch | \(\frac {\left (-12 x^{3} \ln \left (2\right )-\ln \left (2\right ) {\mathrm e}^{1-x}+2 x^{3} \ln \left (2\right ) {\mathrm e}^{1-x}\right ) {\mathrm e}^{-1+x}}{x^{2}}\) | \(44\) |
derivativedivides | \(-\frac {2 \ln \left (2\right )}{x^{2}}+24 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{-1+x} \left (-1-x \right )}{2 x^{2}}+\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )+6 \ln \left (2\right ) \left (\frac {1}{2 x^{2}}-\frac {1}{x}\right )-6 \ln \left (2\right ) \left (-\ln \left (-x \right )+\frac {1}{2 x^{2}}-\frac {2}{x}\right )+2 \ln \left (2\right ) \left (-1+x -3 \ln \left (-x \right )+\frac {1}{2 x^{2}}-\frac {3}{x}\right )-\frac {42 \ln \left (2\right ) {\mathrm e}^{-1+x} \left (1-x \right )}{x^{2}}+42 \ln \left (2\right ) {\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )+108 \ln \left (2\right ) \left (\frac {{\mathrm e}^{-1+x} \left (1-3 x \right )}{2 x^{2}}-\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )-60 \ln \left (2\right ) \left ({\mathrm e}^{-1+x}+\frac {{\mathrm e}^{-1+x} \left (-5 x +1\right )}{2 x^{2}}+\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )+12 \ln \left (2\right ) \left ({\mathrm e}^{-1+x} \left (5-x \right )+\frac {{\mathrm e}^{-1+x} \left (1-7 x \right )}{2 x^{2}}+\frac {5 \,{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )\) | \(221\) |
default | \(-\frac {2 \ln \left (2\right )}{x^{2}}+24 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{-1+x} \left (-1-x \right )}{2 x^{2}}+\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )+6 \ln \left (2\right ) \left (\frac {1}{2 x^{2}}-\frac {1}{x}\right )-6 \ln \left (2\right ) \left (-\ln \left (-x \right )+\frac {1}{2 x^{2}}-\frac {2}{x}\right )+2 \ln \left (2\right ) \left (-1+x -3 \ln \left (-x \right )+\frac {1}{2 x^{2}}-\frac {3}{x}\right )-\frac {42 \ln \left (2\right ) {\mathrm e}^{-1+x} \left (1-x \right )}{x^{2}}+42 \ln \left (2\right ) {\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )+108 \ln \left (2\right ) \left (\frac {{\mathrm e}^{-1+x} \left (1-3 x \right )}{2 x^{2}}-\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )-60 \ln \left (2\right ) \left ({\mathrm e}^{-1+x}+\frac {{\mathrm e}^{-1+x} \left (-5 x +1\right )}{2 x^{2}}+\frac {{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )+12 \ln \left (2\right ) \left ({\mathrm e}^{-1+x} \left (5-x \right )+\frac {{\mathrm e}^{-1+x} \left (1-7 x \right )}{2 x^{2}}+\frac {5 \,{\mathrm e}^{-1} \operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )\) | \(221\) |
orering | \(\frac {\left (2 x^{5}+6 x^{4}+4 x^{3}-x^{2}-2 x +6\right ) \left (2 \left (x^{3}+1\right ) \ln \left (2\right ) {\mathrm e}^{1-x}+2 \left (-6 x^{4}-6 x^{3}\right ) \ln \left (2\right )\right ) {\mathrm e}^{-1+x}}{2 x^{2} \left (x^{3}-x +3\right ) \left (1+x \right )^{2}}-\frac {x^{2} \left (2 x^{3}+2 x -3\right ) \left (\frac {\left (6 x^{2} \ln \left (2\right ) {\mathrm e}^{1-x}-2 \left (x^{3}+1\right ) \ln \left (2\right ) {\mathrm e}^{1-x}+2 \left (-24 x^{3}-18 x^{2}\right ) \ln \left (2\right )\right ) {\mathrm e}^{-1+x}}{x^{3}}-\frac {3 \left (2 \left (x^{3}+1\right ) \ln \left (2\right ) {\mathrm e}^{1-x}+2 \left (-6 x^{4}-6 x^{3}\right ) \ln \left (2\right )\right ) {\mathrm e}^{-1+x}}{x^{4}}+\frac {\left (2 \left (x^{3}+1\right ) \ln \left (2\right ) {\mathrm e}^{1-x}+2 \left (-6 x^{4}-6 x^{3}\right ) \ln \left (2\right )\right ) {\mathrm e}^{-1+x}}{x^{3}}\right )}{2 \left (x^{3}-x +3\right ) \left (1+x \right )}\) | \(260\) |
Input:
int((2*(x^3+1)*ln(2)*exp(1-x)+2*(-6*x^4-6*x^3)*ln(2))/x^3/exp(1-x),x,metho d=_RETURNVERBOSE)
Output:
2*x*ln(2)-ln(2)/x^2-12*x*ln(2)*exp(-1+x)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=-\frac {12 \, x^{3} e^{\left (x - 1\right )} \log \left (2\right ) - {\left (2 \, x^{3} - 1\right )} \log \left (2\right )}{x^{2}} \] Input:
integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x) ,x, algorithm="fricas")
Output:
-(12*x^3*e^(x - 1)*log(2) - (2*x^3 - 1)*log(2))/x^2
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=- 12 x e^{x - 1} \log {\left (2 \right )} + 2 x \log {\left (2 \right )} - \frac {\log {\left (2 \right )}}{x^{2}} \] Input:
integrate((2*(x**3+1)*ln(2)*exp(1-x)+2*(-6*x**4-6*x**3)*ln(2))/x**3/exp(1- x),x)
Output:
-12*x*exp(x - 1)*log(2) + 2*x*log(2) - log(2)/x**2
Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=-12 \, {\left (x - 1\right )} e^{\left (x - 1\right )} \log \left (2\right ) + 2 \, x \log \left (2\right ) - 12 \, e^{\left (x - 1\right )} \log \left (2\right ) - \frac {\log \left (2\right )}{x^{2}} \] Input:
integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x) ,x, algorithm="maxima")
Output:
-12*(x - 1)*e^(x - 1)*log(2) + 2*x*log(2) - 12*e^(x - 1)*log(2) - log(2)/x ^2
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=\frac {{\left (2 \, x^{3} e \log \left (2\right ) - 12 \, x^{3} e^{x} \log \left (2\right ) - e \log \left (2\right )\right )} e^{\left (-1\right )}}{x^{2}} \] Input:
integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x) ,x, algorithm="giac")
Output:
(2*x^3*e*log(2) - 12*x^3*e^x*log(2) - e*log(2))*e^(-1)/x^2
Time = 3.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=x\,{\mathrm {e}}^{-1}\,\ln \left (2\right )\,\left (2\,\mathrm {e}-12\,{\mathrm {e}}^x\right )-\frac {\ln \left (2\right )}{x^2} \] Input:
int(-(exp(x - 1)*(2*log(2)*(6*x^3 + 6*x^4) - 2*exp(1 - x)*log(2)*(x^3 + 1) ))/x^3,x)
Output:
x*exp(-1)*log(2)*(2*exp(1) - 12*exp(x)) - log(2)/x^2
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-1+x} \left (e^{1-x} \left (1+x^3\right ) \log (4)+\left (-6 x^3-6 x^4\right ) \log (4)\right )}{x^3} \, dx=\frac {\mathrm {log}\left (2\right ) \left (-12 e^{x} x^{3}+2 e \,x^{3}-e \right )}{e \,x^{2}} \] Input:
int((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x),x)
Output:
(log(2)*( - 12*e**x*x**3 + 2*e*x**3 - e))/(e*x**2)