Integrand size = 75, antiderivative size = 27 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x} \] Output:
ln(x)-1/2*ln(4*ln(-3+x)/x/exp(exp(x)))/x
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {1}{2} \left (2 \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x}\right ) \] Input:
Integrate[(-x + (-3 - 5*x + 2*x^2 + E^x*(-3*x + x^2))*Log[-3 + x] + (-3 + x)*Log[-3 + x]*Log[(4*Log[-3 + x])/(E^E^x*x)])/((-6*x^2 + 2*x^3)*Log[-3 + x]),x]
Output:
(2*Log[x] - Log[(4*Log[-3 + x])/(E^E^x*x)]/x)/2
Time = 1.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2026, 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.
\(\displaystyle \int \frac {\left (2 x^2+e^x \left (x^2-3 x\right )-5 x-3\right ) \log (x-3)-x+(x-3) \log (x-3) \log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{\left (2 x^3-6 x^2\right ) \log (x-3)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (2 x^2+e^x \left (x^2-3 x\right )-5 x-3\right ) \log (x-3)-x+(x-3) \log (x-3) \log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{x^2 (2 x-6) \log (x-3)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2 \log (x-3)-x-5 x \log (x-3)+x \log (x-3) \log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )-3 \log (x-3)-3 \log (x-3) \log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{2 (x-3) x^2 \log (x-3)}+\frac {e^x}{2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{2 x}\) |
Input:
Int[(-x + (-3 - 5*x + 2*x^2 + E^x*(-3*x + x^2))*Log[-3 + x] + (-3 + x)*Log [-3 + x]*Log[(4*Log[-3 + x])/(E^E^x*x)])/((-6*x^2 + 2*x^3)*Log[-3 + x]),x]
Output:
Log[x] - Log[(4*Log[-3 + x])/(E^E^x*x)]/(2*x)
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 2.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {-12 x \ln \left (x \right )+6 \ln \left (\frac {4 \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{x}\right )}{12 x}\) | \(29\) |
risch | \(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )}{2 x}+\frac {-i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{3}+i \pi \,\operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-i \pi \,\operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{2}+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (-3+x \right )}{x}\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \ln \left (-3+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right )\right ) \operatorname {csgn}\left (i \ln \left (-3+x \right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )+4 x \ln \left (x \right )-4 \ln \left (2\right )+2 \ln \left (x \right )-2 \ln \left (\ln \left (-3+x \right )\right )}{4 x}\) | \(271\) |
Input:
int(((-3+x)*ln(-3+x)*ln(4*ln(-3+x)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x^2- 5*x-3)*ln(-3+x)-x)/(2*x^3-6*x^2)/ln(-3+x),x,method=_RETURNVERBOSE)
Output:
-1/12*(-12*x*ln(x)+6*ln(4*ln(-3+x)/x/exp(exp(x))))/x
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {2 \, x \log \left (x\right ) - \log \left (\frac {4 \, e^{\left (-e^{x}\right )} \log \left (x - 3\right )}{x}\right )}{2 \, x} \] Input:
integrate(((-3+x)*log(-3+x)*log(4*log(-3+x)/x/exp(exp(x)))+((x^2-3*x)*exp( x)+2*x^2-5*x-3)*log(-3+x)-x)/(2*x^3-6*x^2)/log(-3+x),x, algorithm="fricas" )
Output:
1/2*(2*x*log(x) - log(4*e^(-e^x)*log(x - 3)/x))/x
Time = 3.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\log {\left (x \right )} - \frac {\log {\left (\frac {4 e^{- e^{x}} \log {\left (x - 3 \right )}}{x} \right )}}{2 x} \] Input:
integrate(((-3+x)*ln(-3+x)*ln(4*ln(-3+x)/x/exp(exp(x)))+((x**2-3*x)*exp(x) +2*x**2-5*x-3)*ln(-3+x)-x)/(2*x**3-6*x**2)/ln(-3+x),x)
Output:
log(x) - log(4*exp(-exp(x))*log(x - 3)/x)/(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {x \log \left (x - 3\right ) + {\left (5 \, x + 3\right )} \log \left (x\right ) + 3 \, e^{x} - 6 \, \log \left (2\right ) - 3 \, \log \left (\log \left (x - 3\right )\right ) + 3}{6 \, x} - \frac {1}{2 \, x} - \frac {1}{6} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x\right ) \] Input:
integrate(((-3+x)*log(-3+x)*log(4*log(-3+x)/x/exp(exp(x)))+((x^2-3*x)*exp( x)+2*x^2-5*x-3)*log(-3+x)-x)/(2*x^3-6*x^2)/log(-3+x),x, algorithm="maxima" )
Output:
1/6*(x*log(x - 3) + (5*x + 3)*log(x) + 3*e^x - 6*log(2) - 3*log(log(x - 3) ) + 3)/x - 1/2/x - 1/6*log(x - 3) + 1/6*log(x)
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {2 \, x \log \left (x\right ) + e^{x} + \log \left (x\right ) - \log \left (4 \, \log \left (x - 3\right )\right )}{2 \, x} \] Input:
integrate(((-3+x)*log(-3+x)*log(4*log(-3+x)/x/exp(exp(x)))+((x^2-3*x)*exp( x)+2*x^2-5*x-3)*log(-3+x)-x)/(2*x^3-6*x^2)/log(-3+x),x, algorithm="giac")
Output:
1/2*(2*x*log(x) + e^x + log(x) - log(4*log(x - 3)))/x
Time = 3.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\ln \left (x\right )+\frac {{\mathrm {e}}^x}{2\,x}-\frac {\ln \left (\frac {\ln \left (x-3\right )}{x}\right )}{2\,x}-\frac {\ln \left (2\right )}{x} \] Input:
int((x + log(x - 3)*(5*x + exp(x)*(3*x - x^2) - 2*x^2 + 3) - log(x - 3)*lo g((4*log(x - 3)*exp(-exp(x)))/x)*(x - 3))/(log(x - 3)*(6*x^2 - 2*x^3)),x)
Output:
log(x) + exp(x)/(2*x) - log(log(x - 3)/x)/(2*x) - log(2)/x
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{\left (-6 x^2+2 x^3\right ) \log (-3+x)} \, dx=\frac {-\mathrm {log}\left (\frac {4 \,\mathrm {log}\left (x -3\right )}{e^{e^{x}} x}\right )+2 \,\mathrm {log}\left (x \right ) x}{2 x} \] Input:
int(((-3+x)*log(-3+x)*log(4*log(-3+x)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x ^2-5*x-3)*log(-3+x)-x)/(2*x^3-6*x^2)/log(-3+x),x)
Output:
( - log((4*log(x - 3))/(e**(e**x)*x)) + 2*log(x)*x)/(2*x)