Integrand size = 59, antiderivative size = 27 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {4+x+\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x} \] Output:
(ln(2*x/exp(x)/(4*exp(49)^2-4*x))+4+x)/x
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x} \] Input:
Integrate[(E^98*(-3 - x) + 4*x + x^2 + (-E^98 + x)*Log[x/(E^x*(2*E^98 - 2* x))])/(E^98*x^2 - x^3),x]
Output:
4/x + Log[x/(E^x*(2*E^98 - 2*x))]/x
Time = 0.53 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, 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.
\(\displaystyle \int \frac {x^2+4 x+e^{98} (-x-3)+\left (x-e^{98}\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^2+4 x+e^{98} (-x-3)+\left (x-e^{98}\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{\left (e^{98}-x\right ) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2+\left (4-e^{98}\right ) x-3 e^{98}}{\left (e^{98}-x\right ) x^2}-\frac {\log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{x}+\frac {\log \left (\frac {e^{-x} x}{2 \left (e^{98}-x\right )}\right )}{x}\) |
Input:
Int[(E^98*(-3 - x) + 4*x + x^2 + (-E^98 + x)*Log[x/(E^x*(2*E^98 - 2*x))])/ (E^98*x^2 - x^3),x]
Output:
4/x + Log[x/(2*E^x*(E^98 - x))]/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 = 0.58 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {4+\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}\) | \(26\) |
parallelrisch | \(\frac {4+\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}\) | \(26\) |
default | \(\frac {\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}+\ln \left ({\mathrm e}^{98}-x \right )-{\mathrm e}^{98} \left (-\frac {{\mathrm e}^{-196} {\mathrm e}^{98}}{x}+{\mathrm e}^{-196} \ln \left (x \right )-{\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )\right )+\frac {1}{{\mathrm e}^{98}-x}+{\mathrm e}^{-98} \ln \left (x \right )-2 \,{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )-{\mathrm e}^{98} \left ({\mathrm e}^{-196} \ln \left (x \right )-\frac {{\mathrm e}^{-196} {\mathrm e}^{98}}{-{\mathrm e}^{98}+x}-{\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )\right )+{\mathrm e}^{98} \left ({\mathrm e}^{-98} \ln \left (x \right )-{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )\right )+\frac {3 \,{\mathrm e}^{-98} {\mathrm e}^{98}}{x}-\left ({\mathrm e}^{98}-1\right ) {\mathrm e}^{-98} \ln \left (x \right )\) | \(182\) |
parts | \(\frac {\ln \left (\frac {x \,{\mathrm e}^{-x}}{2 \,{\mathrm e}^{98}-2 x}\right )}{x}+\ln \left ({\mathrm e}^{98}-x \right )-{\mathrm e}^{98} \left (-\frac {{\mathrm e}^{-196} {\mathrm e}^{98}}{x}+{\mathrm e}^{-196} \ln \left (x \right )-{\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )\right )+\frac {1}{{\mathrm e}^{98}-x}+{\mathrm e}^{-98} \ln \left (x \right )-2 \,{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )-{\mathrm e}^{98} \left ({\mathrm e}^{-196} \ln \left (x \right )-\frac {{\mathrm e}^{-196} {\mathrm e}^{98}}{-{\mathrm e}^{98}+x}-{\mathrm e}^{-196} \ln \left (-{\mathrm e}^{98}+x \right )\right )+{\mathrm e}^{98} \left ({\mathrm e}^{-98} \ln \left (x \right )-{\mathrm e}^{-98} \ln \left (-{\mathrm e}^{98}+x \right )\right )+\frac {3 \,{\mathrm e}^{-98} {\mathrm e}^{98}}{x}-\left ({\mathrm e}^{98}-1\right ) {\mathrm e}^{-98} \ln \left (x \right )\) | \(182\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{x}\right )}{x}-\frac {i \pi \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}-i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{3}+i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )-i \pi \operatorname {csgn}\left (\frac {i x \,{\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{98}-x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-x}}{{\mathrm e}^{98}-x}\right )^{2}-8+2 \ln \left (2\right )-2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{98}-x \right )}{2 x}\) | \(292\) |
Input:
int(((-exp(49)^2+x)*ln(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4* x)/(x^2*exp(49)^2-x^3),x,method=_RETURNVERBOSE)
Output:
(4+ln(x/(2*exp(49)^2-2*x)/exp(x)))/x
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {\log \left (-\frac {x e^{\left (-x\right )}}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \] Input:
integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2 +x^2+4*x)/(x^2*exp(49)^2-x^3),x, algorithm="fricas")
Output:
(log(-1/2*x*e^(-x)/(x - e^98)) + 4)/x
Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {\log {\left (\frac {x e^{- x}}{- 2 x + 2 e^{98}} \right )}}{x} + \frac {4}{x} \] Input:
integrate(((-exp(49)**2+x)*ln(x/(2*exp(49)**2-2*x)/exp(x))+(-3-x)*exp(49)* *2+x**2+4*x)/(x**2*exp(49)**2-x**3),x)
Output:
log(x*exp(-x)/(-2*x + 2*exp(98)))/x + 4/x
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (23) = 46\).
Time = 0.10 (sec) , antiderivative size = 264, normalized size of antiderivative = 9.78 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx={\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \left (x\right ) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \left (x\right )\right )} e^{98} + 3 \, {\left (e^{\left (-196\right )} \log \left (x - e^{98}\right ) - e^{\left (-196\right )} \log \left (x\right ) + \frac {e^{\left (-98\right )}}{x}\right )} e^{98} - \frac {1}{2} \, {\left (2 \, {\left (x - e^{98} - \log \left (x\right )\right )} \log \left (x - e^{98}\right ) + \log \left (x - e^{98}\right )^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}\right )} e^{\left (-98\right )} - 4 \, e^{\left (-98\right )} \log \left (x - e^{98}\right ) + 4 \, e^{\left (-98\right )} \log \left (x\right ) - {\left (e^{\left (-98\right )} \log \left (x - e^{98}\right ) - e^{\left (-98\right )} \log \left (x\right )\right )} \log \left (-\frac {x}{2 \, {\left (x e^{x} - e^{\left (x + 98\right )}\right )}}\right ) + \frac {{\left (x \log \left (x - e^{98}\right )^{2} + x \log \left (x\right )^{2} + 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )} - x \log \left (x\right )\right )} \log \left (x - e^{98}\right ) - 2 \, {\left (x^{2} - x {\left (e^{98} - 1\right )}\right )} \log \left (x\right ) + 2 \, e^{98}\right )} e^{\left (-98\right )}}{2 \, x} - \log \left (x - e^{98}\right ) \] Input:
integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2 +x^2+4*x)/(x^2*exp(49)^2-x^3),x, algorithm="maxima")
Output:
(e^(-196)*log(x - e^98) - e^(-196)*log(x) + e^(-98)/x)*e^98*log(-1/2*x/(x* e^x - e^(x + 98))) + (e^(-98)*log(x - e^98) - e^(-98)*log(x))*e^98 + 3*(e^ (-196)*log(x - e^98) - e^(-196)*log(x) + e^(-98)/x)*e^98 - 1/2*(2*(x - e^9 8 - log(x))*log(x - e^98) + log(x - e^98)^2 - 2*x*log(x) + log(x)^2)*e^(-9 8) - 4*e^(-98)*log(x - e^98) + 4*e^(-98)*log(x) - (e^(-98)*log(x - e^98) - e^(-98)*log(x))*log(-1/2*x/(x*e^x - e^(x + 98))) + 1/2*(x*log(x - e^98)^2 + x*log(x)^2 + 2*(x^2 - x*(e^98 - 1) - x*log(x))*log(x - e^98) - 2*(x^2 - x*(e^98 - 1))*log(x) + 2*e^98)*e^(-98)/x - log(x - e^98)
Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {\log \left (-\frac {x}{2 \, {\left (x - e^{98}\right )}}\right ) + 4}{x} \] Input:
integrate(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2 +x^2+4*x)/(x^2*exp(49)^2-x^3),x, algorithm="giac")
Output:
(log(-1/2*x/(x - e^98)) + 4)/x
Time = 2.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {\ln \left (-\frac {x\,{\mathrm {e}}^{-x}}{2\,\left (x-{\mathrm {e}}^{98}\right )}\right )+4}{x} \] Input:
int((4*x - exp(98)*(x + 3) + x^2 + log(-(x*exp(-x))/(2*x - 2*exp(98)))*(x - exp(98)))/(x^2*exp(98) - x^3),x)
Output:
(log(-(x*exp(-x))/(2*(x - exp(98)))) + 4)/x
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{98} (-3-x)+4 x+x^2+\left (-e^{98}+x\right ) \log \left (\frac {e^{-x} x}{2 e^{98}-2 x}\right )}{e^{98} x^2-x^3} \, dx=\frac {\mathrm {log}\left (\frac {x}{2 e^{x} e^{98}-2 e^{x} x}\right )+4}{x} \] Input:
int(((-exp(49)^2+x)*log(x/(2*exp(49)^2-2*x)/exp(x))+(-3-x)*exp(49)^2+x^2+4 *x)/(x^2*exp(49)^2-x^3),x)
Output:
(log(x/(2*e**x*e**98 - 2*e**x*x)) + 4)/x