Integrand size = 52, antiderivative size = 32 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=\frac {-3-e+x}{x}-(5+x) \log \left (\frac {e^{e^4}}{x}\right )-\log (2+x) \] Output:
(x-exp(1)-3)/x-ln(2+x)-ln(exp(exp(4))/x)*(5+x)
Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=-\frac {3}{x}-\frac {e}{x}-e^4 x-x \log \left (\frac {1}{x}\right )+5 \log (x)-\log (2+x) \] Input:
Integrate[(6 + 13*x + 6*x^2 + x^3 + E*(2 + x) + (-2*x^2 - x^3)*Log[E^E^4/x ])/(2*x^2 + x^3),x]
Output:
-3/x - E/x - E^4*x - x*Log[x^(-1)] + 5*Log[x] - Log[2 + x]
Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, 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.
\(\displaystyle \int \frac {x^3+6 x^2+\left (-x^3-2 x^2\right ) \log \left (\frac {e^{e^4}}{x}\right )+13 x+e (x+2)+6}{x^3+2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^3+6 x^2+\left (-x^3-2 x^2\right ) \log \left (\frac {e^{e^4}}{x}\right )+13 x+e (x+2)+6}{x^2 (x+2)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^3+6 x^2+(13+e) x+2 (3+e)}{x^2 (x+2)}-\log \left (\frac {e^{e^4}}{x}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3+e}{x}-x \log \left (\frac {e^{e^4}}{x}\right )+5 \log (x)-\log (x+2)\) |
Input:
Int[(6 + 13*x + 6*x^2 + x^3 + E*(2 + x) + (-2*x^2 - x^3)*Log[E^E^4/x])/(2* x^2 + x^3),x]
Output:
-((3 + E)/x) - x*Log[E^E^4/x] + 5*Log[x] - Log[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 = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00
method | result | size |
parts | \(-\frac {{\mathrm e}+3}{x}+5 \ln \left (x \right )-\ln \left (2+x \right )-x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )\) | \(32\) |
risch | \(-x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+\frac {5 x \ln \left (x \right )-x \ln \left (2+x \right )-{\mathrm e}-3}{x}\) | \(35\) |
norman | \(\frac {-5 x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-{\mathrm e}-3}{x}-\ln \left (2+x \right )\) | \(42\) |
parallelrisch | \(\frac {-3-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-x \ln \left (2+x \right )-5 x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-{\mathrm e}}{x}\) | \(42\) |
derivativedivides | \(-{\mathrm e}^{{\mathrm e}^{4}} \left ({\mathrm e}^{-2 \,{\mathrm e}^{4}} \left (\frac {{\mathrm e}^{{\mathrm e}^{4}} {\mathrm e}}{x}+\frac {3 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}+\frac {\left (14 \,{\mathrm e}^{3 \,{\mathrm e}^{4}}-12 \,{\mathrm e}^{2 \,{\mathrm e}^{4}} {\mathrm e}^{{\mathrm e}^{4}}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )}{2}-{\mathrm e}^{4 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} x \,{\mathrm e}^{-{\mathrm e}^{4}}+4 \,{\mathrm e}^{3 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )\right )+x \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+{\mathrm e}^{-{\mathrm e}^{4}} x \right )\) | \(136\) |
default | \(-{\mathrm e}^{{\mathrm e}^{4}} \left ({\mathrm e}^{-2 \,{\mathrm e}^{4}} \left (\frac {{\mathrm e}^{{\mathrm e}^{4}} {\mathrm e}}{x}+\frac {3 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}+\frac {\left (14 \,{\mathrm e}^{3 \,{\mathrm e}^{4}}-12 \,{\mathrm e}^{2 \,{\mathrm e}^{4}} {\mathrm e}^{{\mathrm e}^{4}}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )}{2}-{\mathrm e}^{4 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} x \,{\mathrm e}^{-{\mathrm e}^{4}}+4 \,{\mathrm e}^{3 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )\right )+x \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+{\mathrm e}^{-{\mathrm e}^{4}} x \right )\) | \(136\) |
Input:
int(((-x^3-2*x^2)*ln(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2* x^2),x,method=_RETURNVERBOSE)
Output:
-(exp(1)+3)/x+5*ln(x)-ln(2+x)-x*ln(exp(exp(4))/x)
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=-\frac {x^{2} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right ) + x \log \left (x + 2\right ) - 5 \, x \log \left (x\right ) + e + 3}{x} \] Input:
integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/ (x^3+2*x^2),x, algorithm="fricas")
Output:
-(x^2*log(e^(e^4)/x) + x*log(x + 2) - 5*x*log(x) + e + 3)/x
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=- x \log {\left (\frac {e^{e^{4}}}{x} \right )} + 5 \log {\left (x \right )} - \log {\left (x + 2 \right )} + \frac {-3 - e}{x} \] Input:
integrate(((-x**3-2*x**2)*ln(exp(exp(4))/x)+(2+x)*exp(1)+x**3+6*x**2+13*x+ 6)/(x**3+2*x**2),x)
Output:
-x*log(exp(exp(4))/x) + 5*log(x) - log(x + 2) + (-3 - E)/x
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=-x e^{4} - \frac {1}{2} \, {\left (\frac {2}{x} - \log \left (x + 2\right ) + \log \left (x\right )\right )} e - \frac {1}{2} \, {\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} e + x \log \left (x\right ) - \frac {3}{x} - \log \left (x + 2\right ) + 5 \, \log \left (x\right ) \] Input:
integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/ (x^3+2*x^2),x, algorithm="maxima")
Output:
-x*e^4 - 1/2*(2/x - log(x + 2) + log(x))*e - 1/2*(log(x + 2) - log(x))*e + x*log(x) - 3/x - log(x + 2) + 5*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (31) = 62\).
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 4.38 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=-\frac {1}{2} \, {\left (2 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right ) + \frac {e^{\left (2 \, e^{4} + 1\right )} \log \left (\frac {2 \, e^{\left (e^{4} + 1\right )}}{x} + e^{\left (e^{4} + 1\right )}\right )}{x} + \frac {2 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {2 \, e^{\left (e^{4}\right )}}{x} + e^{\left (e^{4}\right )}\right )}{x} - \frac {e^{\left (2 \, e^{4} + 1\right )} \log \left (\frac {2 \, e^{\left (e^{4}\right )}}{x} + e^{\left (e^{4}\right )}\right )}{x} + \frac {8 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right )}{x} + \frac {6 \, e^{\left (2 \, e^{4}\right )}}{x^{2}} + \frac {2 \, e^{\left (2 \, e^{4} + 1\right )}}{x^{2}}\right )} x e^{\left (-2 \, e^{4}\right )} \] Input:
integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/ (x^3+2*x^2),x, algorithm="giac")
Output:
-1/2*(2*e^(2*e^4)*log(e^(e^4)/x) + e^(2*e^4 + 1)*log(2*e^(e^4 + 1)/x + e^( e^4 + 1))/x + 2*e^(2*e^4)*log(2*e^(e^4)/x + e^(e^4))/x - e^(2*e^4 + 1)*log (2*e^(e^4)/x + e^(e^4))/x + 8*e^(2*e^4)*log(e^(e^4)/x)/x + 6*e^(2*e^4)/x^2 + 2*e^(2*e^4 + 1)/x^2)*x*e^(-2*e^4)
Time = 3.58 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=-\ln \left (x+2\right )-5\,\ln \left (\frac {1}{x}\right )-x\,\left (\ln \left (\frac {1}{x}\right )+{\mathrm {e}}^4\right )-\frac {\mathrm {e}+3}{x} \] Input:
int((13*x - log(exp(exp(4))/x)*(2*x^2 + x^3) + exp(1)*(x + 2) + 6*x^2 + x^ 3 + 6)/(2*x^2 + x^3),x)
Output:
- log(x + 2) - 5*log(1/x) - x*(log(1/x) + exp(4)) - (exp(1) + 3)/x
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{2 x^2+x^3} \, dx=\frac {-\mathrm {log}\left (x +2\right ) x -\mathrm {log}\left (\frac {e^{e^{4}}}{x}\right ) x^{2}+5 \,\mathrm {log}\left (x \right ) x -e -3}{x} \] Input:
int(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2 *x^2),x)
Output:
( - log(x + 2)*x - log(e**(e**4)/x)*x**2 + 5*log(x)*x - e - 3)/x