Integrand size = 63, antiderivative size = 24 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=2-x+\frac {3 \log \left (\frac {3 e^x (-3+x)}{x^2}\right )}{1+x} \] Output:
2-x+3*ln((3*x-9)*exp(x)/x^2)/(1+x)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=-x+\frac {3 \log \left (-\frac {3 e^x (3-x)}{x^2}\right )}{1+x} \] Input:
Integrate[(18 + 9*x - 4*x^2 + 4*x^3 - x^4 + (9*x - 3*x^2)*Log[(E^x*(-9 + 3 *x))/x^2])/(-3*x - 5*x^2 - x^3 + x^4),x]
Output:
-x + (3*Log[(-3*E^x*(3 - x))/x^2])/(1 + x)
Time = 1.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2026, 2463, 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.
\(\displaystyle \int \frac {-x^4+4 x^3-4 x^2+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (3 x-9)}{x^2}\right )+9 x+18}{x^4-x^3-5 x^2-3 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^4+4 x^3-4 x^2+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (3 x-9)}{x^2}\right )+9 x+18}{x \left (x^3-x^2-5 x-3\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {-x^4+4 x^3-4 x^2+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (3 x-9)}{x^2}\right )+9 x+18}{16 (x-3) x}-\frac {-x^4+4 x^3-4 x^2+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (3 x-9)}{x^2}\right )+9 x+18}{16 x (x+1)}-\frac {-x^4+4 x^3-4 x^2+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (3 x-9)}{x^2}\right )+9 x+18}{4 x (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \log \left (-\frac {3 e^x (3-x)}{x^2}\right )}{x+1}-x\) |
Input:
Int[(18 + 9*x - 4*x^2 + 4*x^3 - x^4 + (9*x - 3*x^2)*Log[(E^x*(-9 + 3*x))/x ^2])/(-3*x - 5*x^2 - x^3 + x^4),x]
Output:
-x + (3*Log[(-3*E^x*(3 - x))/x^2])/(1 + 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])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {3 \ln \left (\frac {\left (3 x -9\right ) {\mathrm e}^{x}}{x^{2}}\right )}{1+x}-x\) | \(24\) |
parts | \(\frac {3 \ln \left (\frac {\left (3 x -9\right ) {\mathrm e}^{x}}{x^{2}}\right )}{1+x}-x\) | \(24\) |
norman | \(\frac {-x^{2}+3 \ln \left (\frac {\left (3 x -9\right ) {\mathrm e}^{x}}{x^{2}}\right )+1}{1+x}\) | \(28\) |
parallelrisch | \(-\frac {5+x^{2}+6 x -3 \ln \left (\frac {\left (3 x -9\right ) {\mathrm e}^{x}}{x^{2}}\right )}{1+x}\) | \(30\) |
risch | \(\frac {3 \ln \left ({\mathrm e}^{x}\right )}{1+x}+\frac {3 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-3 i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right )^{3}+3 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right ) {\mathrm e}^{x}}{x^{2}}\right )^{2}-3 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right ) {\mathrm e}^{x}}{x^{2}}\right )-3 i \pi \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right )+3 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right ) {\mathrm e}^{x}}{x^{2}}\right )^{2}+3 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right )^{2}-3 i \pi \operatorname {csgn}\left (\frac {i \left (-3+x \right ) {\mathrm e}^{x}}{x^{2}}\right )^{3}-6 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+3 i \pi \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (-3+x \right )\right )^{2}+3 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 x^{2}+6 \ln \left (3\right )-2 x -12 \ln \left (x \right )+6 \ln \left (-3+x \right )}{2+2 x}\) | \(277\) |
Input:
int(((-3*x^2+9*x)*ln((3*x-9)*exp(x)/x^2)-x^4+4*x^3-4*x^2+9*x+18)/(x^4-x^3- 5*x^2-3*x),x,method=_RETURNVERBOSE)
Output:
3*ln((3*x-9)*exp(x)/x^2)/(1+x)-x
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=-\frac {x^{2} + x - 3 \, \log \left (\frac {3 \, {\left (x - 3\right )} e^{x}}{x^{2}}\right )}{x + 1} \] Input:
integrate(((-3*x^2+9*x)*log((3*x-9)*exp(x)/x^2)-x^4+4*x^3-4*x^2+9*x+18)/(x ^4-x^3-5*x^2-3*x),x, algorithm="fricas")
Output:
-(x^2 + x - 3*log(3*(x - 3)*e^x/x^2))/(x + 1)
Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=- x + \frac {3 \log {\left (\frac {\left (3 x - 9\right ) e^{x}}{x^{2}} \right )}}{x + 1} \] Input:
integrate(((-3*x**2+9*x)*ln((3*x-9)*exp(x)/x**2)-x**4+4*x**3-4*x**2+9*x+18 )/(x**4-x**3-5*x**2-3*x),x)
Output:
-x + 3*log((3*x - 9)*exp(x)/x**2)/(x + 1)
Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=-x - \frac {3 \, {\left ({\left (x - 3\right )} \log \left (x - 3\right ) - 8 \, x \log \left (x\right ) - 4 \, \log \left (3\right ) + 4\right )}}{4 \, {\left (x + 1\right )}} + \frac {3}{4} \, \log \left (x - 3\right ) - 6 \, \log \left (x\right ) \] Input:
integrate(((-3*x^2+9*x)*log((3*x-9)*exp(x)/x^2)-x^4+4*x^3-4*x^2+9*x+18)/(x ^4-x^3-5*x^2-3*x),x, algorithm="maxima")
Output:
-x - 3/4*((x - 3)*log(x - 3) - 8*x*log(x) - 4*log(3) + 4)/(x + 1) + 3/4*lo g(x - 3) - 6*log(x)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=-x + \frac {3 \, \log \left (\frac {3 \, {\left (x - 3\right )}}{x^{2}}\right )}{x + 1} - \frac {3}{x + 1} \] Input:
integrate(((-3*x^2+9*x)*log((3*x-9)*exp(x)/x^2)-x^4+4*x^3-4*x^2+9*x+18)/(x ^4-x^3-5*x^2-3*x),x, algorithm="giac")
Output:
-x + 3*log(3*(x - 3)/x^2)/(x + 1) - 3/(x + 1)
Time = 0.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=\frac {3\,\ln \left (\frac {{\mathrm {e}}^x\,\left (3\,x-9\right )}{x^2}\right )}{x+1}-x \] Input:
int(-(9*x + log((exp(x)*(3*x - 9))/x^2)*(9*x - 3*x^2) - 4*x^2 + 4*x^3 - x^ 4 + 18)/(3*x + 5*x^2 + x^3 - x^4),x)
Output:
(3*log((exp(x)*(3*x - 9))/x^2))/(x + 1) - x
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {18+9 x-4 x^2+4 x^3-x^4+\left (9 x-3 x^2\right ) \log \left (\frac {e^x (-9+3 x)}{x^2}\right )}{-3 x-5 x^2-x^3+x^4} \, dx=\frac {3 \,\mathrm {log}\left (\frac {3 e^{x} x -9 e^{x}}{x^{2}}\right )-x^{2}-x}{x +1} \] Input:
int(((-3*x^2+9*x)*log((3*x-9)*exp(x)/x^2)-x^4+4*x^3-4*x^2+9*x+18)/(x^4-x^3 -5*x^2-3*x),x)
Output:
(3*log((3*e**x*x - 9*e**x)/x**2) - x**2 - x)/(x + 1)