Integrand size = 52, antiderivative size = 26 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=-x^2+e^{9+x} \left (\frac {1}{x^2}+x-5 (e+x-\log (x))\right ) \] Output:
exp(x+9)*(1/x^2-5*exp(1)-4*x+5*ln(x))-x^2
Time = 0.49 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=-x^2+e^x \left (-5 e^{10}+\frac {e^9}{x^2}-4 e^9 x\right )+5 e^{9+x} \log (x) \] Input:
Integrate[(-2*x^4 + E^(9 + x)*(-2 + x + 5*x^2 - 4*x^3 - 5*E*x^3 - 4*x^4) + 5*E^(9 + x)*x^3*Log[x])/x^3,x]
Output:
-x^2 + E^x*(-5*E^10 + E^9/x^2 - 4*E^9*x) + 5*E^(9 + x)*Log[x]
Time = 0.55 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2010, 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.
\(\displaystyle \int \frac {-2 x^4+5 e^{x+9} x^3 \log (x)+e^{x+9} \left (-4 x^4-5 e x^3-4 x^3+5 x^2+x-2\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {e^{x+9} \left (-4 x^4-4 \left (1+\frac {5 e}{4}\right ) x^3+5 x^3 \log (x)+5 x^2+x-2\right )}{x^3}-2 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^2+\frac {e^{x+9}}{x^2}-4 e^{x+9} x+4 e^{x+9}-(4+5 e) e^{x+9}+5 e^{x+9} \log (x)\) |
Input:
Int[(-2*x^4 + E^(9 + x)*(-2 + x + 5*x^2 - 4*x^3 - 5*E*x^3 - 4*x^4) + 5*E^( 9 + x)*x^3*Log[x])/x^3,x]
Output:
4*E^(9 + x) - E^(9 + x)*(4 + 5*E) + E^(9 + x)/x^2 - 4*E^(9 + x)*x - x^2 + 5*E^(9 + x)*Log[x]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.53 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65
method | result | size |
risch | \(5 \,{\mathrm e}^{x +9} \ln \left (x \right )-\frac {5 x^{2} {\mathrm e}^{x +10}+x^{4}+4 \,{\mathrm e}^{x +9} x^{3}-{\mathrm e}^{x +9}}{x^{2}}\) | \(43\) |
norman | \(\frac {-x^{4}-4 \,{\mathrm e}^{x +9} x^{3}+5 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )-5 \,{\mathrm e} x^{2} {\mathrm e}^{x +9}+{\mathrm e}^{x +9}}{x^{2}}\) | \(46\) |
default | \(\frac {-4 \,{\mathrm e}^{x +9} x^{3}+5 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )-5 \,{\mathrm e} x^{2} {\mathrm e}^{x +9}+{\mathrm e}^{x +9}}{x^{2}}-x^{2}\) | \(47\) |
parallelrisch | \(-\frac {5 \,{\mathrm e} x^{2} {\mathrm e}^{x +9}+x^{4}+4 \,{\mathrm e}^{x +9} x^{3}-5 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )-{\mathrm e}^{x +9}}{x^{2}}\) | \(47\) |
parts | \(\frac {-4 \,{\mathrm e}^{x +9} x^{3}+5 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )-5 \,{\mathrm e} x^{2} {\mathrm e}^{x +9}+{\mathrm e}^{x +9}}{x^{2}}-x^{2}\) | \(47\) |
orering | \(\frac {\left (4 x^{7}-5 x^{6}-30 x^{5}+42 x^{4}-82 x^{3}+64 x^{2}+84 x -144\right ) \left (5 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+\left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-2 x^{4}\right )}{2 \left (4 x^{4}-5 x^{3}+6 x^{2}-6 x -24\right ) \left (-1+x \right )^{2} x^{3}}-\frac {\left (8 x^{7}-34 x^{6}+13 x^{5}+4 x^{4}-70 x^{3}+84 x^{2}+24 x -96\right ) \left (\frac {15 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )+5 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+5 \,{\mathrm e}^{x +9} x^{2}+\left (-15 x^{2} {\mathrm e}-16 x^{3}-12 x^{2}+10 x +1\right ) {\mathrm e}^{x +9}+\left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-8 x^{3}}{x^{3}}-\frac {3 \left (5 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+\left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-2 x^{4}\right )}{x^{4}}\right )}{2 \left (4 x^{5}-9 x^{4}+11 x^{3}-12 x^{2}-18 x +24\right ) \left (-1+x \right )}+\frac {x \left (4 x^{5}-17 x^{4}+2 x^{2}-32 x -24\right ) \left (\frac {30 x \ln \left (x \right ) {\mathrm e}^{x +9}+30 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )+25 x \,{\mathrm e}^{x +9}+5 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+10 \,{\mathrm e}^{x +9} x^{2}+\left (-30 x \,{\mathrm e}-48 x^{2}-24 x +10\right ) {\mathrm e}^{x +9}+2 \left (-15 x^{2} {\mathrm e}-16 x^{3}-12 x^{2}+10 x +1\right ) {\mathrm e}^{x +9}+\left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-24 x^{2}}{x^{3}}-\frac {6 \left (15 x^{2} {\mathrm e}^{x +9} \ln \left (x \right )+5 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+5 \,{\mathrm e}^{x +9} x^{2}+\left (-15 x^{2} {\mathrm e}-16 x^{3}-12 x^{2}+10 x +1\right ) {\mathrm e}^{x +9}+\left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-8 x^{3}\right )}{x^{4}}+\frac {60 x^{3} {\mathrm e}^{x +9} \ln \left (x \right )+12 \left (-5 x^{3} {\mathrm e}-4 x^{4}-4 x^{3}+5 x^{2}+x -2\right ) {\mathrm e}^{x +9}-24 x^{4}}{x^{5}}\right )}{2 \left (4 x^{4}-5 x^{3}+6 x^{2}-6 x -24\right ) \left (-1+x \right )}\) | \(673\) |
Input:
int((5*x^3*exp(x+9)*ln(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)-2 *x^4)/x^3,x,method=_RETURNVERBOSE)
Output:
5*exp(x+9)*ln(x)-(5*x^2*exp(x+10)+x^4+4*exp(x+9)*x^3-exp(x+9))/x^2
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=-\frac {x^{4} - 5 \, x^{2} e^{\left (x + 9\right )} \log \left (x\right ) + {\left (4 \, x^{3} + 5 \, x^{2} e - 1\right )} e^{\left (x + 9\right )}}{x^{2}} \] Input:
integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp (x+9)-2*x^4)/x^3,x, algorithm="fricas")
Output:
-(x^4 - 5*x^2*e^(x + 9)*log(x) + (4*x^3 + 5*x^2*e - 1)*e^(x + 9))/x^2
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=- x^{2} + \frac {\left (- 4 x^{3} + 5 x^{2} \log {\left (x \right )} - 5 e x^{2} + 1\right ) e^{x + 9}}{x^{2}} \] Input:
integrate((5*x**3*exp(x+9)*ln(x)+(-5*x**3*exp(1)-4*x**4-4*x**3+5*x**2+x-2) *exp(x+9)-2*x**4)/x**3,x)
Output:
-x**2 + (-4*x**3 + 5*x**2*log(x) - 5*E*x**2 + 1)*exp(x + 9)/x**2
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=-x^{2} - 4 \, {\left (x e^{9} - e^{9}\right )} e^{x} + e^{9} \Gamma \left (-1, -x\right ) + 2 \, e^{9} \Gamma \left (-2, -x\right ) + 5 \, e^{\left (x + 9\right )} \log \left (x\right ) - 5 \, e^{\left (x + 10\right )} - 4 \, e^{\left (x + 9\right )} \] Input:
integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp (x+9)-2*x^4)/x^3,x, algorithm="maxima")
Output:
-x^2 - 4*(x*e^9 - e^9)*e^x + e^9*gamma(-1, -x) + 2*e^9*gamma(-2, -x) + 5*e ^(x + 9)*log(x) - 5*e^(x + 10) - 4*e^(x + 9)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=-\frac {x^{4} + 4 \, x^{3} e^{\left (x + 9\right )} - 5 \, x^{2} e^{\left (x + 9\right )} \log \left (x\right ) + 5 \, x^{2} e^{\left (x + 10\right )} - e^{\left (x + 9\right )}}{x^{2}} \] Input:
integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp (x+9)-2*x^4)/x^3,x, algorithm="giac")
Output:
-(x^4 + 4*x^3*e^(x + 9) - 5*x^2*e^(x + 9)*log(x) + 5*x^2*e^(x + 10) - e^(x + 9))/x^2
Time = 2.61 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=\frac {{\mathrm {e}}^{x+9}}{x^2}-4\,x\,{\mathrm {e}}^{x+9}-5\,{\mathrm {e}}^{x+10}+5\,{\mathrm {e}}^{x+9}\,\ln \left (x\right )-x^2 \] Input:
int(-(exp(x + 9)*(5*x^3*exp(1) - x - 5*x^2 + 4*x^3 + 4*x^4 + 2) + 2*x^4 - 5*x^3*exp(x + 9)*log(x))/x^3,x)
Output:
exp(x + 9)/x^2 - 4*x*exp(x + 9) - 5*exp(x + 10) + 5*exp(x + 9)*log(x) - x^ 2
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x^4+e^{9+x} \left (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4\right )+5 e^{9+x} x^3 \log (x)}{x^3} \, dx=\frac {5 e^{x} \mathrm {log}\left (x \right ) e^{9} x^{2}-5 e^{x} e^{10} x^{2}-4 e^{x} e^{9} x^{3}+e^{x} e^{9}-x^{4}}{x^{2}} \] Input:
int((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)- 2*x^4)/x^3,x)
Output:
(5*e**x*log(x)*e**9*x**2 - 5*e**x*e**10*x**2 - 4*e**x*e**9*x**3 + e**x*e** 9 - x**4)/x**2