Integrand size = 43, antiderivative size = 24 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=-1+x \left (x+\frac {x^2}{4}\right )-\left (e^x+x\right ) (2+\log (x)) \] Output:
(1/4*x^2+x)*x-(exp(x)+x)*(ln(x)+2)-1
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=-2 e^x-2 x+x^2+\frac {x^3}{4}-\left (e^x+x\right ) \log (x) \] Input:
Integrate[(E^x*(-4 - 8*x) - 12*x + 8*x^2 + 3*x^3 + (-4*x - 4*E^x*x)*Log[x] )/(4*x),x]
Output:
-2*E^x - 2*x + x^2 + x^3/4 - (E^x + x)*Log[x]
Time = 0.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {3 x^3+8 x^2-12 x+e^x (-8 x-4)+\left (-4 e^x x-4 x\right ) \log (x)}{4 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {-3 x^3-8 x^2+12 x+4 e^x (2 x+1)+4 \left (e^x x+x\right ) \log (x)}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {-3 x^3-8 x^2+12 x+4 e^x (2 x+1)+4 \left (e^x x+x\right ) \log (x)}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{4} \int \left (-3 x^2-8 x+4 \log (x)+12+\frac {4 e^x (\log (x) x+2 x+1)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (x^3+4 x^2-8 x-4 x \log (x)-\frac {4 e^x (2 x+x \log (x))}{x}\right )\) |
Input:
Int[(E^x*(-4 - 8*x) - 12*x + 8*x^2 + 3*x^3 + (-4*x - 4*E^x*x)*Log[x])/(4*x ),x]
Output:
(-8*x + 4*x^2 + x^3 - 4*x*Log[x] - (4*E^x*(2*x + x*Log[x]))/x)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 = 0.90 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
default | \(-2 x -{\mathrm e}^{x} \ln \left (x \right )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \left (x \right )\) | \(28\) |
norman | \(-2 x -{\mathrm e}^{x} \ln \left (x \right )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \left (x \right )\) | \(28\) |
parallelrisch | \(-2 x -{\mathrm e}^{x} \ln \left (x \right )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \left (x \right )\) | \(28\) |
parts | \(-2 x -{\mathrm e}^{x} \ln \left (x \right )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \left (x \right )\) | \(28\) |
risch | \(\frac {\left (-4 x -4 \,{\mathrm e}^{x}\right ) \ln \left (x \right )}{4}+\frac {x^{3}}{4}+x^{2}-2 x -2 \,{\mathrm e}^{x}\) | \(29\) |
orering | \(\frac {\left (-4 \,{\mathrm e}^{x} x -4 x \right ) \ln \left (x \right )}{4}+\frac {\left (-8 x -4\right ) {\mathrm e}^{x}}{4}+\frac {3 x^{3}}{4}+2 x^{2}-3 x -\frac {\left (x^{8}-x^{7}-11 x^{6}+30 x^{5}-36 x^{4}-6 x^{3}+50 x^{2}+8 x -4\right ) \left (\frac {\left (-4 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-4\right ) \ln \left (x \right )+\frac {-4 \,{\mathrm e}^{x} x -4 x}{x}-8 \,{\mathrm e}^{x}+\left (-8 x -4\right ) {\mathrm e}^{x}+9 x^{2}+16 x -12}{4 x}-\frac {\left (-4 \,{\mathrm e}^{x} x -4 x \right ) \ln \left (x \right )+\left (-8 x -4\right ) {\mathrm e}^{x}+3 x^{3}+8 x^{2}-12 x}{4 x^{2}}\right )}{3 x^{6}-11 x^{5}+19 x^{4}-16 x^{3}-12 x^{2}+4 x +4}+\frac {2 x \left (x^{7}-5 x^{6}+6 x^{5}-x^{3}-11 x^{2}+16 x +4\right ) \left (\frac {\left (-4 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {-8 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}-8}{x}-\frac {-4 \,{\mathrm e}^{x} x -4 x}{x^{2}}-16 \,{\mathrm e}^{x}+\left (-8 x -4\right ) {\mathrm e}^{x}+18 x +16}{4 x}-\frac {\left (-4 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-4\right ) \ln \left (x \right )+\frac {-4 \,{\mathrm e}^{x} x -4 x}{x}-8 \,{\mathrm e}^{x}+\left (-8 x -4\right ) {\mathrm e}^{x}+9 x^{2}+16 x -12}{2 x^{2}}+\frac {\left (-4 \,{\mathrm e}^{x} x -4 x \right ) \ln \left (x \right )+\left (-8 x -4\right ) {\mathrm e}^{x}+3 x^{3}+8 x^{2}-12 x}{2 x^{3}}\right )}{3 x^{6}-11 x^{5}+19 x^{4}-16 x^{3}-12 x^{2}+4 x +4}-\frac {\left (x^{6}-6 x^{5}+9 x^{4}-x^{2}-12 x -2\right ) x^{2} \left (\frac {\left (-4 \,{\mathrm e}^{x} x -12 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {-12 \,{\mathrm e}^{x} x -24 \,{\mathrm e}^{x}}{x}-\frac {3 \left (-4 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-4\right )}{x^{2}}+\frac {-8 \,{\mathrm e}^{x} x -8 x}{x^{3}}-24 \,{\mathrm e}^{x}+\left (-8 x -4\right ) {\mathrm e}^{x}+18}{4 x}-\frac {3 \left (\left (-4 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {-8 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}-8}{x}-\frac {-4 \,{\mathrm e}^{x} x -4 x}{x^{2}}-16 \,{\mathrm e}^{x}+\left (-8 x -4\right ) {\mathrm e}^{x}+18 x +16\right )}{4 x^{2}}+\frac {\frac {3 \left (-4 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-4\right ) \ln \left (x \right )}{2}+\frac {3 \left (-4 \,{\mathrm e}^{x} x -4 x \right )}{2 x}-12 \,{\mathrm e}^{x}+\frac {3 \left (-8 x -4\right ) {\mathrm e}^{x}}{2}+\frac {27 x^{2}}{2}+24 x -18}{x^{3}}-\frac {3 \left (\left (-4 \,{\mathrm e}^{x} x -4 x \right ) \ln \left (x \right )+\left (-8 x -4\right ) {\mathrm e}^{x}+3 x^{3}+8 x^{2}-12 x \right )}{2 x^{4}}\right )}{3 x^{6}-11 x^{5}+19 x^{4}-16 x^{3}-12 x^{2}+4 x +4}\) | \(721\) |
Input:
int(1/4*((-4*exp(x)*x-4*x)*ln(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x,met hod=_RETURNVERBOSE)
Output:
-2*x-exp(x)*ln(x)-2*exp(x)+x^2+1/4*x^3-x*ln(x)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=\frac {1}{4} \, x^{3} + x^{2} - {\left (x + e^{x}\right )} \log \left (x\right ) - 2 \, x - 2 \, e^{x} \] Input:
integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/ x,x, algorithm="fricas")
Output:
1/4*x^3 + x^2 - (x + e^x)*log(x) - 2*x - 2*e^x
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=\frac {x^{3}}{4} + x^{2} - x \log {\left (x \right )} - 2 x + \left (- \log {\left (x \right )} - 2\right ) e^{x} \] Input:
integrate(1/4*((-4*exp(x)*x-4*x)*ln(x)+(-8*x-4)*exp(x)+3*x**3+8*x**2-12*x) /x,x)
Output:
x**3/4 + x**2 - x*log(x) - 2*x + (-log(x) - 2)*exp(x)
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=\frac {1}{4} \, x^{3} + x^{2} - x \log \left (x\right ) - e^{x} \log \left (x\right ) - 2 \, x - 2 \, e^{x} \] Input:
integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/ x,x, algorithm="maxima")
Output:
1/4*x^3 + x^2 - x*log(x) - e^x*log(x) - 2*x - 2*e^x
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=\frac {1}{4} \, x^{3} + x^{2} - x \log \left (x\right ) - e^{x} \log \left (x\right ) - 2 \, x - 2 \, e^{x} \] Input:
integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/ x,x, algorithm="giac")
Output:
1/4*x^3 + x^2 - x*log(x) - e^x*log(x) - 2*x - 2*e^x
Time = 1.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=x^2-2\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (x\right )-x\,\ln \left (x\right )-2\,x+\frac {x^3}{4} \] Input:
int(-(3*x + (log(x)*(4*x + 4*x*exp(x)))/4 + (exp(x)*(8*x + 4))/4 - 2*x^2 - (3*x^3)/4)/x,x)
Output:
x^2 - 2*exp(x) - exp(x)*log(x) - x*log(x) - 2*x + x^3/4
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{4 x} \, dx=-e^{x} \mathrm {log}\left (x \right )-2 e^{x}-\mathrm {log}\left (x \right ) x +\frac {x^{3}}{4}+x^{2}-2 x \] Input:
int(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x)
Output:
( - 4*e**x*log(x) - 8*e**x - 4*log(x)*x + x**3 + 4*x**2 - 8*x)/4