Integrand size = 31, antiderivative size = 23 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=2 \left (3+\frac {x+4 \left (x+e^x x\right ) (x+\log (x))}{x^2}\right ) \] Output:
2*(4*(x+ln(x))*(exp(x)*x+x)+x)/x^2+6
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2+8 e^x x+8 \left (1+e^x\right ) \log (x)}{x} \] Input:
Integrate[(6 + E^x*(8 + 8*x^2) + (-8 + E^x*(-8 + 8*x))*Log[x])/x^2,x]
Output:
(2 + 8*E^x*x + 8*(1 + E^x)*Log[x])/x
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.
\(\displaystyle \int \frac {e^x \left (8 x^2+8\right )+\left (e^x (8 x-8)-8\right ) \log (x)+6}{x^2} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {8 e^x \left (x^2+x \log (x)-\log (x)+1\right )}{x^2}-\frac {2 (4 \log (x)-3)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8 e^x \left (x^2+x \log (x)\right )}{x^2}+\frac {8}{x}-\frac {2 (3-4 \log (x))}{x}\) |
Input:
Int[(6 + E^x*(8 + 8*x^2) + (-8 + E^x*(-8 + 8*x))*Log[x])/x^2,x]
Output:
8/x - (2*(3 - 4*Log[x]))/x + (8*E^x*(x^2 + x*Log[x]))/x^2
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 = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {2+8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )+8 \ln \left (x \right )}{x}\) | \(22\) |
parallelrisch | \(-\frac {-8 \,{\mathrm e}^{x} x -2-8 \,{\mathrm e}^{x} \ln \left (x \right )-8 \ln \left (x \right )}{x}\) | \(23\) |
risch | \(\frac {8 \left ({\mathrm e}^{x}+1\right ) \ln \left (x \right )}{x}+\frac {8 \,{\mathrm e}^{x} x +2}{x}\) | \(25\) |
default | \(\frac {8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )}{x}+\frac {2}{x}+\frac {8 \ln \left (x \right )}{x}\) | \(30\) |
parts | \(\frac {8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )}{x}+\frac {2}{x}+\frac {8 \ln \left (x \right )}{x}\) | \(30\) |
orering | \(-\frac {\left (3 x^{5}+7 x^{4}-11 x^{3}-22 x^{2}-14 x +4\right ) \left (\left (\left (8 x -8\right ) {\mathrm e}^{x}-8\right ) \ln \left (x \right )+\left (8 x^{2}+8\right ) {\mathrm e}^{x}+6\right )}{\left (x^{4}+5 x^{3}+7 x^{2}+2 x -2\right ) x^{2}}-\frac {x \left (x^{5}-5 x^{4}-24 x^{3}-6 x^{2}+6 x +14\right ) \left (\frac {\left (8 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {\left (8 x -8\right ) {\mathrm e}^{x}-8}{x}+16 \,{\mathrm e}^{x} x +\left (8 x^{2}+8\right ) {\mathrm e}^{x}}{x^{2}}-\frac {2 \left (\left (\left (8 x -8\right ) {\mathrm e}^{x}-8\right ) \ln \left (x \right )+\left (8 x^{2}+8\right ) {\mathrm e}^{x}+6\right )}{x^{3}}\right )}{x^{4}+5 x^{3}+7 x^{2}+2 x -2}+\frac {x^{2} \left (2 x^{4}+x^{3}-14 x^{2}-12 x -8\right ) \left (\frac {\left (16 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {16 \,{\mathrm e}^{x}+2 \left (8 x -8\right ) {\mathrm e}^{x}}{x}-\frac {\left (8 x -8\right ) {\mathrm e}^{x}-8}{x^{2}}+32 \,{\mathrm e}^{x} x +16 \,{\mathrm e}^{x}+\left (8 x^{2}+8\right ) {\mathrm e}^{x}}{x^{2}}-\frac {4 \left (\left (8 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {\left (8 x -8\right ) {\mathrm e}^{x}-8}{x}+16 \,{\mathrm e}^{x} x +\left (8 x^{2}+8\right ) {\mathrm e}^{x}\right )}{x^{3}}+\frac {6 \left (\left (8 x -8\right ) {\mathrm e}^{x}-8\right ) \ln \left (x \right )+6 \left (8 x^{2}+8\right ) {\mathrm e}^{x}+36}{x^{4}}\right )}{x^{4}+5 x^{3}+7 x^{2}+2 x -2}-\frac {\left (x^{3}+3 x^{2}+2 x +1\right ) x^{3} \left (\frac {\left (24 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {48 \,{\mathrm e}^{x}+3 \left (8 x -8\right ) {\mathrm e}^{x}}{x}-\frac {3 \left (8 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right )}{x^{2}}+\frac {2 \left (8 x -8\right ) {\mathrm e}^{x}-16}{x^{3}}+48 \,{\mathrm e}^{x} x +48 \,{\mathrm e}^{x}+\left (8 x^{2}+8\right ) {\mathrm e}^{x}}{x^{2}}-\frac {6 \left (\left (16 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {16 \,{\mathrm e}^{x}+2 \left (8 x -8\right ) {\mathrm e}^{x}}{x}-\frac {\left (8 x -8\right ) {\mathrm e}^{x}-8}{x^{2}}+32 \,{\mathrm e}^{x} x +16 \,{\mathrm e}^{x}+\left (8 x^{2}+8\right ) {\mathrm e}^{x}\right )}{x^{3}}+\frac {18 \left (8 \,{\mathrm e}^{x}+\left (8 x -8\right ) {\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {18 \left (\left (8 x -8\right ) {\mathrm e}^{x}-8\right )}{x}+288 \,{\mathrm e}^{x} x +18 \left (8 x^{2}+8\right ) {\mathrm e}^{x}}{x^{4}}-\frac {24 \left (\left (\left (8 x -8\right ) {\mathrm e}^{x}-8\right ) \ln \left (x \right )+\left (8 x^{2}+8\right ) {\mathrm e}^{x}+6\right )}{x^{5}}\right )}{x^{4}+5 x^{3}+7 x^{2}+2 x -2}\) | \(686\) |
Input:
int((((8*x-8)*exp(x)-8)*ln(x)+(8*x^2+8)*exp(x)+6)/x^2,x,method=_RETURNVERB OSE)
Output:
(2+8*exp(x)*x+8*exp(x)*ln(x)+8*ln(x))/x
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2 \, {\left (4 \, x e^{x} + 4 \, {\left (e^{x} + 1\right )} \log \left (x\right ) + 1\right )}}{x} \] Input:
integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm= "fricas")
Output:
2*(4*x*e^x + 4*(e^x + 1)*log(x) + 1)/x
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {\left (8 x + 8 \log {\left (x \right )}\right ) e^{x}}{x} + \frac {8 \log {\left (x \right )}}{x} + \frac {2}{x} \] Input:
integrate((((8*x-8)*exp(x)-8)*ln(x)+(8*x**2+8)*exp(x)+6)/x**2,x)
Output:
(8*x + 8*log(x))*exp(x)/x + 8*log(x)/x + 2/x
\[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\int { \frac {2 \, {\left (4 \, {\left (x^{2} + 1\right )} e^{x} + 4 \, {\left ({\left (x - 1\right )} e^{x} - 1\right )} \log \left (x\right ) + 3\right )}}{x^{2}} \,d x } \] Input:
integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm= "maxima")
Output:
8*e^x*log(x)/x + 8*log(x)/x + 2/x + 8*e^x + 8*gamma(-1, -x) - 8*integrate( e^x/x^2, x)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2 \, {\left (4 \, x e^{x} + 4 \, e^{x} \log \left (x\right ) + 4 \, \log \left (x\right ) + 1\right )}}{x} \] Input:
integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm= "giac")
Output:
2*(4*x*e^x + 4*e^x*log(x) + 4*log(x) + 1)/x
Time = 2.92 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=8\,{\mathrm {e}}^x+\frac {8\,\ln \left (x\right )+8\,{\mathrm {e}}^x\,\ln \left (x\right )+2}{x} \] Input:
int((log(x)*(exp(x)*(8*x - 8) - 8) + exp(x)*(8*x^2 + 8) + 6)/x^2,x)
Output:
8*exp(x) + (8*log(x) + 8*exp(x)*log(x) + 2)/x
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {8 e^{x} \mathrm {log}\left (x \right )+8 e^{x} x +8 \,\mathrm {log}\left (x \right )+2}{x} \] Input:
int((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x)
Output:
(2*(4*e**x*log(x) + 4*e**x*x + 4*log(x) + 1))/x