\(\int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} (7 x+8 x^2))}{x^2} \, dx\) [2349]

Optimal result

Integrand size = 105, antiderivative size = 38 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\frac {e^{-x+\left (e^{3-x+5 \left (x+\frac {\log (x)}{2}\right )}+x\right )^2}}{x}-x+x^2 \] Output:

exp((x+exp(5/2*ln(x)+4*x+3))^2-x)/x-x+x^2
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\frac {e^{-x+x^2+2 e^{3+4 x} x^{7/2}+e^{6+8 x} x^5}}{x}-x+x^2 \] Input:

Integrate[(-x^2 + 2*x^3 + E^(-x + 2*E^((6 + 8*x + 5*Log[x])/2)*x + x^2 + E 
^(6 + 8*x)*x^5)*(-1 - x + 2*x^2 + E^(6 + 8*x)*x^5*(5 + 8*x) + E^((6 + 8*x 
+ 5*Log[x])/2)*(7*x + 8*x^2)))/x^2,x]
 

Output:

E^(-x + x^2 + 2*E^(3 + 4*x)*x^(7/2) + E^(6 + 8*x)*x^5)/x - x + x^2
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (e^{8 x+6} (8 x+5) x^5+2 x^2+\left (8 x^2+7 x\right ) e^{\frac {1}{2} (8 x+5 \log (x)+6)}-x-1\right ) \exp \left (e^{8 x+6} x^5+x^2-x+2 x e^{\frac {1}{2} (8 x+5 \log (x)+6)}\right )+2 x^3-x^2}{x^2} \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left ((8 x+7) x^{3/2} \exp \left (2 e^{4 x+3} x^{7/2}+e^{8 x+6} x^5+x^2+3 x+3\right )+\frac {e^{-x} \left (5 x^5 \exp \left (2 e^{4 x+3} x^{7/2}+e^{8 x+6} x^5+x^2+8 x+6\right )+8 x^6 \exp \left (2 e^{4 x+3} x^{7/2}+e^{8 x+6} x^5+x^2+8 x+6\right )-e^{\left (e^{4 x+3} x^{5/2}+x\right )^2} x-e^{\left (e^{4 x+3} x^{5/2}+x\right )^2}+2 e^x x^3-e^x x^2+2 e^{\left (e^{4 x+3} x^{5/2}+x\right )^2} x^2\right )}{x^2}\right )dx\)

\(\Big \downarrow \) 7267

\(\displaystyle 2 \int \left (\exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+7 x+6\right ) (8 x+5) x^{7/2}+\exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+3 x+3\right ) (8 x+7) x^2-(1-2 x) \sqrt {x}-\frac {e^{\left (e^{4 x+3} x^{5/2}+x\right )^2-x} \left (-2 x^2+x+1\right )}{x^{3/2}}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (7 \int \exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+3 x+3\right ) x^2d\sqrt {x}+5 \int \exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+7 x+6\right ) x^{7/2}d\sqrt {x}+8 \int \exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+7 x+6\right ) x^{9/2}d\sqrt {x}+8 \int \exp \left (e^{8 x+6} x^5+2 e^{4 x+3} x^{7/2}+x^2+3 x+3\right ) x^3d\sqrt {x}-\int \frac {e^{\left (e^{4 x+3} x^{5/2}+x\right )^2-x}}{x^{3/2}}d\sqrt {x}-\int \frac {e^{\left (e^{4 x+3} x^{5/2}+x\right )^2-x}}{\sqrt {x}}d\sqrt {x}+2 \int e^{\left (e^{4 x+3} x^{5/2}+x\right )^2-x} \sqrt {x}d\sqrt {x}+\frac {x^2}{2}-\frac {x}{2}\right )\)

Input:

Int[(-x^2 + 2*x^3 + E^(-x + 2*E^((6 + 8*x + 5*Log[x])/2)*x + x^2 + E^(6 + 
8*x)*x^5)*(-1 - x + 2*x^2 + E^(6 + 8*x)*x^5*(5 + 8*x) + E^((6 + 8*x + 5*Lo 
g[x])/2)*(7*x + 8*x^2)))/x^2,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08

method	result	size
risch	\(x^{2}-x +\frac {{\mathrm e}^{x^{5} {\mathrm e}^{6+8 x}+2 x^{\frac {7}{2}} {\mathrm e}^{3+4 x}+x^{2}-x}}{x}\)	\(41\)
parallelrisch	\(-\frac {-x^{3}+x^{2}-{\mathrm e}^{{\mathrm e}^{5 \ln \left (x \right )+8 x +6}+2 x \,{\mathrm e}^{\frac {5 \ln \left (x \right )}{2}+4 x +3}+x^{2}-x}}{x}\)	\(50\)

Input:

int((((8*x+5)*exp(5/2*ln(x)+4*x+3)^2+(8*x^2+7*x)*exp(5/2*ln(x)+4*x+3)+2*x^ 
2-x-1)*exp(exp(5/2*ln(x)+4*x+3)^2+2*x*exp(5/2*ln(x)+4*x+3)+x^2-x)+2*x^3-x^ 
2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

x^2-x+1/x*exp(x^5*exp(6+8*x)+2*x^(7/2)*exp(3+4*x)+x^2-x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\frac {x^{3} - x^{2} + e^{\left (x^{2} + 2 \, x e^{\left (4 \, x + \frac {5}{2} \, \log \left (x\right ) + 3\right )} - x + e^{\left (8 \, x + 5 \, \log \left (x\right ) + 6\right )}\right )}}{x} \] Input:

integrate((((8*x+5)*exp(5/2*log(x)+4*x+3)^2+(8*x^2+7*x)*exp(5/2*log(x)+4*x 
+3)+2*x^2-x-1)*exp(exp(5/2*log(x)+4*x+3)^2+2*x*exp(5/2*log(x)+4*x+3)+x^2-x 
)+2*x^3-x^2)/x^2,x, algorithm="fricas")
 

Output:

(x^3 - x^2 + e^(x^2 + 2*x*e^(4*x + 5/2*log(x) + 3) - x + e^(8*x + 5*log(x) 
 + 6)))/x
 

Sympy [F(-1)]

Timed out. \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\text {Timed out} \] Input:

integrate((((8*x+5)*exp(5/2*ln(x)+4*x+3)**2+(8*x**2+7*x)*exp(5/2*ln(x)+4*x 
+3)+2*x**2-x-1)*exp(exp(5/2*ln(x)+4*x+3)**2+2*x*exp(5/2*ln(x)+4*x+3)+x**2- 
x)+2*x**3-x**2)/x**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=x^{2} - x + \frac {e^{\left (x^{5} e^{\left (8 \, x + 6\right )} + 2 \, x^{\frac {7}{2}} e^{\left (4 \, x + 3\right )} + x^{2} - x\right )}}{x} \] Input:

integrate((((8*x+5)*exp(5/2*log(x)+4*x+3)^2+(8*x^2+7*x)*exp(5/2*log(x)+4*x 
+3)+2*x^2-x-1)*exp(exp(5/2*log(x)+4*x+3)^2+2*x*exp(5/2*log(x)+4*x+3)+x^2-x 
)+2*x^3-x^2)/x^2,x, algorithm="maxima")
 

Output:

x^2 - x + e^(x^5*e^(8*x + 6) + 2*x^(7/2)*e^(4*x + 3) + x^2 - x)/x
 

Giac [F]

\[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\int { \frac {2 \, x^{3} - x^{2} + {\left (2 \, x^{2} + {\left (8 \, x + 5\right )} e^{\left (8 \, x + 5 \, \log \left (x\right ) + 6\right )} + {\left (8 \, x^{2} + 7 \, x\right )} e^{\left (4 \, x + \frac {5}{2} \, \log \left (x\right ) + 3\right )} - x - 1\right )} e^{\left (x^{2} + 2 \, x e^{\left (4 \, x + \frac {5}{2} \, \log \left (x\right ) + 3\right )} - x + e^{\left (8 \, x + 5 \, \log \left (x\right ) + 6\right )}\right )}}{x^{2}} \,d x } \] Input:

integrate((((8*x+5)*exp(5/2*log(x)+4*x+3)^2+(8*x^2+7*x)*exp(5/2*log(x)+4*x 
+3)+2*x^2-x-1)*exp(exp(5/2*log(x)+4*x+3)^2+2*x*exp(5/2*log(x)+4*x+3)+x^2-x 
)+2*x^3-x^2)/x^2,x, algorithm="giac")
 

Output:

integrate((2*x^3 - x^2 + (2*x^2 + (8*x + 5)*e^(8*x + 5*log(x) + 6) + (8*x^ 
2 + 7*x)*e^(4*x + 5/2*log(x) + 3) - x - 1)*e^(x^2 + 2*x*e^(4*x + 5/2*log(x 
) + 3) - x + e^(8*x + 5*log(x) + 6)))/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 1.93 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=x^2-x+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{2\,x^{7/2}\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^3}\,{\mathrm {e}}^{x^5\,{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^6}}{x} \] Input:

int((2*x^3 - x^2 + exp(exp(8*x + 5*log(x) + 6) - x + 2*x*exp(4*x + (5*log( 
x))/2 + 3) + x^2)*(exp(8*x + 5*log(x) + 6)*(8*x + 5) - x + exp(4*x + (5*lo 
g(x))/2 + 3)*(7*x + 8*x^2) + 2*x^2 - 1))/x^2,x)
 

Output:

x^2 - x + (exp(-x)*exp(x^2)*exp(2*x^(7/2)*exp(4*x)*exp(3))*exp(x^5*exp(8*x 
)*exp(6)))/x
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {-x^2+2 x^3+e^{-x+2 e^{\frac {1}{2} (6+8 x+5 \log (x))} x+x^2+e^{6+8 x} x^5} \left (-1-x+2 x^2+e^{6+8 x} x^5 (5+8 x)+e^{\frac {1}{2} (6+8 x+5 \log (x))} \left (7 x+8 x^2\right )\right )}{x^2} \, dx=\frac {e^{e^{8 x} e^{6} x^{5}+2 \sqrt {x}\, e^{4 x} e^{3} x^{3}+x^{2}}+e^{x} x^{3}-e^{x} x^{2}}{e^{x} x} \] Input:

int((((8*x+5)*exp(5/2*log(x)+4*x+3)^2+(8*x^2+7*x)*exp(5/2*log(x)+4*x+3)+2* 
x^2-x-1)*exp(exp(5/2*log(x)+4*x+3)^2+2*x*exp(5/2*log(x)+4*x+3)+x^2-x)+2*x^ 
3-x^2)/x^2,x)
 

Output:

(e**(e**(8*x)*e**6*x**5 + 2*sqrt(x)*e**(4*x)*e**3*x**3 + x**2) + e**x*x**3 
 - e**x*x**2)/(e**x*x)