\(\int \frac {-192+64 x-96 x^3+32 x^4+e^x (-128 x+64 x^2+32 x^4-16 x^5)+e^{x^2} (-30 x^3+10 x^4+e^x (10 x^2-5 x^3))+(-64 x+5 e^{x^2} x^2+16 x^4) \log (\frac {64-5 e^{x^2} x-16 x^3}{16 x})}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx\) [782]

Optimal result

Integrand size = 135, antiderivative size = 33 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=(3-x) \left (e^x-\log \left (-\frac {5 e^{x^2}}{16}+\frac {4}{x}-x^2\right )\right ) \] Output:

(exp(x)-ln(4/x-5/16*exp(x^2)-x^2))*(3-x)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=e^x (3-x)+3 \log (x)+x \log \left (-\frac {5 e^{x^2}}{16}+\frac {4}{x}-x^2\right )-3 \log \left (64-5 e^{x^2} x-16 x^3\right ) \] Input:

Integrate[(-192 + 64*x - 96*x^3 + 32*x^4 + E^x*(-128*x + 64*x^2 + 32*x^4 - 
 16*x^5) + E^x^2*(-30*x^3 + 10*x^4 + E^x*(10*x^2 - 5*x^3)) + (-64*x + 5*E^ 
x^2*x^2 + 16*x^4)*Log[(64 - 5*E^x^2*x - 16*x^3)/(16*x)])/(-64*x + 5*E^x^2* 
x^2 + 16*x^4),x]
 

Output:

E^x*(3 - x) + 3*Log[x] + x*Log[(-5*E^x^2)/16 + 4/x - x^2] - 3*Log[64 - 5*E 
^x^2*x - 16*x^3]
 

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.

\(\displaystyle \int \frac {32 x^4-96 x^3+e^x \left (-16 x^5+32 x^4+64 x^2-128 x\right )+e^{x^2} \left (10 x^4-30 x^3+e^x \left (10 x^2-5 x^3\right )\right )+\left (16 x^4+5 e^{x^2} x^2-64 x\right ) \log \left (\frac {-16 x^3-5 e^{x^2} x+64}{16 x}\right )+64 x-192}{16 x^4+5 e^{x^2} x^2-64 x} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (2 x^2+\log \left (-\frac {16 x^3+5 e^{x^2} x-64}{16 x}\right )-\frac {32 \left (x^6-3 x^5-x^4-x^3+12 x^2-2 x+6\right )}{\left (16 x^3+5 e^{x^2} x-64\right ) x}-e^x x-6 x+2 e^x\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -192 \int \frac {1}{x \left (16 x^3+5 e^{x^2} x-64\right )}dx-384 \int \frac {x}{16 x^3+5 e^{x^2} x-64}dx-96 \int \frac {x^2}{16 x^3+5 e^{x^2} x-64}dx+96 \int \frac {x^4}{16 x^3+5 e^{x^2} x-64}dx-3 x^2+x \log \left (\frac {-16 x^3-5 e^{x^2} x+64}{16 x}\right )-e^x x+3 e^x\)

Input:

Int[(-192 + 64*x - 96*x^3 + 32*x^4 + E^x*(-128*x + 64*x^2 + 32*x^4 - 16*x^ 
5) + E^x^2*(-30*x^3 + 10*x^4 + E^x*(10*x^2 - 5*x^3)) + (-64*x + 5*E^x^2*x^ 
2 + 16*x^4)*Log[(64 - 5*E^x^2*x - 16*x^3)/(16*x)])/(-64*x + 5*E^x^2*x^2 + 
16*x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 4.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58

method	result	size
parallelrisch	\(-{\mathrm e}^{x} x +\ln \left (-\frac {16 x^{3}+5 \,{\mathrm e}^{x^{2}} x -64}{16 x}\right ) x +3 \ln \left (x \right )-3 \ln \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )+3 \,{\mathrm e}^{x}\)	\(52\)
default	\(-3 \ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right )+\ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right ) x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}\)	\(55\)
parts	\(-3 \ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right )+\ln \left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right ) x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}\)	\(55\)
risch	\(x \ln \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )-x \ln \left (x \right )-i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{2}}{2}+\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{3}+\frac {5 \,{\mathrm e}^{x^{2}} x}{16}-4\right )}{x}\right )}^{3}}{2}+i \pi x -{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}-3 \ln \left ({\mathrm e}^{x^{2}}+\frac {\frac {16 x^{3}}{5}-\frac {64}{5}}{x}\right )\)	\(228\)

Input:

int(((5*x^2*exp(x^2)+16*x^4-64*x)*ln(1/16*(-5*exp(x^2)*x-16*x^3+64)/x)+((- 
5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2-128*x) 
*exp(x)+32*x^4-96*x^3+64*x-192)/(5*x^2*exp(x^2)+16*x^4-64*x),x,method=_RET 
URNVERBOSE)
 

Output:

-exp(x)*x+ln(-1/16*(16*x^3+5*exp(x^2)*x-64)/x)*x+3*ln(x)-3*ln(x^3+5/16*exp 
(x^2)*x-4)+3*exp(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-{\left (x - 3\right )} e^{x} + {\left (x - 3\right )} \log \left (-\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{16 \, x}\right ) \] Input:

integrate(((5*exp(x^2)*x^2+16*x^4-64*x)*log(1/16*(-5*exp(x^2)*x-16*x^3+64) 
/x)+((-5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2 
-128*x)*exp(x)+32*x^4-96*x^3+64*x-192)/(5*exp(x^2)*x^2+16*x^4-64*x),x, alg 
orithm="fricas")
 

Output:

-(x - 3)*e^x + (x - 3)*log(-1/16*(16*x^3 + 5*x*e^(x^2) - 64)/x)
 

Sympy [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=x \log {\left (\frac {- x^{3} - \frac {5 x e^{x^{2}}}{16} + 4}{x} \right )} + \left (3 - x\right ) e^{x} - 3 \log {\left (e^{x^{2}} + \frac {16 x^{3} - 64}{5 x} \right )} \] Input:

integrate(((5*exp(x**2)*x**2+16*x**4-64*x)*ln(1/16*(-5*exp(x**2)*x-16*x**3 
+64)/x)+((-5*x**3+10*x**2)*exp(x)+10*x**4-30*x**3)*exp(x**2)+(-16*x**5+32* 
x**4+64*x**2-128*x)*exp(x)+32*x**4-96*x**3+64*x-192)/(5*exp(x**2)*x**2+16* 
x**4-64*x),x)
 

Output:

x*log((-x**3 - 5*x*exp(x**2)/16 + 4)/x) + (3 - x)*exp(x) - 3*log(exp(x**2) 
 + (16*x**3 - 64)/(5*x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).

Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-{\left (x - 3\right )} e^{x} - 4 \, x \log \left (2\right ) + x \log \left (-16 \, x^{3} - 5 \, x e^{\left (x^{2}\right )} + 64\right ) - x \log \left (x\right ) - 3 \, \log \left (\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{5 \, x}\right ) \] Input:

integrate(((5*exp(x^2)*x^2+16*x^4-64*x)*log(1/16*(-5*exp(x^2)*x-16*x^3+64) 
/x)+((-5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2 
-128*x)*exp(x)+32*x^4-96*x^3+64*x-192)/(5*exp(x^2)*x^2+16*x^4-64*x),x, alg 
orithm="maxima")
 

Output:

-(x - 3)*e^x - 4*x*log(2) + x*log(-16*x^3 - 5*x*e^(x^2) + 64) - x*log(x) - 
 3*log(1/5*(16*x^3 + 5*x*e^(x^2) - 64)/x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=-x e^{x} + x \log \left (-\frac {16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64}{16 \, x}\right ) + 3 \, e^{x} - 3 \, \log \left (16 \, x^{3} + 5 \, x e^{\left (x^{2}\right )} - 64\right ) + 3 \, \log \left (x\right ) \] Input:

integrate(((5*exp(x^2)*x^2+16*x^4-64*x)*log(1/16*(-5*exp(x^2)*x-16*x^3+64) 
/x)+((-5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2 
-128*x)*exp(x)+32*x^4-96*x^3+64*x-192)/(5*exp(x^2)*x^2+16*x^4-64*x),x, alg 
orithm="giac")
 

Output:

-x*e^x + x*log(-1/16*(16*x^3 + 5*x*e^(x^2) - 64)/x) + 3*e^x - 3*log(16*x^3 
 + 5*x*e^(x^2) - 64) + 3*log(x)
 

Mupad [B] (verification not implemented)

Time = 2.71 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=x\,\ln \left (-\frac {\frac {5\,x\,{\mathrm {e}}^{x^2}}{16}+x^3-4}{x}\right )-{\mathrm {e}}^x\,\left (x-3\right )-3\,\ln \left (\frac {5\,x\,{\mathrm {e}}^{x^2}+16\,x^3-64}{x}\right ) \] Input:

int((64*x - exp(x)*(128*x - 64*x^2 - 32*x^4 + 16*x^5) + log(-((5*x*exp(x^2 
))/16 + x^3 - 4)/x)*(5*x^2*exp(x^2) - 64*x + 16*x^4) + exp(x^2)*(exp(x)*(1 
0*x^2 - 5*x^3) - 30*x^3 + 10*x^4) - 96*x^3 + 32*x^4 - 192)/(5*x^2*exp(x^2) 
 - 64*x + 16*x^4),x)
 

Output:

x*log(-((5*x*exp(x^2))/16 + x^3 - 4)/x) - exp(x)*(x - 3) - 3*log((5*x*exp( 
x^2) + 16*x^3 - 64)/x)
 

Reduce [F]

\[ \int \frac {-192+64 x-96 x^3+32 x^4+e^x \left (-128 x+64 x^2+32 x^4-16 x^5\right )+e^{x^2} \left (-30 x^3+10 x^4+e^x \left (10 x^2-5 x^3\right )\right )+\left (-64 x+5 e^{x^2} x^2+16 x^4\right ) \log \left (\frac {64-5 e^{x^2} x-16 x^3}{16 x}\right )}{-64 x+5 e^{x^2} x^2+16 x^4} \, dx=\int \frac {\left (5 \,{\mathrm e}^{x^{2}} x^{2}+16 x^{4}-64 x \right ) \mathrm {log}\left (\frac {-5 \,{\mathrm e}^{x^{2}} x -16 x^{3}+64}{16 x}\right )+\left (\left (-5 x^{3}+10 x^{2}\right ) {\mathrm e}^{x}+10 x^{4}-30 x^{3}\right ) {\mathrm e}^{x^{2}}+\left (-16 x^{5}+32 x^{4}+64 x^{2}-128 x \right ) {\mathrm e}^{x}+32 x^{4}-96 x^{3}+64 x -192}{5 \,{\mathrm e}^{x^{2}} x^{2}+16 x^{4}-64 x}d x \] Input:

int(((5*exp(x^2)*x^2+16*x^4-64*x)*log(1/16*(-5*exp(x^2)*x-16*x^3+64)/x)+(( 
-5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2-128*x 
)*exp(x)+32*x^4-96*x^3+64*x-192)/(5*exp(x^2)*x^2+16*x^4-64*x),x)
 

Output:

int(((5*exp(x^2)*x^2+16*x^4-64*x)*log(1/16*(-5*exp(x^2)*x-16*x^3+64)/x)+(( 
-5*x^3+10*x^2)*exp(x)+10*x^4-30*x^3)*exp(x^2)+(-16*x^5+32*x^4+64*x^2-128*x 
)*exp(x)+32*x^4-96*x^3+64*x-192)/(5*exp(x^2)*x^2+16*x^4-64*x),x)