\(\int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} (-50 x+20 x^2-2 x^3)}{(-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} (25-10 x+x^2)) \log (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x})} \, dx\) [803]

Optimal result

Integrand size = 117, antiderivative size = 33 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=2+\log \left (\log \left (-3+e^{2-x^2}-2 x-\frac {x}{5-x}+4 x^2\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\log \left (\log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )\right ) \] Input:

Integrate[(-55 + 220*x - 82*x^2 + 8*x^3 + E^(2 - x^2)*(-50*x + 20*x^2 - 2* 
x^3))/((-75 - 25*x + 118*x^2 - 42*x^3 + 4*x^4 + E^(2 - x^2)*(25 - 10*x + x 
^2))*Log[(15 + E^(2 - x^2)*(-5 + x) + 8*x - 22*x^2 + 4*x^3)/(-5 + x)]),x]
 

Output:

Log[Log[(15 + E^(2 - x^2)*(-5 + x) + 8*x - 22*x^2 + 4*x^3)/(-5 + x)]]
 

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.

\(\displaystyle \int \frac {8 x^3-82 x^2+e^{2-x^2} \left (-2 x^3+20 x^2-50 x\right )+220 x-55}{\left (4 x^4-42 x^3+118 x^2+e^{2-x^2} \left (x^2-10 x+25\right )-25 x-75\right ) \log \left (\frac {4 x^3-22 x^2+e^{2-x^2} (x-5)+8 x+15}{x-5}\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {8 x^3-82 x^2+220 x-55}{(x-5) \left (4 x^3-22 x^2+8 x+15\right ) \log \left (\frac {4 x^3-22 x^2+e^{2-x^2} (x-5)+8 x+15}{x-5}\right )}-\frac {e^2 \left (8 x^5-84 x^4+244 x^3-132 x^2+70 x-55\right )}{\left (4 x^3-22 x^2+8 x+15\right ) \left (-22 e^{x^2} x^2+8 e^{x^2} x+15 e^{x^2}+4 e^{x^2} x^3+e^2 x-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+e^{2-x^2} (x-5)+8 x+15}{x-5}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {1}{(x-5) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx+8 \int \frac {1}{\left (4 x^3-22 x^2+8 x+15\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx-44 \int \frac {x}{\left (4 x^3-22 x^2+8 x+15\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx+12 \int \frac {x^2}{\left (4 x^3-22 x^2+8 x+15\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx-2 e^2 \int \frac {1}{\left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx+10 e^2 \int \frac {x}{\left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx-2 e^2 \int \frac {x^2}{\left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx+85 e^2 \int \frac {1}{\left (4 x^3-22 x^2+8 x+15\right ) \left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx-204 e^2 \int \frac {x}{\left (4 x^3-22 x^2+8 x+15\right ) \left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx+38 e^2 \int \frac {x^2}{\left (4 x^3-22 x^2+8 x+15\right ) \left (4 e^{x^2} x^3-22 e^{x^2} x^2+8 e^{x^2} x+e^2 x+15 e^{x^2}-5 e^2\right ) \log \left (\frac {4 x^3-22 x^2+8 x+e^{2-x^2} (x-5)+15}{x-5}\right )}dx\)

Input:

Int[(-55 + 220*x - 82*x^2 + 8*x^3 + E^(2 - x^2)*(-50*x + 20*x^2 - 2*x^3))/ 
((-75 - 25*x + 118*x^2 - 42*x^3 + 4*x^4 + E^(2 - x^2)*(25 - 10*x + x^2))*L 
og[(15 + E^(2 - x^2)*(-5 + x) + 8*x - 22*x^2 + 4*x^3)/(-5 + x)]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
norman	\(\ln \left (\ln \left (\frac {\left (-5+x \right ) {\mathrm e}^{-x^{2}+2}+4 x^{3}-22 x^{2}+8 x +15}{-5+x}\right )\right )\)	\(36\)
parallelrisch	\(\ln \left (\ln \left (\frac {\left (-5+x \right ) {\mathrm e}^{-x^{2}+2}+4 x^{3}-22 x^{2}+8 x +15}{-5+x}\right )\right )\)	\(36\)
risch	\(\ln \left (2 \ln \left (2\right )-\ln \left (-5+x \right )+\ln \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )}{-5+x}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{-5+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )}{-5+x}\right )}^{2}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )}{-5+x}\right )}^{2}}{2}-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (x^{3}-\frac {11 x^{2}}{2}+\left (\frac {{\mathrm e}^{-x^{2}+2}}{4}+2\right ) x -\frac {5 \,{\mathrm e}^{-x^{2}+2}}{4}+\frac {15}{4}\right )}{-5+x}\right )}^{3}}{2}\right )\)	\(336\)

Input:

int(((-2*x^3+20*x^2-50*x)*exp(-x^2+2)+8*x^3-82*x^2+220*x-55)/((x^2-10*x+25 
)*exp(-x^2+2)+4*x^4-42*x^3+118*x^2-25*x-75)/ln(((-5+x)*exp(-x^2+2)+4*x^3-2 
2*x^2+8*x+15)/(-5+x)),x,method=_RETURNVERBOSE)
 

Output:

ln(ln(((-5+x)*exp(-x^2+2)+4*x^3-22*x^2+8*x+15)/(-5+x)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\log \left (\log \left (\frac {4 \, x^{3} - 22 \, x^{2} + {\left (x - 5\right )} e^{\left (-x^{2} + 2\right )} + 8 \, x + 15}{x - 5}\right )\right ) \] Input:

integrate(((-2*x^3+20*x^2-50*x)*exp(-x^2+2)+8*x^3-82*x^2+220*x-55)/((x^2-1 
0*x+25)*exp(-x^2+2)+4*x^4-42*x^3+118*x^2-25*x-75)/log(((-5+x)*exp(-x^2+2)+ 
4*x^3-22*x^2+8*x+15)/(-5+x)),x, algorithm="fricas")
 

Output:

log(log((4*x^3 - 22*x^2 + (x - 5)*e^(-x^2 + 2) + 8*x + 15)/(x - 5)))
 

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\log {\left (\log {\left (\frac {4 x^{3} - 22 x^{2} + 8 x + \left (x - 5\right ) e^{2 - x^{2}} + 15}{x - 5} \right )} \right )} \] Input:

integrate(((-2*x**3+20*x**2-50*x)*exp(-x**2+2)+8*x**3-82*x**2+220*x-55)/(( 
x**2-10*x+25)*exp(-x**2+2)+4*x**4-42*x**3+118*x**2-25*x-75)/ln(((-5+x)*exp 
(-x**2+2)+4*x**3-22*x**2+8*x+15)/(-5+x)),x)
 

Output:

log(log((4*x**3 - 22*x**2 + 8*x + (x - 5)*exp(2 - x**2) + 15)/(x - 5)))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\log \left (-x^{2} + \log \left (x e^{2} + {\left (4 \, x^{3} - 22 \, x^{2} + 8 \, x + 15\right )} e^{\left (x^{2}\right )} - 5 \, e^{2}\right ) - \log \left (x - 5\right )\right ) \] Input:

integrate(((-2*x^3+20*x^2-50*x)*exp(-x^2+2)+8*x^3-82*x^2+220*x-55)/((x^2-1 
0*x+25)*exp(-x^2+2)+4*x^4-42*x^3+118*x^2-25*x-75)/log(((-5+x)*exp(-x^2+2)+ 
4*x^3-22*x^2+8*x+15)/(-5+x)),x, algorithm="maxima")
 

Output:

log(-x^2 + log(x*e^2 + (4*x^3 - 22*x^2 + 8*x + 15)*e^(x^2) - 5*e^2) - log( 
x - 5))
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\log \left (\log \left (\frac {4 \, x^{3} - 22 \, x^{2} + x e^{\left (-x^{2} + 2\right )} + 8 \, x - 5 \, e^{\left (-x^{2} + 2\right )} + 15}{x - 5}\right )\right ) \] Input:

integrate(((-2*x^3+20*x^2-50*x)*exp(-x^2+2)+8*x^3-82*x^2+220*x-55)/((x^2-1 
0*x+25)*exp(-x^2+2)+4*x^4-42*x^3+118*x^2-25*x-75)/log(((-5+x)*exp(-x^2+2)+ 
4*x^3-22*x^2+8*x+15)/(-5+x)),x, algorithm="giac")
 

Output:

log(log((4*x^3 - 22*x^2 + x*e^(-x^2 + 2) + 8*x - 5*e^(-x^2 + 2) + 15)/(x - 
 5)))
 

Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\ln \left (\ln \left (\frac {8\,x-22\,x^2+4\,x^3+{\mathrm {e}}^2\,{\mathrm {e}}^{-x^2}\,\left (x-5\right )+15}{x-5}\right )\right ) \] Input:

int((82*x^2 - 220*x - 8*x^3 + exp(2 - x^2)*(50*x - 20*x^2 + 2*x^3) + 55)/( 
log((8*x + exp(2 - x^2)*(x - 5) - 22*x^2 + 4*x^3 + 15)/(x - 5))*(25*x - ex 
p(2 - x^2)*(x^2 - 10*x + 25) - 118*x^2 + 42*x^3 - 4*x^4 + 75)),x)
 

Output:

log(log((8*x - 22*x^2 + 4*x^3 + exp(2)*exp(-x^2)*(x - 5) + 15)/(x - 5)))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {-55+220 x-82 x^2+8 x^3+e^{2-x^2} \left (-50 x+20 x^2-2 x^3\right )}{\left (-75-25 x+118 x^2-42 x^3+4 x^4+e^{2-x^2} \left (25-10 x+x^2\right )\right ) \log \left (\frac {15+e^{2-x^2} (-5+x)+8 x-22 x^2+4 x^3}{-5+x}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {4 e^{x^{2}} x^{3}-22 e^{x^{2}} x^{2}+8 e^{x^{2}} x +15 e^{x^{2}}+e^{2} x -5 e^{2}}{e^{x^{2}} x -5 e^{x^{2}}}\right )\right ) \] Input:

int(((-2*x^3+20*x^2-50*x)*exp(-x^2+2)+8*x^3-82*x^2+220*x-55)/((x^2-10*x+25 
)*exp(-x^2+2)+4*x^4-42*x^3+118*x^2-25*x-75)/log(((-5+x)*exp(-x^2+2)+4*x^3- 
22*x^2+8*x+15)/(-5+x)),x)
 

Output:

log(log((4*e**(x**2)*x**3 - 22*e**(x**2)*x**2 + 8*e**(x**2)*x + 15*e**(x** 
2) + e**2*x - 5*e**2)/(e**(x**2)*x - 5*e**(x**2))))