\(\int \frac {-30 x+6 x^2+e^x (-45+54 x-9 x^2) \log (4)+(-90 x^2+18 x^3) \log (4)+(-10 x+2 x^2+e^x (-30+36 x-6 x^2) \log (4)+(-60 x^2+12 x^3) \log (4)) \log (5-x)+(e^x (-5+6 x-x^2) \log (4)+(-10 x^2+2 x^3) \log (4)) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+(-30 x^2+6 x^3) \log (5-x)+(-5 x^2+x^3) \log ^2(5-x)} \, dx\) [1557]

Optimal result

Integrand size = 182, antiderivative size = 30 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=-\left (\left (\frac {e^x}{x}-2 x\right ) \log (4)\right )+\frac {2 \log (x)}{3+\log (5-x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=-\frac {e^x \log (4)}{x}+2 x \log (4)+\frac {2 \log (x)}{3+\log (5-x)} \] Input:

Integrate[(-30*x + 6*x^2 + E^x*(-45 + 54*x - 9*x^2)*Log[4] + (-90*x^2 + 18 
*x^3)*Log[4] + (-10*x + 2*x^2 + E^x*(-30 + 36*x - 6*x^2)*Log[4] + (-60*x^2 
 + 12*x^3)*Log[4])*Log[5 - x] + (E^x*(-5 + 6*x - x^2)*Log[4] + (-10*x^2 + 
2*x^3)*Log[4])*Log[5 - x]^2 - 2*x^2*Log[x])/(-45*x^2 + 9*x^3 + (-30*x^2 + 
6*x^3)*Log[5 - x] + (-5*x^2 + x^3)*Log[5 - x]^2),x]
 

Output:

-((E^x*Log[4])/x) + 2*x*Log[4] + (2*Log[x])/(3 + Log[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.

\(\displaystyle \int \frac {6 x^2-2 x^2 \log (x)+e^x \left (-9 x^2+54 x-45\right ) \log (4)+\left (e^x \left (-x^2+6 x-5\right ) \log (4)+\left (2 x^3-10 x^2\right ) \log (4)\right ) \log ^2(5-x)+\left (2 x^2+e^x \left (-6 x^2+36 x-30\right ) \log (4)+\left (12 x^3-60 x^2\right ) \log (4)-10 x\right ) \log (5-x)+\left (18 x^3-90 x^2\right ) \log (4)-30 x}{9 x^3-45 x^2+\left (x^3-5 x^2\right ) \log ^2(5-x)+\left (6 x^3-30 x^2\right ) \log (5-x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-6 x^2+2 x^2 \log (x)-e^x \left (-9 x^2+54 x-45\right ) \log (4)-\left (e^x \left (-x^2+6 x-5\right ) \log (4)+\left (2 x^3-10 x^2\right ) \log (4)\right ) \log ^2(5-x)-\left (2 x^2+e^x \left (-6 x^2+36 x-30\right ) \log (4)+\left (12 x^3-60 x^2\right ) \log (4)-10 x\right ) \log (5-x)-\left (18 x^3-90 x^2\right ) \log (4)+30 x}{(5-x) x^2 (\log (5-x)+3)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^x (x-1) \log (4)}{x^2}+\frac {2 \log (4) \log ^2(5-x)}{(\log (5-x)+3)^2}+\frac {2 \log (5-x)}{(x-5) (\log (5-x)+3)^2}-\frac {10 \log (5-x)}{(x-5) x (\log (5-x)+3)^2}+\frac {12 \log (4) \log (5-x)}{(\log (5-x)+3)^2}-\frac {2 \log (x)}{(x-5) (\log (5-x)+3)^2}+\frac {6}{(x-5) (\log (5-x)+3)^2}-\frac {30}{(x-5) x (\log (5-x)+3)^2}+\frac {18 \log (4)}{(\log (5-x)+3)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \text {Subst}\left (\int \frac {\log (5-x)}{x (\log (x)+3)^2}dx,x,5-x\right )+2 \int \frac {1}{x (\log (5-x)+3)}dx+2 x \log (4)-\frac {e^x \log (4)}{x}\)

Input:

Int[(-30*x + 6*x^2 + E^x*(-45 + 54*x - 9*x^2)*Log[4] + (-90*x^2 + 18*x^3)* 
Log[4] + (-10*x + 2*x^2 + E^x*(-30 + 36*x - 6*x^2)*Log[4] + (-60*x^2 + 12* 
x^3)*Log[4])*Log[5 - x] + (E^x*(-5 + 6*x - x^2)*Log[4] + (-10*x^2 + 2*x^3) 
*Log[4])*Log[5 - x]^2 - 2*x^2*Log[x])/(-45*x^2 + 9*x^3 + (-30*x^2 + 6*x^3) 
*Log[5 - x] + (-5*x^2 + x^3)*Log[5 - x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 151.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {2 \ln \left (2\right ) \left (2 x^{2}-{\mathrm e}^{x}\right )}{x}+\frac {2 \ln \left (x \right )}{\ln \left (5-x \right )+3}\)	\(33\)
parallelrisch	\(\frac {4 \ln \left (2\right ) \ln \left (5-x \right ) x^{2}+12 x^{2} \ln \left (2\right )+40 \ln \left (5-x \right ) x \ln \left (2\right )-2 \ln \left (2\right ) {\mathrm e}^{x} \ln \left (5-x \right )+120 x \ln \left (2\right )-6 \,{\mathrm e}^{x} \ln \left (2\right )+2 x \ln \left (x \right )}{x \left (\ln \left (5-x \right )+3\right )}\)	\(75\)

Input:

int((-2*x^2*ln(x)+(2*(-x^2+6*x-5)*ln(2)*exp(x)+2*(2*x^3-10*x^2)*ln(2))*ln( 
5-x)^2+(2*(-6*x^2+36*x-30)*ln(2)*exp(x)+2*(12*x^3-60*x^2)*ln(2)+2*x^2-10*x 
)*ln(5-x)+2*(-9*x^2+54*x-45)*ln(2)*exp(x)+2*(18*x^3-90*x^2)*ln(2)+6*x^2-30 
*x)/((x^3-5*x^2)*ln(5-x)^2+(6*x^3-30*x^2)*ln(5-x)+9*x^3-45*x^2),x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=\frac {2 \, {\left (6 \, x^{2} \log \left (2\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right ) + {\left (2 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right )\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \] Input:

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*lo 
g(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2 
)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2)* 
log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-4 
5*x^2),x, algorithm="fricas")
 

Output:

2*(6*x^2*log(2) - 3*e^x*log(2) + x*log(x) + (2*x^2*log(2) - e^x*log(2))*lo 
g(-x + 5))/(x*log(-x + 5) + 3*x)
 

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=4 x \log {\left (2 \right )} + \frac {2 \log {\left (x \right )}}{\log {\left (5 - x \right )} + 3} - \frac {2 e^{x} \log {\left (2 \right )}}{x} \] Input:

integrate((-2*x**2*ln(x)+(2*(-x**2+6*x-5)*ln(2)*exp(x)+2*(2*x**3-10*x**2)* 
ln(2))*ln(5-x)**2+(2*(-6*x**2+36*x-30)*ln(2)*exp(x)+2*(12*x**3-60*x**2)*ln 
(2)+2*x**2-10*x)*ln(5-x)+2*(-9*x**2+54*x-45)*ln(2)*exp(x)+2*(18*x**3-90*x* 
*2)*ln(2)+6*x**2-30*x)/((x**3-5*x**2)*ln(5-x)**2+(6*x**3-30*x**2)*ln(5-x)+ 
9*x**3-45*x**2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=\frac {2 \, {\left (6 \, x^{2} \log \left (2\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right ) + {\left (2 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right )\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \] Input:

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*lo 
g(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2 
)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2)* 
log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-4 
5*x^2),x, algorithm="maxima")
 

Output:

2*(6*x^2*log(2) - 3*e^x*log(2) + x*log(x) + (2*x^2*log(2) - e^x*log(2))*lo 
g(-x + 5))/(x*log(-x + 5) + 3*x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.97 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=\frac {2 \, {\left (2 \, x^{2} \log \left (2\right ) \log \left (-x + 5\right ) + 6 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right ) \log \left (-x + 5\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \] Input:

integrate((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*lo 
g(2))*log(5-x)^2+(2*(-6*x^2+36*x-30)*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2 
)+2*x^2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2)* 
log(2)+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-4 
5*x^2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=4\,x\,\ln \left (2\right )-\frac {10}{x}+\frac {\frac {2\,\left (x\,\ln \left (x\right )-3\,x+15\right )}{x}-\frac {2\,\ln \left (5-x\right )\,\left (x-5\right )}{x}}{\ln \left (5-x\right )+3}-\frac {2\,{\mathrm {e}}^x\,\ln \left (2\right )}{x} \] Input:

int((30*x + log(5 - x)^2*(2*log(2)*(10*x^2 - 2*x^3) + 2*exp(x)*log(2)*(x^2 
 - 6*x + 5)) + 2*x^2*log(x) + log(5 - x)*(10*x + 2*log(2)*(60*x^2 - 12*x^3 
) - 2*x^2 + 2*exp(x)*log(2)*(6*x^2 - 36*x + 30)) + 2*log(2)*(90*x^2 - 18*x 
^3) - 6*x^2 + 2*exp(x)*log(2)*(9*x^2 - 54*x + 45))/(log(5 - x)^2*(5*x^2 - 
x^3) + log(5 - x)*(30*x^2 - 6*x^3) + 45*x^2 - 9*x^3),x)
 

Output:

4*x*log(2) - 10/x + ((2*(x*log(x) - 3*x + 15))/x - (2*log(5 - x)*(x - 5))/ 
x)/(log(5 - x) + 3) - (2*exp(x)*log(2))/x
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx=\frac {-2 e^{x} \mathrm {log}\left (-x +5\right ) \mathrm {log}\left (2\right )-6 e^{x} \mathrm {log}\left (2\right )+4 \,\mathrm {log}\left (-x +5\right ) \mathrm {log}\left (2\right ) x^{2}+2 \,\mathrm {log}\left (x \right ) x +12 \,\mathrm {log}\left (2\right ) x^{2}}{x \left (\mathrm {log}\left (-x +5\right )+3\right )} \] Input:

int((-2*x^2*log(x)+(2*(-x^2+6*x-5)*log(2)*exp(x)+2*(2*x^3-10*x^2)*log(2))* 
log(5-x)^2+(2*(-6*x^2+36*x-30)*log(2)*exp(x)+2*(12*x^3-60*x^2)*log(2)+2*x^ 
2-10*x)*log(5-x)+2*(-9*x^2+54*x-45)*log(2)*exp(x)+2*(18*x^3-90*x^2)*log(2) 
+6*x^2-30*x)/((x^3-5*x^2)*log(5-x)^2+(6*x^3-30*x^2)*log(5-x)+9*x^3-45*x^2) 
,x)
 

Output:

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