\(\int \frac {e^x (2 x+2 x^2+(-8 x^4-2 x^5) \log ^2(5))+e^x (-2 x-x^2+(2 x^4+x^5) \log ^2(5)) \log (e^x (-x+x^4 \log ^2(5)))}{(-1+x^3 \log ^2(5)) \log ^3(e^x (-x+x^4 \log ^2(5)))} \, dx\) [2605]

Optimal result

Integrand size = 106, antiderivative size = 26 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {e^x x^2}{\log ^2\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {e^x x^2}{\log ^2\left (e^x x \left (-1+x^3 \log ^2(5)\right )\right )} \] Input:

Integrate[(E^x*(2*x + 2*x^2 + (-8*x^4 - 2*x^5)*Log[5]^2) + E^x*(-2*x - x^2 
 + (2*x^4 + x^5)*Log[5]^2)*Log[E^x*(-x + x^4*Log[5]^2)])/((-1 + x^3*Log[5] 
^2)*Log[E^x*(-x + x^4*Log[5]^2)]^3),x]
 

Output:

(E^x*x^2)/Log[E^x*x*(-1 + x^3*Log[5]^2)]^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.

\(\displaystyle \int \frac {e^x \left (2 x^2+\left (-2 x^5-8 x^4\right ) \log ^2(5)+2 x\right )+e^x \left (-x^2+\left (x^5+2 x^4\right ) \log ^2(5)-2 x\right ) \log \left (e^x \left (x^4 \log ^2(5)-x\right )\right )}{\left (x^3 \log ^2(5)-1\right ) \log ^3\left (e^x \left (x^4 \log ^2(5)-x\right )\right )} \, dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (-\frac {2 e^x x}{\left (x^3 \log ^2(5)-1\right ) \log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}+\frac {2 e^x x}{\left (x^3 \log ^2(5)-1\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}+\frac {e^x x^5 \log ^2(5)}{\left (x^3 \log ^2(5)-1\right ) \log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}-\frac {2 e^x x^5 \log ^2(5)}{\left (x^3 \log ^2(5)-1\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}+\frac {2 e^x x^4 \log ^2(5)}{\left (x^3 \log ^2(5)-1\right ) \log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}-\frac {8 e^x x^4 \log ^2(5)}{\left (x^3 \log ^2(5)-1\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}-\frac {e^x x^2}{\left (x^3 \log ^2(5)-1\right ) \log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}+\frac {2 e^x x^2}{\left (x^3 \log ^2(5)-1\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \frac {e^x x}{\log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx-8 \int \frac {e^x x}{\log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx-\frac {2 \sqrt [3]{-1} \int \frac {e^x}{\left (1-x (-\log (5))^{2/3}\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx}{\log ^{\frac {2}{3}}(5)}+\frac {2 \int \frac {e^x}{\left (1-x \log ^{\frac {2}{3}}(5)\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx}{\log ^{\frac {2}{3}}(5)}+2 \left (-\frac {1}{\log (5)}\right )^{2/3} \int \frac {e^x}{\left (\sqrt [3]{-1} \log ^{\frac {2}{3}}(5) x+1\right ) \log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx+\int \frac {e^x x^2}{\log ^2\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx-2 \int \frac {e^x x^2}{\log ^3\left (e^x x \left (x^3 \log ^2(5)-1\right )\right )}dx\)

Input:

Int[(E^x*(2*x + 2*x^2 + (-8*x^4 - 2*x^5)*Log[5]^2) + E^x*(-2*x - x^2 + (2* 
x^4 + x^5)*Log[5]^2)*Log[E^x*(-x + x^4*Log[5]^2)])/((-1 + x^3*Log[5]^2)*Lo 
g[E^x*(-x + x^4*Log[5]^2)]^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x} x^{2}}{{\ln \left (\left (x^{4} \ln \left (5\right )^{2}-x \right ) {\mathrm e}^{x}\right )}^{2}}\)	\(25\)
risch	\(-\frac {4 x^{2} {\mathrm e}^{x}}{{\left (\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \left (x^{3} \ln \left (5\right )^{2}-1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{3} \ln \left (5\right )^{2}-1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right ) {\operatorname {csgn}\left (i x \left (x^{3} \ln \left (5\right )^{2}-1\right ) {\mathrm e}^{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{3} \ln \left (5\right )^{2}-1\right )\right ) \operatorname {csgn}\left (i x \left (x^{3} \ln \left (5\right )^{2}-1\right ) {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i x \right )+\pi {\operatorname {csgn}\left (i x \left (x^{3} \ln \left (5\right )^{2}-1\right ) {\mathrm e}^{x}\right )}^{3}-\pi {\operatorname {csgn}\left (i x \left (x^{3} \ln \left (5\right )^{2}-1\right ) {\mathrm e}^{x}\right )}^{2} \operatorname {csgn}\left (i x \right )+2 i \ln \left (x \right )+2 i \ln \left (x^{3} \ln \left (5\right )^{2}-1\right )+2 i \ln \left ({\mathrm e}^{x}\right )\right )}^{2}}\)	\(282\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {x^{2} e^{x}}{\log \left ({\left (x^{4} \log \left (5\right )^{2} - x\right )} e^{x}\right )^{2}} \] Input:

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

Output:

x^2*e^x/log((x^4*log(5)^2 - x)*e^x)^2
 

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {x^{2} e^{x}}{\log {\left (\left (x^{4} \log {\left (5 \right )}^{2} - x\right ) e^{x} \right )}^{2}} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (24) = 48\).

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {x^{2} e^{x}}{x^{2} + 2 \, {\left (x + \log \left (x\right )\right )} \log \left (x^{3} \log \left (5\right )^{2} - 1\right ) + \log \left (x^{3} \log \left (5\right )^{2} - 1\right )^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {x^{2} e^{x}}{x^{2} + 2 \, x \log \left (x^{4} \log \left (5\right )^{2} - x\right ) + \log \left (x^{4} \log \left (5\right )^{2} - x\right )^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {x^2\,{\mathrm {e}}^x}{{\ln \left (-{\mathrm {e}}^x\,\left (x-x^4\,{\ln \left (5\right )}^2\right )\right )}^2} \] Input:

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

Output:

(x^2*exp(x))/log(-exp(x)*(x - x^4*log(5)^2))^2
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (2 x+2 x^2+\left (-8 x^4-2 x^5\right ) \log ^2(5)\right )+e^x \left (-2 x-x^2+\left (2 x^4+x^5\right ) \log ^2(5)\right ) \log \left (e^x \left (-x+x^4 \log ^2(5)\right )\right )}{\left (-1+x^3 \log ^2(5)\right ) \log ^3\left (e^x \left (-x+x^4 \log ^2(5)\right )\right )} \, dx=\frac {e^{x} x^{2}}{\mathrm {log}\left (e^{x} \mathrm {log}\left (5\right )^{2} x^{4}-e^{x} x \right )^{2}} \] Input:

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

Output:

(e**x*x**2)/log(e**x*log(5)**2*x**4 - e**x*x)**2