\(\int \frac {-5 e^{2 x}+e^x (-45+15 x+10 x^2-3 x^3)+e^x (-5+5 x+3 x^2) \log (x)+e^x (-5+5 x+3 x^2) \log (81 x^4)}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x (8 x^2-2 x^3)+(8 x^2+2 e^x x^2-2 x^3) \log (x)+x^2 \log ^2(x)+(8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)) \log (81 x^4)+x^2 \log ^2(81 x^4)} \, dx\) [271]

Optimal result

Integrand size = 179, antiderivative size = 29 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=\frac {e^x \left (3+\frac {5}{x}\right )}{4+e^x-x+\log (x)+\log \left (81 x^4\right )} \] Output:

(3+5/x)/(ln(81*x^4)+4+exp(x)-x+ln(x))*exp(x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=\frac {e^x (5+3 x)}{x \left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )} \] Input:

Integrate[(-5*E^(2*x) + E^x*(-45 + 15*x + 10*x^2 - 3*x^3) + E^x*(-5 + 5*x 
+ 3*x^2)*Log[x] + E^x*(-5 + 5*x + 3*x^2)*Log[81*x^4])/(16*x^2 + E^(2*x)*x^ 
2 - 8*x^3 + x^4 + E^x*(8*x^2 - 2*x^3) + (8*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x] 
 + x^2*Log[x]^2 + (8*x^2 + 2*E^x*x^2 - 2*x^3 + 2*x^2*Log[x])*Log[81*x^4] + 
 x^2*Log[81*x^4]^2),x]
 

Output:

(E^x*(5 + 3*x))/(x*(4 + E^x - x + Log[x] + Log[81*x^4]))
 

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.

\(\displaystyle \int \frac {e^x \left (3 x^2+5 x-5\right ) \log (x)+e^x \left (3 x^2+5 x-5\right ) \log \left (81 x^4\right )+e^x \left (-3 x^3+10 x^2+15 x-45\right )-5 e^{2 x}}{x^4-8 x^3+e^{2 x} x^2+16 x^2+x^2 \log ^2(x)+x^2 \log ^2\left (81 x^4\right )+e^x \left (8 x^2-2 x^3\right )+\left (-2 x^3+2 e^x x^2+8 x^2\right ) \log (x)+\left (-2 x^3+2 e^x x^2+8 x^2+2 x^2 \log (x)\right ) \log \left (81 x^4\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^x \left (-3 x^3+10 x^2-\left (-3 x^2-5 x+5\right ) \log (x)-\left (-3 x^2-5 x+5\right ) \log \left (81 x^4\right )+15 x-5 e^x-45\right )}{x^2 \left (\log \left (81 x^4\right )-x+e^x+\log (x)+4\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {5 e^x}{x^2 \left (-\log \left (x^4\right )+x-e^x-\log (81 x)-4\right )}-\frac {e^x (3 x+5) \left (-x \log \left (81 x^4\right )+x^2-5 x-x \log (x)+5\right )}{x^2 \left (-\log \left (81 x^4\right )+x-e^x-\log (x)-4\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 10 \int \frac {e^x}{x \left (x-e^x-\log (x)-\log \left (81 x^4\right )-4\right )^2}dx-3 \int \frac {e^x x}{\left (x-e^x-\log (x)-\log \left (81 x^4\right )-4\right )^2}dx+5 \int \frac {e^x \log (x)}{x \left (x-e^x-\log (x)-\log \left (81 x^4\right )-4\right )^2}dx+5 \int \frac {e^x \log \left (81 x^4\right )}{x \left (x-e^x-\log (x)-\log \left (81 x^4\right )-4\right )^2}dx+10 \int \frac {e^x}{\left (-x+e^x+\log (x)+\log \left (81 x^4\right )+4\right )^2}dx+3 \int \frac {e^x \log (x)}{\left (-x+e^x+\log (x)+\log \left (81 x^4\right )+4\right )^2}dx+3 \int \frac {e^x \log \left (81 x^4\right )}{\left (-x+e^x+\log (x)+\log \left (81 x^4\right )+4\right )^2}dx+5 \int \frac {e^x}{x^2 \left (x-e^x-\log (81 x)-\log \left (x^4\right )-4\right )}dx-25 \int \frac {e^x}{x^2 \left (x-e^x-\log (x)-\log \left (81 x^4\right )-4\right )^2}dx\)

Input:

Int[(-5*E^(2*x) + E^x*(-45 + 15*x + 10*x^2 - 3*x^3) + E^x*(-5 + 5*x + 3*x^ 
2)*Log[x] + E^x*(-5 + 5*x + 3*x^2)*Log[81*x^4])/(16*x^2 + E^(2*x)*x^2 - 8* 
x^3 + x^4 + E^x*(8*x^2 - 2*x^3) + (8*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x] + x^2 
*Log[x]^2 + (8*x^2 + 2*E^x*x^2 - 2*x^3 + 2*x^2*Log[x])*Log[81*x^4] + x^2*L 
og[81*x^4]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24

method	result	size
parallelrisch	\(\frac {-3 \,{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}}{x \left (x -{\mathrm e}^{x}-\ln \left (x \right )-\ln \left (81 x^{4}\right )-4\right )}\)	\(36\)
risch	\(-\frac {2 \left (3 x +5\right ) {\mathrm e}^{x}}{x \left (i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}-i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+i \pi \operatorname {csgn}\left (i x^{4}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{4}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-8-8 \ln \left (3\right )+2 x -2 \,{\mathrm e}^{x}-10 \ln \left (x \right )\right )}\)	\(229\)

Input:

int(((3*x^2+5*x-5)*exp(x)*ln(81*x^4)+(3*x^2+5*x-5)*exp(x)*ln(x)-5*exp(x)^2 
+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^2*ln(81*x^4)^2+(2*x^2*ln(x)+2*exp(x)*x 
^2-2*x^3+8*x^2)*ln(81*x^4)+x^2*ln(x)^2+(2*exp(x)*x^2-2*x^3+8*x^2)*ln(x)+ex 
p(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x,method=_RETURNVERBOSE 
)
 

Output:

1/x*(-3*exp(x)*x-5*exp(x))/(x-exp(x)-ln(x)-ln(81*x^4)-4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=-\frac {{\left (3 \, x + 5\right )} e^{x}}{x^{2} - x e^{x} - 4 \, x \log \left (3\right ) - 5 \, x \log \left (x\right ) - 4 \, x} \] Input:

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5* 
exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+ 
2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8*x 
^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorit 
hm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=\frac {3 x^{2} - 15 x \log {\left (x \right )} - 12 x \log {\left (3 \right )} - 7 x - 25 \log {\left (x \right )} - 20 \log {\left (3 \right )} - 20}{- x^{2} + x e^{x} + 5 x \log {\left (x \right )} + 4 x + 4 x \log {\left (3 \right )}} + \frac {5}{x} \] Input:

integrate(((3*x**2+5*x-5)*exp(x)*ln(81*x**4)+(3*x**2+5*x-5)*exp(x)*ln(x)-5 
*exp(x)**2+(-3*x**3+10*x**2+15*x-45)*exp(x))/(x**2*ln(81*x**4)**2+(2*x**2* 
ln(x)+2*exp(x)*x**2-2*x**3+8*x**2)*ln(81*x**4)+x**2*ln(x)**2+(2*exp(x)*x** 
2-2*x**3+8*x**2)*ln(x)+exp(x)**2*x**2+(-2*x**3+8*x**2)*exp(x)+x**4-8*x**3+ 
16*x**2),x)
 

Output:

(3*x**2 - 15*x*log(x) - 12*x*log(3) - 7*x - 25*log(x) - 20*log(3) - 20)/(- 
x**2 + x*exp(x) + 5*x*log(x) + 4*x + 4*x*log(3)) + 5/x
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=-\frac {3 \, x^{2} - 12 \, x {\left (\log \left (3\right ) + 1\right )} - 15 \, x \log \left (x\right ) + 5 \, e^{x}}{x^{2} - 4 \, x {\left (\log \left (3\right ) + 1\right )} - x e^{x} - 5 \, x \log \left (x\right )} \] Input:

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5* 
exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+ 
2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8*x 
^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorit 
hm="maxima")
 

Output:

-(3*x^2 - 12*x*(log(3) + 1) - 15*x*log(x) + 5*e^x)/(x^2 - 4*x*(log(3) + 1) 
 - x*e^x - 5*x*log(x))
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=-\frac {3 \, x e^{x} + 5 \, e^{x}}{x^{2} - x e^{x} - 4 \, x \log \left (3\right ) - 5 \, x \log \left (x\right ) - 4 \, x} \] Input:

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5* 
exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+ 
2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8*x 
^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorit 
hm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=-\frac {12\,x-5\,{\mathrm {e}}^x+3\,x\,\ln \left (81\,x^4\right )+3\,x\,\ln \left (x\right )-3\,x^2}{x\,\left (\ln \left (81\,x^4\right )-x+{\mathrm {e}}^x+\ln \left (x\right )+4\right )} \] Input:

int((exp(x)*(15*x + 10*x^2 - 3*x^3 - 45) - 5*exp(2*x) + exp(x)*log(81*x^4) 
*(5*x + 3*x^2 - 5) + exp(x)*log(x)*(5*x + 3*x^2 - 5))/(exp(x)*(8*x^2 - 2*x 
^3) + log(x)*(2*x^2*exp(x) + 8*x^2 - 2*x^3) + x^2*log(81*x^4)^2 + x^2*exp( 
2*x) + x^2*log(x)^2 + 16*x^2 - 8*x^3 + x^4 + log(81*x^4)*(2*x^2*exp(x) + 2 
*x^2*log(x) + 8*x^2 - 2*x^3)),x)
 

Output:

-(12*x - 5*exp(x) + 3*x*log(81*x^4) + 3*x*log(x) - 3*x^2)/(x*(log(81*x^4) 
- x + exp(x) + log(x) + 4))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx=\frac {e^{x} \left (3 x +5\right )}{x \left (e^{x}+\mathrm {log}\left (81 x^{4}\right )+\mathrm {log}\left (x \right )-x +4\right )} \] Input:

int(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5*exp(x) 
^2+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+2*exp( 
x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8*x^2)*lo 
g(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x)
 

Output:

(e**x*(3*x + 5))/(x*(e**x + log(81*x**4) + log(x) - x + 4))