\(\int \frac {e^x (130+80 x)+e^x (50 x^2+80 x^3+e^5 (169 x^2+208 x^3+64 x^4)) \log (x)+e^x (-130-30 x+80 x^2) \log (x) \log (\log (x))}{(169 x^2+208 x^3+64 x^4) \log (x)} \, dx\) [1308]

Optimal result

Integrand size = 89, antiderivative size = 30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=e^x \left (e^5+\frac {x+\log (\log (x))}{\frac {x}{2}+\frac {4}{5} x (1+x)}\right ) \] Output:

exp(x)*(exp(5)+(ln(ln(x))+x)/(1/2*x+4*x*(1/5*x+1/5)))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {e^x \left (x \left (10+e^5 (13+8 x)\right )+10 \log (\log (x))\right )}{x (13+8 x)} \] Input:

Integrate[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^ 
3 + 64*x^4))*Log[x] + E^x*(-130 - 30*x + 80*x^2)*Log[x]*Log[Log[x]])/((169 
*x^2 + 208*x^3 + 64*x^4)*Log[x]),x]
 

Output:

(E^x*(x*(10 + E^5*(13 + 8*x)) + 10*Log[Log[x]]))/(x*(13 + 8*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.

\(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{\left (64 x^4+208 x^3+169 x^2\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{x^2 \left (64 x^2+208 x+169\right ) \log (x)}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{x^2 (8 x+13)^2 \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {16 e^x \left (-64 e^5 x^4 \log (x)-80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)-50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)-80 x^2 \log (x) \log (\log (x))-80 x+30 x \log (x) \log (\log (x))+130 \log (x) \log (\log (x))-130\right )}{2197 x \log (x)}+\frac {128 e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{2197 (8 x+13) \log (x)}+\frac {e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{169 x^2 \log (x)}+\frac {64 e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{169 (8 x+13)^2 \log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {10}{13} \int \frac {e^x}{x^2 \log (x)}dx-\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2}dx-\frac {80}{169} \int \frac {e^x}{x \log (x)}dx+\frac {640}{169} \int \frac {e^x}{(8 x+13) \log (x)}dx+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x}dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(8 x+13)^2}dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{8 x+13}dx-\frac {3 \left (50+169 e^5\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{832 e^{13/8}}+\frac {11 \left (5+13 e^5\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{32 e^{13/8}}-\frac {247}{64} e^{27/8} \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )-\frac {20 \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{13 e^{13/8}}+\frac {128}{169} e^{x+5} x^2+\frac {128 \left (10+13 e^5\right ) e^x x^2}{2197}-\frac {256 \left (5+13 e^5\right ) e^x x^2}{2197}-\frac {1280 e^x x}{2197}-\frac {464}{169} e^{x+5} x-\frac {16 \left (50+169 e^5\right ) e^x x}{2197}-\frac {256 \left (10+13 e^5\right ) e^x x}{2197}+\frac {928 \left (5+13 e^5\right ) e^x x}{2197}+\frac {3360 e^x}{2197}+\frac {971 e^{x+5}}{169}-\frac {169 e^{x+5}}{8 (8 x+13)}-\frac {\left (50+169 e^5\right ) e^x}{8 (8 x+13)}+\frac {13 \left (5+13 e^5\right ) e^x}{4 (8 x+13)}+\frac {42 \left (50+169 e^5\right ) e^x}{2197}+\frac {256 \left (10+13 e^5\right ) e^x}{2197}-\frac {1604 \left (5+13 e^5\right ) e^x}{2197}\)

Input:

Int[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^3 + 64 
*x^4))*Log[x] + E^x*(-130 - 30*x + 80*x^2)*Log[x]*Log[Log[x]])/((169*x^2 + 
 208*x^3 + 64*x^4)*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {10 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{\left (8 x +13\right ) x}+\frac {\left (8 x \,{\mathrm e}^{5}+13 \,{\mathrm e}^{5}+10\right ) {\mathrm e}^{x}}{8 x +13}\)	\(40\)
parallelrisch	\(-\frac {-64 \,{\mathrm e}^{5} {\mathrm e}^{x} x^{2}-104 x \,{\mathrm e}^{5} {\mathrm e}^{x}-80 \,{\mathrm e}^{x} x -80 \ln \left (\ln \left (x \right )\right ) {\mathrm e}^{x}}{8 x \left (8 x +13\right )}\)	\(42\)

Input:

int(((80*x^2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x^4+208*x^3+169*x^2)*ex 
p(5)+80*x^3+50*x^2)*exp(x)*ln(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+169*x^ 
2)/ln(x),x,method=_RETURNVERBOSE)
 

Output:

10/(8*x+13)/x*exp(x)*ln(ln(x))+(8*x*exp(5)+13*exp(5)+10)/(8*x+13)*exp(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {{\left ({\left (8 \, x^{2} + 13 \, x\right )} e^{5} + 10 \, x\right )} e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \] Input:

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 
9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* 
x^3+169*x^2)/log(x),x, algorithm="fricas")
 

Output:

(((8*x^2 + 13*x)*e^5 + 10*x)*e^x + 10*e^x*log(log(x)))/(8*x^2 + 13*x)
 

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {\left (8 x^{2} e^{5} + 10 x + 13 x e^{5} + 10 \log {\left (\log {\left (x \right )} \right )}\right ) e^{x}}{8 x^{2} + 13 x} \] Input:

integrate(((80*x**2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x**4+208*x**3+16 
9*x**2)*exp(5)+80*x**3+50*x**2)*exp(x)*ln(x)+(80*x+130)*exp(x))/(64*x**4+2 
08*x**3+169*x**2)/ln(x),x)
 

Output:

(8*x**2*exp(5) + 10*x + 13*x*exp(5) + 10*log(log(x)))*exp(x)/(8*x**2 + 13* 
x)
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {{\left (8 \, x^{2} e^{5} + x {\left (13 \, e^{5} + 10\right )}\right )} e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \] Input:

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 
9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* 
x^3+169*x^2)/log(x),x, algorithm="maxima")
 

Output:

((8*x^2*e^5 + x*(13*e^5 + 10))*e^x + 10*e^x*log(log(x)))/(8*x^2 + 13*x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {8 \, x^{2} e^{\left (x + 5\right )} + 13 \, x e^{\left (x + 5\right )} + 10 \, x e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \] Input:

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 
9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* 
x^3+169*x^2)/log(x),x, algorithm="giac")
 

Output:

(8*x^2*e^(x + 5) + 13*x*e^(x + 5) + 10*x*e^x + 10*e^x*log(log(x)))/(8*x^2 
+ 13*x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (80\,x+130\right )+{\mathrm {e}}^x\,\ln \left (x\right )\,\left ({\mathrm {e}}^5\,\left (64\,x^4+208\,x^3+169\,x^2\right )+50\,x^2+80\,x^3\right )-\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^x\,\ln \left (x\right )\,\left (-80\,x^2+30\,x+130\right )}{\ln \left (x\right )\,\left (64\,x^4+208\,x^3+169\,x^2\right )} \,d x \] Input:

int((exp(x)*(80*x + 130) + exp(x)*log(x)*(exp(5)*(169*x^2 + 208*x^3 + 64*x 
^4) + 50*x^2 + 80*x^3) - log(log(x))*exp(x)*log(x)*(30*x - 80*x^2 + 130))/ 
(log(x)*(169*x^2 + 208*x^3 + 64*x^4)),x)
 

Output:

int((exp(x)*(80*x + 130) + exp(x)*log(x)*(exp(5)*(169*x^2 + 208*x^3 + 64*x 
^4) + 50*x^2 + 80*x^3) - log(log(x))*exp(x)*log(x)*(30*x - 80*x^2 + 130))/ 
(log(x)*(169*x^2 + 208*x^3 + 64*x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {e^{x} \left (10 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+8 e^{5} x^{2}+13 e^{5} x +10 x \right )}{x \left (8 x +13\right )} \] Input:

int(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+169*x^2) 
*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+16 
9*x^2)/log(x),x)
 

Output:

(e**x*(10*log(log(x)) + 8*e**5*x**2 + 13*e**5*x + 10*x))/(x*(8*x + 13))