\(\int \frac {10+15 x^3+e^x (10 x+5 x^2)+5 \log (\frac {e^{9+e^x (-1-x)-x^3}}{x^2})}{\log ^2(\frac {e^{9+e^x (-1-x)-x^3}}{x^2})} \, dx\) [373]

Optimal result

Integrand size = 71, antiderivative size = 30 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=\frac {5 x}{\log \left (\frac {e^{9-e^x-e^x x-x^3}}{x^2}\right )} \] Output:

5*x/ln(exp(9-exp(x)*x-x^3-exp(x))/x^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=\frac {5 x}{\log \left (\frac {e^{9-x^3-e^x (1+x)}}{x^2}\right )} \] Input:

Integrate[(10 + 15*x^3 + E^x*(10*x + 5*x^2) + 5*Log[E^(9 + E^x*(-1 - x) - 
x^3)/x^2])/Log[E^(9 + E^x*(-1 - x) - x^3)/x^2]^2,x]
 

Output:

(5*x)/Log[E^(9 - x^3 - E^x*(1 + x))/x^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.

\(\displaystyle \int \frac {15 x^3+e^x \left (5 x^2+10 x\right )+5 \log \left (\frac {e^{-x^3+e^x (-x-1)+9}}{x^2}\right )+10}{\log ^2\left (\frac {e^{-x^3+e^x (-x-1)+9}}{x^2}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {5 \left (3 x^3+e^x x^2+\log \left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )+2 e^x x+2\right )}{\log ^2\left (\frac {e^{-x^3+e^x (-x-1)+9}}{x^2}\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \int \frac {3 x^3+e^x x^2+2 e^x x+\log \left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )+2}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 5 \int \left (\frac {e^x x (x+2)}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}+\frac {3 x^3+\log \left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )+2}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 5 \left (2 \int \frac {1}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx+2 \int \frac {e^x x}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx+\int \frac {e^x x^2}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx+3 \int \frac {x^3}{\log ^2\left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx+\int \frac {1}{\log \left (\frac {e^{-x^3-e^x (x+1)+9}}{x^2}\right )}dx\right )\)

Input:

Int[(10 + 15*x^3 + E^x*(10*x + 5*x^2) + 5*Log[E^(9 + E^x*(-1 - x) - x^3)/x 
^2])/Log[E^(9 + E^x*(-1 - x) - x^3)/x^2]^2,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(\frac {5 x}{\ln \left (\frac {{\mathrm e}^{\left (-1-x \right ) {\mathrm e}^{x}-x^{3}+9}}{x^{2}}\right )}\)	\(27\)
risch	\(-\frac {10 i x}{\pi \,\operatorname {csgn}\left (i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )-\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{3}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}}{x^{2}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )-2 i \ln \left ({\mathrm e}^{9-{\mathrm e}^{x} x -x^{3}-{\mathrm e}^{x}}\right )}\)	\(246\)

Input:

int((5*ln(exp((-1-x)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/ln( 
exp((-1-x)*exp(x)-x^3+9)/x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

5*x/ln(exp((-1-x)*exp(x)-x^3+9)/x^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=\frac {5 \, x}{\log \left (\frac {e^{\left (-x^{3} - {\left (x + 1\right )} e^{x} + 9\right )}}{x^{2}}\right )} \] Input:

integrate((5*log(exp((-1-x)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+ 
10)/log(exp((-1-x)*exp(x)-x^3+9)/x^2)^2,x, algorithm="fricas")
 

Output:

5*x/log(e^(-x^3 - (x + 1)*e^x + 9)/x^2)
 

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=\frac {5 x}{\log {\left (\frac {e^{- x^{3} + \left (- x - 1\right ) e^{x} + 9}}{x^{2}} \right )}} \] Input:

integrate((5*ln(exp((-1-x)*exp(x)-x**3+9)/x**2)+(5*x**2+10*x)*exp(x)+15*x* 
*3+10)/ln(exp((-1-x)*exp(x)-x**3+9)/x**2)**2,x)
 

Output:

5*x/log(exp(-x**3 + (-x - 1)*exp(x) + 9)/x**2)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=-\frac {5 \, x}{x^{3} + {\left (x + 1\right )} e^{x} + 2 \, \log \left (x\right ) - 9} \] Input:

integrate((5*log(exp((-1-x)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+ 
10)/log(exp((-1-x)*exp(x)-x^3+9)/x^2)^2,x, algorithm="maxima")
 

Output:

-5*x/(x^3 + (x + 1)*e^x + 2*log(x) - 9)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=-\frac {5 \, x}{x^{3} + x e^{x} + e^{x} + \log \left (x^{2}\right ) - 9} \] Input:

integrate((5*log(exp((-1-x)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+ 
10)/log(exp((-1-x)*exp(x)-x^3+9)/x^2)^2,x, algorithm="giac")
 

Output:

-5*x/(x^3 + x*e^x + e^x + log(x^2) - 9)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=\int \frac {5\,\ln \left (\frac {{\mathrm {e}}^{9-x^3-{\mathrm {e}}^x\,\left (x+1\right )}}{x^2}\right )+{\mathrm {e}}^x\,\left (5\,x^2+10\,x\right )+15\,x^3+10}{{\ln \left (\frac {{\mathrm {e}}^{9-x^3-{\mathrm {e}}^x\,\left (x+1\right )}}{x^2}\right )}^2} \,d x \] Input:

int((5*log(exp(9 - x^3 - exp(x)*(x + 1))/x^2) + exp(x)*(10*x + 5*x^2) + 15 
*x^3 + 10)/log(exp(9 - x^3 - exp(x)*(x + 1))/x^2)^2,x)
 

Output:

int((5*log(exp(9 - x^3 - exp(x)*(x + 1))/x^2) + exp(x)*(10*x + 5*x^2) + 15 
*x^3 + 10)/log(exp(9 - x^3 - exp(x)*(x + 1))/x^2)^2, x)
 

Reduce [F]

\[ \int \frac {10+15 x^3+e^x \left (10 x+5 x^2\right )+5 \log \left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )}{\log ^2\left (\frac {e^{9+e^x (-1-x)-x^3}}{x^2}\right )} \, dx=15 \left (\int \frac {x^{3}}{\mathrm {log}\left (\frac {e^{9}}{e^{e^{x} x +e^{x}+x^{3}} x^{2}}\right )^{2}}d x \right )+5 \left (\int \frac {e^{x} x^{2}}{\mathrm {log}\left (\frac {e^{9}}{e^{e^{x} x +e^{x}+x^{3}} x^{2}}\right )^{2}}d x \right )+10 \left (\int \frac {e^{x} x}{\mathrm {log}\left (\frac {e^{9}}{e^{e^{x} x +e^{x}+x^{3}} x^{2}}\right )^{2}}d x \right )+10 \left (\int \frac {1}{\mathrm {log}\left (\frac {e^{9}}{e^{e^{x} x +e^{x}+x^{3}} x^{2}}\right )^{2}}d x \right )+5 \left (\int \frac {1}{\mathrm {log}\left (\frac {e^{9}}{e^{e^{x} x +e^{x}+x^{3}} x^{2}}\right )}d x \right ) \] Input:

int((5*log(exp((-1-x)*exp(x)-x^3+9)/x^2)+(5*x^2+10*x)*exp(x)+15*x^3+10)/lo 
g(exp((-1-x)*exp(x)-x^3+9)/x^2)^2,x)
 

Output:

5*(3*int(x**3/log(e**9/(e**(e**x*x + e**x + x**3)*x**2))**2,x) + int((e**x 
*x**2)/log(e**9/(e**(e**x*x + e**x + x**3)*x**2))**2,x) + 2*int((e**x*x)/l 
og(e**9/(e**(e**x*x + e**x + x**3)*x**2))**2,x) + 2*int(1/log(e**9/(e**(e* 
*x*x + e**x + x**3)*x**2))**2,x) + int(1/log(e**9/(e**(e**x*x + e**x + x** 
3)*x**2)),x))