\(\int \frac {-700+e^{\frac {1}{25} (-125+145 x-16 x^2+25 x \log (3 x))} (25-170 x+32 x^2-25 x \log (3 x))}{19600-1400 e^{\frac {1}{25} (-125+145 x-16 x^2+25 x \log (3 x))}+25 e^{\frac {2}{25} (-125+145 x-16 x^2+25 x \log (3 x))}} \, dx\) [1072]

Optimal result

Integrand size = 98, antiderivative size = 28 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\frac {x}{-28+e^{4-\left (-3+\frac {4 x}{5}\right )^2+x+x \log (3 x)}} \] Output:

x/(exp(x+4+x*ln(3*x)-(4/5*x-3)^2)-28)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx \] Input:

Integrate[(-700 + E^((-125 + 145*x - 16*x^2 + 25*x*Log[3*x])/25)*(25 - 170 
*x + 32*x^2 - 25*x*Log[3*x]))/(19600 - 1400*E^((-125 + 145*x - 16*x^2 + 25 
*x*Log[3*x])/25) + 25*E^((2*(-125 + 145*x - 16*x^2 + 25*x*Log[3*x]))/25)), 
x]
 

Output:

Integrate[(-700 + E^((-125 + 145*x - 16*x^2 + 25*x*Log[3*x])/25)*(25 - 170 
*x + 32*x^2 - 25*x*Log[3*x]))/(19600 - 1400*E^((-125 + 145*x - 16*x^2 + 25 
*x*Log[3*x])/25) + 25*E^((2*(-125 + 145*x - 16*x^2 + 25*x*Log[3*x]))/25)), 
 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.

\(\displaystyle \int \frac {e^{\frac {1}{25} \left (-16 x^2+145 x+25 x \log (3 x)-125\right )} \left (32 x^2-170 x-25 x \log (3 x)+25\right )-700}{-1400 e^{\frac {1}{25} \left (-16 x^2+145 x+25 x \log (3 x)-125\right )}+25 e^{\frac {2}{25} \left (-16 x^2+145 x+25 x \log (3 x)-125\right )}+19600} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {32 x^2}{25}+10} \left (e^{\frac {1}{25} \left (-16 x^2+145 x+25 x \log (3 x)-125\right )} \left (32 x^2-170 x-25 x \log (3 x)+25\right )-700\right )}{25 \left (28 e^{\frac {16 x^2}{25}+5}-3^x e^{29 x/5} x^x\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{25} \int -\frac {e^{\frac {32 x^2}{25}+10} \left (700-3^x e^{\frac {1}{25} \left (-16 x^2+145 x-125\right )} x^x \left (32 x^2-25 \log (3 x) x-170 x+25\right )\right )}{\left (28 e^{\frac {16 x^2}{25}+5}-3^x e^{29 x/5} x^x\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{25} \int \frac {e^{\frac {32 x^2}{25}+10} \left (700-3^x e^{\frac {1}{25} \left (-16 x^2+145 x-125\right )} x^x \left (32 x^2-25 \log (3 x) x-170 x+25\right )\right )}{\left (28 e^{\frac {16 x^2}{25}+5}-3^x e^{29 x/5} x^x\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{25} \int \left (\frac {e^{\frac {16 x^2}{25}+5} \left (32 x^2-25 \log (3 x) x-170 x+25\right )}{28 e^{\frac {16 x^2}{25}+5}-3^x e^{29 x/5} x^x}-\frac {28 e^{\frac {32 x^2}{25}+10} x (32 x-25 \log (3 x)-170)}{\left (3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{25} \left (-4760 \int \frac {e^{\frac {32 x^2}{25}+10} x}{\left (3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}\right )^2}dx+896 \int \frac {e^{\frac {32 x^2}{25}+10} x^2}{\left (3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}\right )^2}dx+25 \int \frac {e^{\frac {16 x^2}{25}+5}}{3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}}dx-170 \int \frac {e^{\frac {16 x^2}{25}+5} x}{3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}}dx+32 \int \frac {e^{\frac {16 x^2}{25}+5} x^2}{3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}}dx+700 \int \frac {\int \frac {e^{\frac {32 x^2}{25}+10} x}{\left (3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}\right )^2}dx}{x}dx+25 \int \frac {\int \frac {e^{\frac {16 x^2}{25}+5} x}{3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}}dx}{x}dx-700 \log (3 x) \int \frac {e^{\frac {32 x^2}{25}+10} x}{\left (3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}\right )^2}dx-25 \log (3 x) \int \frac {e^{\frac {16 x^2}{25}+5} x}{3^x e^{29 x/5} x^x-28 e^{\frac {16 x^2}{25}+5}}dx\right )\)

Input:

Int[(-700 + E^((-125 + 145*x - 16*x^2 + 25*x*Log[3*x])/25)*(25 - 170*x + 3 
2*x^2 - 25*x*Log[3*x]))/(19600 - 1400*E^((-125 + 145*x - 16*x^2 + 25*x*Log 
[3*x])/25) + 25*E^((2*(-125 + 145*x - 16*x^2 + 25*x*Log[3*x]))/25)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
norman	\(\frac {x}{{\mathrm e}^{x \ln \left (3 x \right )-\frac {16 x^{2}}{25}+\frac {29 x}{5}-5}-28}\)	\(24\)
risch	\(\frac {x}{\left (3 x \right )^{x} {\mathrm e}^{-5-\frac {16}{25} x^{2}+\frac {29}{5} x}-28}\)	\(24\)
parallelrisch	\(\frac {x}{{\mathrm e}^{x \ln \left (3 x \right )-\frac {16 x^{2}}{25}+\frac {29 x}{5}-5}-28}\)	\(24\)

Input:

int(((-25*x*ln(3*x)+32*x^2-170*x+25)*exp(x*ln(3*x)-16/25*x^2+29/5*x-5)-700 
)/(25*exp(x*ln(3*x)-16/25*x^2+29/5*x-5)^2-1400*exp(x*ln(3*x)-16/25*x^2+29/ 
5*x-5)+19600),x,method=_RETURNVERBOSE)
 

Output:

x/(exp(x*ln(3*x)-16/25*x^2+29/5*x-5)-28)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\frac {x}{e^{\left (-\frac {16}{25} \, x^{2} + x \log \left (3 \, x\right ) + \frac {29}{5} \, x - 5\right )} - 28} \] Input:

integrate(((-25*x*log(3*x)+32*x^2-170*x+25)*exp(x*log(3*x)-16/25*x^2+29/5* 
x-5)-700)/(25*exp(x*log(3*x)-16/25*x^2+29/5*x-5)^2-1400*exp(x*log(3*x)-16/ 
25*x^2+29/5*x-5)+19600),x, algorithm="fricas")
 

Output:

x/(e^(-16/25*x^2 + x*log(3*x) + 29/5*x - 5) - 28)
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\frac {x}{e^{- \frac {16 x^{2}}{25} + x \log {\left (3 x \right )} + \frac {29 x}{5} - 5} - 28} \] Input:

integrate(((-25*x*ln(3*x)+32*x**2-170*x+25)*exp(x*ln(3*x)-16/25*x**2+29/5* 
x-5)-700)/(25*exp(x*ln(3*x)-16/25*x**2+29/5*x-5)**2-1400*exp(x*ln(3*x)-16/ 
25*x**2+29/5*x-5)+19600),x)
 

Output:

x/(exp(-16*x**2/25 + x*log(3*x) + 29*x/5 - 5) - 28)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=-\frac {x e^{\left (\frac {16}{25} \, x^{2} + 5\right )}}{28 \, e^{\left (\frac {16}{25} \, x^{2} + 5\right )} - e^{\left (x \log \left (3\right ) + x \log \left (x\right ) + \frac {29}{5} \, x\right )}} \] Input:

integrate(((-25*x*log(3*x)+32*x^2-170*x+25)*exp(x*log(3*x)-16/25*x^2+29/5* 
x-5)-700)/(25*exp(x*log(3*x)-16/25*x^2+29/5*x-5)^2-1400*exp(x*log(3*x)-16/ 
25*x^2+29/5*x-5)+19600),x, algorithm="maxima")
 

Output:

-x*e^(16/25*x^2 + 5)/(28*e^(16/25*x^2 + 5) - e^(x*log(3) + x*log(x) + 29/5 
*x))
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=-\frac {x e^{5}}{28 \, e^{5} - e^{\left (-\frac {16}{25} \, x^{2} + x \log \left (3 \, x\right ) + \frac {29}{5} \, x\right )}} \] Input:

integrate(((-25*x*log(3*x)+32*x^2-170*x+25)*exp(x*log(3*x)-16/25*x^2+29/5* 
x-5)-700)/(25*exp(x*log(3*x)-16/25*x^2+29/5*x-5)^2-1400*exp(x*log(3*x)-16/ 
25*x^2+29/5*x-5)+19600),x, algorithm="giac")
 

Output:

-x*e^5/(28*e^5 - e^(-16/25*x^2 + x*log(3*x) + 29/5*x))
 

Mupad [B] (verification not implemented)

Time = 7.94 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\frac {170\,x+25\,x\,\ln \left (3\,x\right )-32\,x^2}{\left ({\mathrm {e}}^{-\frac {16\,x^2}{25}+\frac {29\,x}{5}-5}\,{\left (3\,x\right )}^x-28\right )\,\left (25\,\ln \left (3\,x\right )-32\,x+170\right )} \] Input:

int(-(exp((29*x)/5 + x*log(3*x) - (16*x^2)/25 - 5)*(170*x + 25*x*log(3*x) 
- 32*x^2 - 25) + 700)/(25*exp((58*x)/5 + 2*x*log(3*x) - (32*x^2)/25 - 10) 
- 1400*exp((29*x)/5 + x*log(3*x) - (16*x^2)/25 - 5) + 19600),x)
 

Output:

(170*x + 25*x*log(3*x) - 32*x^2)/((exp((29*x)/5 - (16*x^2)/25 - 5)*(3*x)^x 
 - 28)*(25*log(3*x) - 32*x + 170))
 

Reduce [F]

\[ \int \frac {-700+e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )} \left (25-170 x+32 x^2-25 x \log (3 x)\right )}{19600-1400 e^{\frac {1}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}+25 e^{\frac {2}{25} \left (-125+145 x-16 x^2+25 x \log (3 x)\right )}} \, dx=\frac {e^{5} \left (-700 \left (\int \frac {e^{\frac {32 x^{2}}{25}}}{784 e^{\frac {32 x^{2}}{25}} e^{10}-56 x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} e^{5}+x^{2 x} e^{\frac {58 x}{5}} 3^{2 x}}d x \right ) e^{5}+32 \left (\int \frac {x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} x^{2}}{784 e^{\frac {32 x^{2}}{25}} e^{10}-56 x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} e^{5}+x^{2 x} e^{\frac {58 x}{5}} 3^{2 x}}d x \right )-25 \left (\int \frac {x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} \mathrm {log}\left (3 x \right ) x}{784 e^{\frac {32 x^{2}}{25}} e^{10}-56 x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} e^{5}+x^{2 x} e^{\frac {58 x}{5}} 3^{2 x}}d x \right )-170 \left (\int \frac {x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} x}{784 e^{\frac {32 x^{2}}{25}} e^{10}-56 x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} e^{5}+x^{2 x} e^{\frac {58 x}{5}} 3^{2 x}}d x \right )+25 \left (\int \frac {x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x}}{784 e^{\frac {32 x^{2}}{25}} e^{10}-56 x^{x} e^{\frac {16}{25} x^{2}+\frac {29}{5} x} 3^{x} e^{5}+x^{2 x} e^{\frac {58 x}{5}} 3^{2 x}}d x \right )\right )}{25} \] Input:

int(((-25*x*log(3*x)+32*x^2-170*x+25)*exp(x*log(3*x)-16/25*x^2+29/5*x-5)-7 
00)/(25*exp(x*log(3*x)-16/25*x^2+29/5*x-5)^2-1400*exp(x*log(3*x)-16/25*x^2 
+29/5*x-5)+19600),x)
 

Output:

(e**5*( - 700*int(e**((32*x**2)/25)/(784*e**((32*x**2)/25)*e**10 - 56*x**x 
*e**((16*x**2 + 145*x)/25)*3**x*e**5 + x**(2*x)*e**((58*x)/5)*3**(2*x)),x) 
*e**5 + 32*int((x**x*e**((16*x**2 + 145*x)/25)*3**x*x**2)/(784*e**((32*x** 
2)/25)*e**10 - 56*x**x*e**((16*x**2 + 145*x)/25)*3**x*e**5 + x**(2*x)*e**( 
(58*x)/5)*3**(2*x)),x) - 25*int((x**x*e**((16*x**2 + 145*x)/25)*3**x*log(3 
*x)*x)/(784*e**((32*x**2)/25)*e**10 - 56*x**x*e**((16*x**2 + 145*x)/25)*3* 
*x*e**5 + x**(2*x)*e**((58*x)/5)*3**(2*x)),x) - 170*int((x**x*e**((16*x**2 
 + 145*x)/25)*3**x*x)/(784*e**((32*x**2)/25)*e**10 - 56*x**x*e**((16*x**2 
+ 145*x)/25)*3**x*e**5 + x**(2*x)*e**((58*x)/5)*3**(2*x)),x) + 25*int((x** 
x*e**((16*x**2 + 145*x)/25)*3**x)/(784*e**((32*x**2)/25)*e**10 - 56*x**x*e 
**((16*x**2 + 145*x)/25)*3**x*e**5 + x**(2*x)*e**((58*x)/5)*3**(2*x)),x))) 
/25