\(\int \frac {(e^{-3+x} (-2 e^5-2 x)-10 e^5 x-10 x^2+(10 e^5 x+e^{-3+x} (-2 x+2 e^5 x+2 x^2)) \log (x)) \log (\frac {e^{-3+x}+5 x}{(e^5+x) \log (x)})}{(5 e^5 x^2+5 x^3+e^{-3+x} (e^5 x+x^2)) \log (x)} \, dx\) [902]

Optimal result

Integrand size = 116, antiderivative size = 24 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=\log ^2\left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=\log ^2\left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 38.97 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\ln \left (\frac {{\mathrm e}^{-3+x}+5 x}{\left ({\mathrm e}^{5}+x \right ) \ln \left (x \right )}\right )^{2}\)	\(23\)
risch	\(\text {Expression too large to display}\)	\(1440\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=\log \left (\frac {5 \, x + e^{\left (x - 3\right )}}{{\left (x + e^{5}\right )} \log \left (x\right )}\right )^{2} \] Input:

integrate((((2*x*exp(5)+2*x^2-2*x)*exp(-3+x)+10*x*exp(5))*log(x)+(-2*exp(5 
)-2*x)*exp(-3+x)-10*x*exp(5)-10*x^2)*log((exp(-3+x)+5*x)/(exp(5)+x)/log(x) 
)/((x*exp(5)+x^2)*exp(-3+x)+5*x^2*exp(5)+5*x^3)/log(x),x, algorithm="frica 
s")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=\log {\left (\frac {5 x + e^{x - 3}}{\left (x + e^{5}\right ) \log {\left (x \right )}} \right )}^{2} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (22) = 44\).

Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.33 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=2 \, {\left (\log \left (x + e^{5}\right ) + \log \left (\log \left (x\right )\right )\right )} \log \left (5 \, x e^{3} + e^{x}\right ) - \log \left (5 \, x e^{3} + e^{x}\right )^{2} - \log \left (x + e^{5}\right )^{2} + 2 \, {\left (\log \left (5 \, x e^{3} + e^{x}\right ) - \log \left (x + e^{5}\right ) - \log \left (\log \left (x\right )\right )\right )} \log \left (\frac {5 \, x + e^{\left (x - 3\right )}}{{\left (x + e^{5}\right )} \log \left (x\right )}\right ) - 2 \, \log \left (x + e^{5}\right ) \log \left (\log \left (x\right )\right ) - \log \left (\log \left (x\right )\right )^{2} \] Input:

integrate((((2*x*exp(5)+2*x^2-2*x)*exp(-3+x)+10*x*exp(5))*log(x)+(-2*exp(5 
)-2*x)*exp(-3+x)-10*x*exp(5)-10*x^2)*log((exp(-3+x)+5*x)/(exp(5)+x)/log(x) 
)/((x*exp(5)+x^2)*exp(-3+x)+5*x^2*exp(5)+5*x^3)/log(x),x, algorithm="maxim 
a")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx={\ln \left (\frac {5\,x+{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{\ln \left (x\right )\,\left (x+{\mathrm {e}}^5\right )}\right )}^2 \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx=\mathrm {log}\left (\frac {e^{x}+5 e^{3} x}{\mathrm {log}\left (x \right ) e^{8}+\mathrm {log}\left (x \right ) e^{3} x}\right )^{2} \] Input:

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

Output:

log((e**x + 5*e**3*x)/(log(x)*e**8 + log(x)*e**3*x))**2