Integrand size = 86, antiderivative size = 31 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=e^5+\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \] Output:
exp(5)-ln(3*x/ln(x/exp(1/4*x)))^2+1/x^2/exp(5)
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \] Input:
Integrate[(-4*Log[x/E^(x/4)] + (E^5*(4*x^2 - x^3) - 4*E^5*x^2*Log[x/E^(x/4 )])*Log[(3*x)/Log[x/E^(x/4)]])/(2*E^5*x^3*Log[x/E^(x/4)]),x]
Output:
1/(E^5*x^2) - Log[(3*x)/Log[x/E^(x/4)]]^2
Time = 0.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )-4 \log \left (e^{-x/4} x\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {4 \log \left (e^{-x/4} x\right )-\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x^3 \log \left (e^{-x/4} x\right )}dx}{2 e^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {4 \log \left (e^{-x/4} x\right )-\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x^3 \log \left (e^{-x/4} x\right )}dx}{2 e^5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {e^5 \left (x+4 \log \left (e^{-x/4} x\right )-4\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x \log \left (e^{-x/4} x\right )}+\frac {4}{x^3}\right )dx}{2 e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^5 \log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )-\frac {2}{x^2}}{2 e^5}\) |
Input:
Int[(-4*Log[x/E^(x/4)] + (E^5*(4*x^2 - x^3) - 4*E^5*x^2*Log[x/E^(x/4)])*Lo g[(3*x)/Log[x/E^(x/4)]])/(2*E^5*x^3*Log[x/E^(x/4)]),x]
Output:
-1/2*(-2/x^2 + 2*E^5*Log[(3*x)/Log[x/E^(x/4)]]^2)/E^5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
\[\int \frac {\left (\left (-4 x^{2} {\mathrm e}^{5} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )+\left (-x^{3}+4 x^{2}\right ) {\mathrm e}^{5}\right ) \ln \left (\frac {3 x}{\ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}\right )-4 \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )\right ) {\mathrm e}^{-5}}{2 x^{3} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}d x\]
Input:
int(1/2*((-4*x^2*exp(5)*ln(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*ln(3*x/ln(x/ exp(1/4*x)))-4*ln(x/exp(1/4*x)))/x^3/exp(5)/ln(x/exp(1/4*x)),x)
Output:
int(1/2*((-4*x^2*exp(5)*ln(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*ln(3*x/ln(x/ exp(1/4*x)))-4*ln(x/exp(1/4*x)))/x^3/exp(5)/ln(x/exp(1/4*x)),x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=-\frac {{\left (x^{2} e^{5} \log \left (\frac {3 \, x}{\log \left (x e^{\left (-\frac {1}{4} \, x\right )}\right )}\right )^{2} - 1\right )} e^{\left (-5\right )}}{x^{2}} \] Input:
integrate(1/2*((-4*x^2*exp(5)*log(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*log(3 *x/log(x/exp(1/4*x)))-4*log(x/exp(1/4*x)))/x^3/exp(5)/log(x/exp(1/4*x)),x, algorithm="fricas")
Output:
-(x^2*e^5*log(3*x/log(x*e^(-1/4*x)))^2 - 1)*e^(-5)/x^2
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=- \log {\left (\frac {3 x}{\log {\left (x e^{- \frac {x}{4}} \right )}} \right )}^{2} + \frac {1}{x^{2} e^{5}} \] Input:
integrate(1/2*((-4*x**2*exp(5)*ln(x/exp(1/4*x))+(-x**3+4*x**2)*exp(5))*ln( 3*x/ln(x/exp(1/4*x)))-4*ln(x/exp(1/4*x)))/x**3/exp(5)/ln(x/exp(1/4*x)),x)
Output:
-log(3*x/log(x*exp(-x/4)))**2 + exp(-5)/x**2
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=-{\left (2 \, {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} e^{5} \log \left (x\right ) + e^{5} \log \left (x\right )^{2} + e^{5} \log \left (-x + 4 \, \log \left (x\right )\right )^{2} - 2 \, {\left ({\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} e^{5} + e^{5} \log \left (x\right )\right )} \log \left (-x + 4 \, \log \left (x\right )\right ) - \frac {1}{x^{2}}\right )} e^{\left (-5\right )} \] Input:
integrate(1/2*((-4*x^2*exp(5)*log(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*log(3 *x/log(x/exp(1/4*x)))-4*log(x/exp(1/4*x)))/x^3/exp(5)/log(x/exp(1/4*x)),x, algorithm="maxima")
Output:
-(2*(log(3) + 2*log(2))*e^5*log(x) + e^5*log(x)^2 + e^5*log(-x + 4*log(x)) ^2 - 2*((log(3) + 2*log(2))*e^5 + e^5*log(x))*log(-x + 4*log(x)) - 1/x^2)* e^(-5)
Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (26) = 52\).
Time = 0.14 (sec) , antiderivative size = 456, normalized size of antiderivative = 14.71 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx =\text {Too large to display} \] Input:
integrate(1/2*((-4*x^2*exp(5)*log(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*log(3 *x/log(x/exp(1/4*x)))-4*log(x/exp(1/4*x)))/x^3/exp(5)/log(x/exp(1/4*x)),x, algorithm="giac")
Output:
-1/4*(2*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x))*sgn(x - 4*log(abs(x)))*sgn(x ) - 10*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x))*sgn(x - 4*log(abs(x))) + 4*pi *x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5*sgn(-2*pi + 2*pi* sgn(x))*sgn(x - 4*log(abs(x))) - 2*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x))*s gn(x) - 6*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x)) - 4*pi*x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5*sgn(-2*pi + 2*pi*sgn(x)) + 2*pi^2*x^2 *e^5*sgn(x - 4*log(abs(x))) + 10*pi^2*x^2*e^5*sgn(x) - 4*pi*x^2*arctan(-2* (pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5*sgn(x) - 12*pi^2*x^2*e^5 + 20*pi *x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5 - 4*x^2*arctan(-2 *(pi - pi*sgn(x))/(x - 4*log(abs(x))))^2*e^5 - 4*x^2*e^5*log(12)*log(-8*pi ^2*sgn(x) + 8*pi^2 + x^2 - 8*x*log(abs(x)) + 16*log(abs(x))^2) + x^2*e^5*l og(-8*pi^2*sgn(x) + 8*pi^2 + x^2 - 8*x*log(abs(x)) + 16*log(abs(x))^2)^2 + 8*x^2*e^5*log(12)*log(abs(x)) - 4*x^2*e^5*log(-8*pi^2*sgn(x) + 8*pi^2 + x ^2 - 8*x*log(abs(x)) + 16*log(abs(x))^2)*log(abs(x)) + 4*x^2*e^5*log(abs(x ))^2 - 4)*e^(-5)/x^2
Time = 3.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {{\mathrm {e}}^{-5}}{x^2}-{\ln \left (-\frac {12\,x}{x-4\,\ln \left (x\right )}\right )}^2 \] Input:
int(-(exp(-5)*(2*log(x*exp(-x/4)) - (log((3*x)/log(x*exp(-x/4)))*(exp(5)*( 4*x^2 - x^3) - 4*x^2*exp(5)*log(x*exp(-x/4))))/2))/(x^3*log(x*exp(-x/4))), x)
Output:
exp(-5)/x^2 - log(-(12*x)/(x - 4*log(x)))^2
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{2 e^5 x^3 \log \left (e^{-x/4} x\right )} \, dx=\frac {-{\mathrm {log}\left (\frac {3 x}{\mathrm {log}\left (\frac {x}{e^{\frac {x}{4}}}\right )}\right )}^{2} e^{5} x^{2}+1}{e^{5} x^{2}} \] Input:
int(1/2*((-4*x^2*exp(5)*log(x/exp(1/4*x))+(-x^3+4*x^2)*exp(5))*log(3*x/log (x/exp(1/4*x)))-4*log(x/exp(1/4*x)))/x^3/exp(5)/log(x/exp(1/4*x)),x)
Output:
( - log((3*x)/log(x/e**(x/4)))**2*e**5*x**2 + 1)/(e**5*x**2)