Integrand size = 142, antiderivative size = 29 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4 x^2 \left (x+\log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right )\right ) \] Output:
4*x^2*(x+ln(ln(x^2/exp(x)^2*(-2+x))/x/ln(x)))
Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4 x^3+4 x^2 \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right ) \] Input:
Integrate[((-16*x + 28*x^2 - 8*x^3)*Log[x] + (8*x - 4*x^2 + (8*x - 28*x^2 + 12*x^3)*Log[x])*Log[(-2*x^2 + x^3)/E^(2*x)] + (-16*x + 8*x^2)*Log[x]*Log [(-2*x^2 + x^3)/E^(2*x)]*Log[Log[(-2*x^2 + x^3)/E^(2*x)]/(x*Log[x])])/((-2 + x)*Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]),x]
Output:
4*x^3 + 4*x^2*Log[Log[((-2 + x)*x^2)/E^(2*x)]/(x*Log[x])]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.92 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {7292, 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.
\(\displaystyle \int \frac {\left (-8 x^3+28 x^2-16 x\right ) \log (x)+\left (8 x^2-16 x\right ) \log \left (e^{-2 x} \left (x^3-2 x^2\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (x^3-2 x^2\right )\right )}{x \log (x)}\right ) \log (x)+\left (-4 x^2+\left (12 x^3-28 x^2+8 x\right ) \log (x)+8 x\right ) \log \left (e^{-2 x} \left (x^3-2 x^2\right )\right )}{(x-2) \log (x) \log \left (e^{-2 x} \left (x^3-2 x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (-8 x^3+28 x^2-16 x\right ) \log (x)\right )-\left (8 x^2-16 x\right ) \log \left (e^{-2 x} \left (x^3-2 x^2\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (x^3-2 x^2\right )\right )}{x \log (x)}\right ) \log (x)-\left (-4 x^2+\left (12 x^3-28 x^2+8 x\right ) \log (x)+8 x\right ) \log \left (e^{-2 x} \left (x^3-2 x^2\right )\right )}{(2-x) \log (x) \log \left (e^{-2 x} (x-2) x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x \left (-2 x^2 \log (x)+3 x^2 \log (x) \log \left (e^{-2 x} (x-2) x^2\right )-7 x \log (x) \log \left (e^{-2 x} (x-2) x^2\right )-x \log \left (e^{-2 x} (x-2) x^2\right )+2 \log (x) \log \left (e^{-2 x} (x-2) x^2\right )+2 \log \left (e^{-2 x} (x-2) x^2\right )+7 x \log (x)-4 \log (x)\right )}{(x-2) \log (x) \log \left (e^{-2 x} (x-2) x^2\right )}+8 x \log \left (\frac {\log \left (e^{-2 x} (x-2) x^2\right )}{x \log (x)}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \operatorname {ExpIntegralEi}(2 \log (x))-4 \log (x) \operatorname {ExpIntegralEi}(2 \log (x))+4 (\log (x)+1) \operatorname {ExpIntegralEi}(2 \log (x))+4 x^3+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )\) |
Input:
Int[((-16*x + 28*x^2 - 8*x^3)*Log[x] + (8*x - 4*x^2 + (8*x - 28*x^2 + 12*x ^3)*Log[x])*Log[(-2*x^2 + x^3)/E^(2*x)] + (-16*x + 8*x^2)*Log[x]*Log[(-2*x ^2 + x^3)/E^(2*x)]*Log[Log[(-2*x^2 + x^3)/E^(2*x)]/(x*Log[x])])/((-2 + x)* Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]),x]
Output:
4*x^3 - 4*ExpIntegralEi[2*Log[x]] - 4*ExpIntegralEi[2*Log[x]]*Log[x] + 4*E xpIntegralEi[2*Log[x]]*(1 + Log[x]) + 4*x^2*Log[Log[-(((2 - x)*x^2)/E^(2*x ))]/(x*Log[x])]
Time = 12.88 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(4 x^{3}+4 \ln \left (\frac {\ln \left (\left (-2+x \right ) x^{2} {\mathrm e}^{-2 x}\right )}{x \ln \left (x \right )}\right ) x^{2}\) | \(33\) |
Input:
int(((8*x^2-16*x)*ln(x)*ln((x^3-2*x^2)/exp(x)^2)*ln(ln((x^3-2*x^2)/exp(x)^ 2)/x/ln(x))+((12*x^3-28*x^2+8*x)*ln(x)-4*x^2+8*x)*ln((x^3-2*x^2)/exp(x)^2) +(-8*x^3+28*x^2-16*x)*ln(x))/(-2+x)/ln(x)/ln((x^3-2*x^2)/exp(x)^2),x,metho d=_RETURNVERBOSE)
Output:
4*x^3+4*ln(ln(x^2/exp(x)^2*(-2+x))/x/ln(x))*x^2
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4 \, x^{3} + 4 \, x^{2} \log \left (\frac {\log \left ({\left (x^{3} - 2 \, x^{2}\right )} e^{\left (-2 \, x\right )}\right )}{x \log \left (x\right )}\right ) \] Input:
integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^ 2)/exp(x)^2)/x/log(x))+((12*x^3-28*x^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x ^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(-2+x)/log(x)/log((x^3-2*x^2)/e xp(x)^2),x, algorithm="fricas")
Output:
4*x^3 + 4*x^2*log(log((x^3 - 2*x^2)*e^(-2*x))/(x*log(x)))
Exception generated. \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((8*x**2-16*x)*ln(x)*ln((x**3-2*x**2)/exp(x)**2)*ln(ln((x**3-2*x **2)/exp(x)**2)/x/ln(x))+((12*x**3-28*x**2+8*x)*ln(x)-4*x**2+8*x)*ln((x**3 -2*x**2)/exp(x)**2)+(-8*x**3+28*x**2-16*x)*ln(x))/(-2+x)/ln(x)/ln((x**3-2* x**2)/exp(x)**2),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4 \, x^{3} - 4 \, x^{2} \log \left (x\right ) + 4 \, x^{2} \log \left (-2 \, x + \log \left (x - 2\right ) + 2 \, \log \left (x\right )\right ) - 4 \, x^{2} \log \left (\log \left (x\right )\right ) \] Input:
integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^ 2)/exp(x)^2)/x/log(x))+((12*x^3-28*x^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x ^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(-2+x)/log(x)/log((x^3-2*x^2)/e xp(x)^2),x, algorithm="maxima")
Output:
4*x^3 - 4*x^2*log(x) + 4*x^2*log(-2*x + log(x - 2) + 2*log(x)) - 4*x^2*log (log(x))
Time = 0.94 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4 \, x^{3} - 4 \, x^{2} \log \left (x\right ) + 4 \, x^{2} \log \left (-2 \, x + \log \left (x - 2\right ) + 2 \, \log \left (x\right )\right ) - 4 \, x^{2} \log \left (\log \left (x\right )\right ) \] Input:
integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^ 2)/exp(x)^2)/x/log(x))+((12*x^3-28*x^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x ^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(-2+x)/log(x)/log((x^3-2*x^2)/e xp(x)^2),x, algorithm="giac")
Output:
4*x^3 - 4*x^2*log(x) + 4*x^2*log(-2*x + log(x - 2) + 2*log(x)) - 4*x^2*log (log(x))
Time = 3.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=4\,x^2\,\left (x+\ln \left (\frac {\ln \left (-{\mathrm {e}}^{-2\,x}\,\left (2\,x^2-x^3\right )\right )}{x\,\ln \left (x\right )}\right )\right ) \] Input:
int(-(log(x)*(16*x - 28*x^2 + 8*x^3) - log(-exp(-2*x)*(2*x^2 - x^3))*(8*x - 4*x^2 + log(x)*(8*x - 28*x^2 + 12*x^3)) + log(log(-exp(-2*x)*(2*x^2 - x^ 3))/(x*log(x)))*log(-exp(-2*x)*(2*x^2 - x^3))*log(x)*(16*x - 8*x^2))/(log( -exp(-2*x)*(2*x^2 - x^3))*log(x)*(x - 2)),x)
Output:
4*x^2*(x + log(log(-exp(-2*x)*(2*x^2 - x^3))/(x*log(x))))
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.34 \[ \int \frac {\left (-16 x+28 x^2-8 x^3\right ) \log (x)+\left (8 x-4 x^2+\left (8 x-28 x^2+12 x^3\right ) \log (x)\right ) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )+\left (-16 x+8 x^2\right ) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right ) \log \left (\frac {\log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )}{x \log (x)}\right )}{(-2+x) \log (x) \log \left (e^{-2 x} \left (-2 x^2+x^3\right )\right )} \, dx=8 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x^{3}-2 x^{2}}{e^{2 x}}\right )\right )-8 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+4 \,\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x^{3}-2 x^{2}}{e^{2 x}}\right )}{\mathrm {log}\left (x \right ) x}\right ) x^{2}-8 \,\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x^{3}-2 x^{2}}{e^{2 x}}\right )}{\mathrm {log}\left (x \right ) x}\right )-8 \,\mathrm {log}\left (x \right )+4 x^{3} \] Input:
int(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^2)/exp (x)^2)/x/log(x))+((12*x^3-28*x^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x^2)/ex p(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(-2+x)/log(x)/log((x^3-2*x^2)/exp(x)^ 2),x)
Output:
4*(2*log(log((x**3 - 2*x**2)/e**(2*x))) - 2*log(log(x)) + log(log((x**3 - 2*x**2)/e**(2*x))/(log(x)*x))*x**2 - 2*log(log((x**3 - 2*x**2)/e**(2*x))/( log(x)*x)) - 2*log(x) + x**3)