Integrand size = 80, antiderivative size = 21 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \]
Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \]
Integrate[(E^(-4 + E^x)*(16 - 16*x) + E^(-4 + E^x)*(E^x*(16*x - 16*x^2) + 4*E^x*x*Log[-x^4])*Log[4 - 4*x + Log[-x^4]])/(4*x - 4*x^2 + x*Log[-x^4]),x ]
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.
\(\displaystyle \int \frac {e^{e^x-4} \left (4 e^x x \log \left (-x^4\right )+e^x \left (16 x-16 x^2\right )\right ) \log \left (\log \left (-x^4\right )-4 x+4\right )+e^{e^x-4} (16-16 x)}{x \log \left (-x^4\right )-4 x^2+4 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {16 e^{e^x-4} (x-1)}{x \left (-\log \left (-x^4\right )+4 x-4\right )}+4 e^{x+e^x-4} \log \left (\log \left (-x^4\right )-4 x+4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 \int \frac {e^{-4+e^x}}{4 x-\log \left (-x^4\right )-4}dx-16 \int \frac {e^{-4+e^x}}{x \left (4 x-\log \left (-x^4\right )-4\right )}dx+4 \int e^{x+e^x-4} \log \left (-4 x+\log \left (-x^4\right )+4\right )dx\) |
Int[(E^(-4 + E^x)*(16 - 16*x) + E^(-4 + E^x)*(E^x*(16*x - 16*x^2) + 4*E^x* x*Log[-x^4])*Log[4 - 4*x + Log[-x^4]])/(4*x - 4*x^2 + x*Log[-x^4]),x]
3.3.5.3.1 Defintions of rubi rules used
Time = 3.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (\ln \left (-x^{4}\right )-4 x +4\right )\) | \(20\) |
risch | \(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (i \pi +4 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{3}\right )\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}+i \pi \operatorname {csgn}\left (i x^{4}\right )^{2} \left (\operatorname {csgn}\left (i x^{4}\right )-1\right )-4 x +4\right )\) | \(158\) |
int(((4*x*exp(x)*ln(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*ln(ln(-x^4) -4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*ln(-x^4)-4*x^2+4*x),x,method=_RETURNV ERBOSE)
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \]
integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo rithm=\
Timed out. \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=\text {Timed out} \]
integrate(((4*x*exp(x)*ln(-x**4)+(-16*x**2+16*x)*exp(x))*exp(exp(x)-4)*ln( ln(-x**4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*ln(-x**4)-4*x**2+4*x),x)
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (i \, \pi - 4 \, x + 4 \, \log \left (x\right ) + 4\right ) \]
integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo rithm=\
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \]
integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo rithm=\
Time = 9.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4\,\ln \left (\ln \left (-x^4\right )-4\,x+4\right )\,{\mathrm {e}}^{{\mathrm {e}}^x-4} \]