Integrand size = 126, antiderivative size = 23 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=-13-x-\frac {1}{81+\frac {\log \left (1+e^{x^4}\right )}{x}} \]
Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=-x-\frac {x}{81 x+\log \left (1+e^{x^4}\right )} \]
Integrate[(-6561*x^2 + E^x^4*(-6561*x^2 + 4*x^4) + (-1 + E^x^4*(-1 - 162*x ) - 162*x)*Log[1 + E^x^4] + (-1 - E^x^4)*Log[1 + E^x^4]^2)/(6561*x^2 + 656 1*E^x^4*x^2 + (162*x + 162*E^x^4*x)*Log[1 + E^x^4] + (1 + E^x^4)*Log[1 + E ^x^4]^2),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.
\(\displaystyle \int \frac {\left (-e^{x^4}-1\right ) \log ^2\left (e^{x^4}+1\right )+\left (e^{x^4} (-162 x-1)-162 x-1\right ) \log \left (e^{x^4}+1\right )-6561 x^2+e^{x^4} \left (4 x^4-6561 x^2\right )}{\left (e^{x^4}+1\right ) \log ^2\left (e^{x^4}+1\right )+\left (162 e^{x^4} x+162 x\right ) \log \left (e^{x^4}+1\right )+6561 x^2+6561 e^{x^4} x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-e^{x^4}-1\right ) \log ^2\left (e^{x^4}+1\right )+\left (e^{x^4} (-162 x-1)-162 x-1\right ) \log \left (e^{x^4}+1\right )-6561 x^2+e^{x^4} \left (4 x^4-6561 x^2\right )}{\left (e^{x^4}+1\right ) \left (\log \left (e^{x^4}+1\right )+81 x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x^4-\log ^2\left (e^{x^4}+1\right )-162 x \log \left (e^{x^4}+1\right )-\log \left (e^{x^4}+1\right )-6561 x^2}{\left (\log \left (e^{x^4}+1\right )+81 x\right )^2}-\frac {4 x^4}{\left (e^{x^4}+1\right ) \left (\log \left (e^{x^4}+1\right )+81 x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{-81 x-\log \left (1+e^{x^4}\right )}dx+81 \int \frac {x}{\left (81 x+\log \left (1+e^{x^4}\right )\right )^2}dx+4 \int \frac {x^4}{\left (81 x+\log \left (1+e^{x^4}\right )\right )^2}dx-4 \int \frac {x^4}{\left (1+e^{x^4}\right ) \left (81 x+\log \left (1+e^{x^4}\right )\right )^2}dx-x\) |
Int[(-6561*x^2 + E^x^4*(-6561*x^2 + 4*x^4) + (-1 + E^x^4*(-1 - 162*x) - 16 2*x)*Log[1 + E^x^4] + (-1 - E^x^4)*Log[1 + E^x^4]^2)/(6561*x^2 + 6561*E^x^ 4*x^2 + (162*x + 162*E^x^4*x)*Log[1 + E^x^4] + (1 + E^x^4)*Log[1 + E^x^4]^ 2),x]
3.20.78.3.1 Defintions of rubi rules used
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-x -\frac {x}{81 x +\ln \left ({\mathrm e}^{x^{4}}+1\right )}\) | \(21\) |
parallelrisch | \(\frac {-81 x^{2}-\ln \left ({\mathrm e}^{x^{4}}+1\right ) x -x}{81 x +\ln \left ({\mathrm e}^{x^{4}}+1\right )}\) | \(34\) |
norman | \(\frac {\frac {\ln \left ({\mathrm e}^{x^{4}}+1\right )}{81}-\ln \left ({\mathrm e}^{x^{4}}+1\right ) x -81 x^{2}}{81 x +\ln \left ({\mathrm e}^{x^{4}}+1\right )}\) | \(40\) |
int(((-exp(x^4)-1)*ln(exp(x^4)+1)^2+((-162*x-1)*exp(x^4)-162*x-1)*ln(exp(x ^4)+1)+(4*x^4-6561*x^2)*exp(x^4)-6561*x^2)/((exp(x^4)+1)*ln(exp(x^4)+1)^2+ (162*x*exp(x^4)+162*x)*ln(exp(x^4)+1)+6561*x^2*exp(x^4)+6561*x^2),x,method =_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=-\frac {81 \, x^{2} + x \log \left (e^{\left (x^{4}\right )} + 1\right ) + x}{81 \, x + \log \left (e^{\left (x^{4}\right )} + 1\right )} \]
integrate(((-exp(x^4)-1)*log(exp(x^4)+1)^2+((-162*x-1)*exp(x^4)-162*x-1)*l og(exp(x^4)+1)+(4*x^4-6561*x^2)*exp(x^4)-6561*x^2)/((exp(x^4)+1)*log(exp(x ^4)+1)^2+(162*x*exp(x^4)+162*x)*log(exp(x^4)+1)+6561*x^2*exp(x^4)+6561*x^2 ),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=- x - \frac {x}{81 x + \log {\left (e^{x^{4}} + 1 \right )}} \]
integrate(((-exp(x**4)-1)*ln(exp(x**4)+1)**2+((-162*x-1)*exp(x**4)-162*x-1 )*ln(exp(x**4)+1)+(4*x**4-6561*x**2)*exp(x**4)-6561*x**2)/((exp(x**4)+1)*l n(exp(x**4)+1)**2+(162*x*exp(x**4)+162*x)*ln(exp(x**4)+1)+6561*x**2*exp(x* *4)+6561*x**2),x)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=-\frac {81 \, x^{2} + x \log \left (e^{\left (x^{4}\right )} + 1\right ) + x}{81 \, x + \log \left (e^{\left (x^{4}\right )} + 1\right )} \]
integrate(((-exp(x^4)-1)*log(exp(x^4)+1)^2+((-162*x-1)*exp(x^4)-162*x-1)*l og(exp(x^4)+1)+(4*x^4-6561*x^2)*exp(x^4)-6561*x^2)/((exp(x^4)+1)*log(exp(x ^4)+1)^2+(162*x*exp(x^4)+162*x)*log(exp(x^4)+1)+6561*x^2*exp(x^4)+6561*x^2 ),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=-\frac {81 \, x^{2} + x \log \left (e^{\left (x^{4}\right )} + 1\right ) + x}{81 \, x + \log \left (e^{\left (x^{4}\right )} + 1\right )} \]
integrate(((-exp(x^4)-1)*log(exp(x^4)+1)^2+((-162*x-1)*exp(x^4)-162*x-1)*l og(exp(x^4)+1)+(4*x^4-6561*x^2)*exp(x^4)-6561*x^2)/((exp(x^4)+1)*log(exp(x ^4)+1)^2+(162*x*exp(x^4)+162*x)*log(exp(x^4)+1)+6561*x^2*exp(x^4)+6561*x^2 ),x, algorithm=\
Timed out. \[ \int \frac {-6561 x^2+e^{x^4} \left (-6561 x^2+4 x^4\right )+\left (-1+e^{x^4} (-1-162 x)-162 x\right ) \log \left (1+e^{x^4}\right )+\left (-1-e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )}{6561 x^2+6561 e^{x^4} x^2+\left (162 x+162 e^{x^4} x\right ) \log \left (1+e^{x^4}\right )+\left (1+e^{x^4}\right ) \log ^2\left (1+e^{x^4}\right )} \, dx=\int -\frac {{\ln \left ({\mathrm {e}}^{x^4}+1\right )}^2\,\left ({\mathrm {e}}^{x^4}+1\right )+{\mathrm {e}}^{x^4}\,\left (6561\,x^2-4\,x^4\right )+\ln \left ({\mathrm {e}}^{x^4}+1\right )\,\left (162\,x+{\mathrm {e}}^{x^4}\,\left (162\,x+1\right )+1\right )+6561\,x^2}{{\ln \left ({\mathrm {e}}^{x^4}+1\right )}^2\,\left ({\mathrm {e}}^{x^4}+1\right )+6561\,x^2\,{\mathrm {e}}^{x^4}+\ln \left ({\mathrm {e}}^{x^4}+1\right )\,\left (162\,x+162\,x\,{\mathrm {e}}^{x^4}\right )+6561\,x^2} \,d x \]
int(-(log(exp(x^4) + 1)^2*(exp(x^4) + 1) + exp(x^4)*(6561*x^2 - 4*x^4) + l og(exp(x^4) + 1)*(162*x + exp(x^4)*(162*x + 1) + 1) + 6561*x^2)/(log(exp(x ^4) + 1)^2*(exp(x^4) + 1) + 6561*x^2*exp(x^4) + log(exp(x^4) + 1)*(162*x + 162*x*exp(x^4)) + 6561*x^2),x)