Integrand size = 109, antiderivative size = 23 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=(-3+x) \log \left (\log \left (x-\left (e^x+e^x x^2\right ) \log (16)\right )\right ) \end {dmath*}
Time = 0.51 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=-3 \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )+x \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \end {dmath*}
Integrate[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2 )*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]*Log[Log[x + E^x*(-1 - x^2)*Log[ 16]]])/((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),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.
\(\displaystyle \int \frac {\left (e^x \left (x^2+1\right ) \log (16)-x\right ) \log \left (e^x \left (-x^2-1\right ) \log (16)+x\right ) \log \left (\log \left (e^x \left (-x^2-1\right ) \log (16)+x\right )\right )+e^x \left (x^3-x^2-5 x-3\right ) \log (16)-x+3}{\left (e^x \left (x^2+1\right ) \log (16)-x\right ) \log \left (e^x \left (-x^2-1\right ) \log (16)+x\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^3-x^2+x^2 \log \left (x-e^x \left (x^2+1\right ) \log (16)\right ) \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )+\log \left (x-e^x \left (x^2+1\right ) \log (16)\right ) \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )-5 x-3}{\left (x^2+1\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}+\frac {x^4-2 x^3-2 x^2-4 x+3}{\left (x^2+1\right ) \left (e^x x^2 \log (16)-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {1}{\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+(3-i) \int \frac {1}{(i-x) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \frac {x}{\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-(3+i) \int \frac {1}{(x+i) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-3 \int \frac {1}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+(1+3 i) \int \frac {1}{(i-x) \left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-2 \int \frac {x}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \frac {x^2}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-(1-3 i) \int \frac {1}{(x+i) \left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )dx\) |
Int[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2)*Log[ 16])*Log[x + E^x*(-1 - x^2)*Log[16]]*Log[Log[x + E^x*(-1 - x^2)*Log[16]]]) /((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),x]
3.2.17.3.1 Defintions of rubi rules used
Time = 25.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74
method | result | size |
risch | \(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right )\) | \(40\) |
parallelrisch | \(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right )\) | \(40\) |
int(((4*(x^2+1)*ln(2)*exp(x)-x)*ln(4*(-x^2-1)*ln(2)*exp(x)+x)*ln(ln(4*(-x^ 2-1)*ln(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*ln(2)*exp(x)+3-x)/(4*(x^2+1)*ln(2) *exp(x)-x)/ln(4*(-x^2-1)*ln(2)*exp(x)+x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx={\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} + 1\right )} e^{x} \log \left (2\right ) + x\right )\right ) \end {dmath*}
integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log (log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* (x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm=\
Time = 1.63 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=\left (x - 1\right ) \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\left (2 \right )} \right )} \right )} - 2 \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\left (2 \right )} \right )} \right )} \end {dmath*}
integrate(((4*(x**2+1)*ln(2)*exp(x)-x)*ln(4*(-x**2-1)*ln(2)*exp(x)+x)*ln(l n(4*(-x**2-1)*ln(2)*exp(x)+x))+4*(x**3-x**2-5*x-3)*ln(2)*exp(x)+3-x)/(4*(x **2+1)*ln(2)*exp(x)-x)/ln(4*(-x**2-1)*ln(2)*exp(x)+x),x)
(x - 1)*log(log(x + (-4*x**2 - 4)*exp(x)*log(2))) - 2*log(log(x + (-4*x**2 - 4)*exp(x)*log(2)))
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx={\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} \log \left (2\right ) + \log \left (2\right )\right )} e^{x} + x\right )\right ) \end {dmath*}
integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log (log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* (x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).
Time = 0.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=x \log \left (\log \left (-4 \, x^{2} e^{x} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + x\right )\right ) - 3 \, \log \left (\log \left (-4 \, x^{2} e^{x} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + x\right )\right ) \end {dmath*}
integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log (log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* (x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm=\
x*log(log(-4*x^2*e^x*log(2) - 4*e^x*log(2) + x)) - 3*log(log(-4*x^2*e^x*lo g(2) - 4*e^x*log(2) + x))
Time = 14.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \begin {dmath*} \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=\ln \left (\ln \left (x-4\,{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^2+1\right )\right )\right )\,\left (x-3\right ) \end {dmath*}