Integrand size = 222, antiderivative size = 27 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x+\frac {1}{-2 x+\log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \]
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x+\frac {1}{-2 x+\log \left (-x+16 e^{-2 x} (1+\log (5))^2\right )} \]
Integrate[(1 - 2*x - 4*x^3 + ((4 + 4*x^2)*(4 + 4*Log[5])^2)/E^(2*x) + (4*x ^2 - (4*x*(4 + 4*Log[5])^2)/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)]^2)/(-4* x^3 + (4*x^2*(4 + 4*Log[5])^2)/E^(2*x) + (4*x^2 - (4*x*(4 + 4*Log[5])^2)/E ^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x ))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)]^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.
| \(\displaystyle \int \frac {-4 x^3+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (e^{-2 x} (4+4 \log (5))^2-x\right )+e^{-2 x} \left (4 x^2+4\right ) (4+4 \log (5))^2-2 x+\left (e^{-2 x} (4+4 \log (5))^2-x\right ) \log ^2\left (e^{-2 x} (4+4 \log (5))^2-x\right )+1}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (e^{-2 x} (4+4 \log (5))^2-x\right )+\left (e^{-2 x} (4+4 \log (5))^2-x\right ) \log ^2\left (e^{-2 x} (4+4 \log (5))^2-x\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 x} \left (4 x^3-\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (e^{-2 x} (4+4 \log (5))^2-x\right )-e^{-2 x} \left (4 x^2+4\right ) (4+4 \log (5))^2+2 x-\left (e^{-2 x} (4+4 \log (5))^2-x\right ) \log ^2\left (e^{-2 x} (4+4 \log (5))^2-x\right )-1\right )}{\left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4 x^2+\log ^2\left (16 e^{-2 x} (1+\log (5))^2-x\right )-4 x \log \left (16 e^{-2 x} (1+\log (5))^2-x\right )+4}{\left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}+\frac {e^{2 x} (-2 x-1)}{\left (e^{2 x} x-16 (1+\log (5) (2+\log (5)))\right ) \left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \int \frac {1}{\left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}dx+\int \frac {e^{2 x}}{\left (16 (1+\log (5) (2+\log (5)))-e^{2 x} x\right ) \left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}dx+2 \int \frac {e^{2 x} x}{\left (16 (1+\log (5) (2+\log (5)))-e^{2 x} x\right ) \left (2 x-\log \left (16 e^{-2 x} (1+\log (5))^2-x\right )\right )^2}dx+x\) |
Int[(1 - 2*x - 4*x^3 + ((4 + 4*x^2)*(4 + 4*Log[5])^2)/E^(2*x) + (4*x^2 - ( 4*x*(4 + 4*Log[5])^2)/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x))*Log[-x + (4 + 4*Log[5])^2/E^(2*x)]^2)/(-4*x^3 + (4*x^2*(4 + 4*Log[5])^2)/E^(2*x) + (4*x^2 - (4*x*(4 + 4*Log[5])^2)/E^(2*x) )*Log[-x + (4 + 4*Log[5])^2/E^(2*x)] + (-x + (4 + 4*Log[5])^2/E^(2*x))*Log [-x + (4 + 4*Log[5])^2/E^(2*x)]^2),x]
3.17.32.3.1 Defintions of rubi rules used
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
| method | result | size |
| risch | \(x -\frac {1}{2 x -\ln \left (\left (4 \ln \left (5\right )+4\right )^{2} {\mathrm e}^{-2 x}-x \right )}\) | \(31\) |
| parallelrisch | \(\frac {-1+2 x^{2}-\ln \left (16 \left (\ln \left (5\right )+1\right )^{2} {\mathrm e}^{-2 x}-x \right ) x}{-\ln \left (16 \left (\ln \left (5\right )+1\right )^{2} {\mathrm e}^{-2 x}-x \right )+2 x}\) | \(58\) |
int(((exp(ln(4*ln(5)+4)-x)^2-x)*ln(exp(ln(4*ln(5)+4)-x)^2-x)^2+(-4*x*exp(l n(4*ln(5)+4)-x)^2+4*x^2)*ln(exp(ln(4*ln(5)+4)-x)^2-x)+(4*x^2+4)*exp(ln(4*l n(5)+4)-x)^2-4*x^3-2*x+1)/((exp(ln(4*ln(5)+4)-x)^2-x)*ln(exp(ln(4*ln(5)+4) -x)^2-x)^2+(-4*x*exp(ln(4*ln(5)+4)-x)^2+4*x^2)*ln(exp(ln(4*ln(5)+4)-x)^2-x )+4*x^2*exp(ln(4*ln(5)+4)-x)^2-4*x^3),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {2 \, x^{2} - x \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \left (5\right ) + 4\right )\right )}\right ) - 1}{2 \, x - \log \left (-x + e^{\left (-2 \, x + 2 \, \log \left (4 \, \log \left (5\right ) + 4\right )\right )}\right )} \]
integrate(((exp(log(4*log(5)+4)-x)^2-x)*log(exp(log(4*log(5)+4)-x)^2-x)^2+ (-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*log(exp(log(4*log(5)+4)-x)^2-x)+(4*x ^2+4)*exp(log(4*log(5)+4)-x)^2-4*x^3-2*x+1)/((exp(log(4*log(5)+4)-x)^2-x)* log(exp(log(4*log(5)+4)-x)^2-x)^2+(-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*lo g(exp(log(4*log(5)+4)-x)^2-x)+4*x^2*exp(log(4*log(5)+4)-x)^2-4*x^3),x, alg orithm=\
(2*x^2 - x*log(-x + e^(-2*x + 2*log(4*log(5) + 4))) - 1)/(2*x - log(-x + e ^(-2*x + 2*log(4*log(5) + 4))))
Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x + \frac {1}{- 2 x + \log {\left (- x + \left (4 + 4 \log {\left (5 \right )}\right )^{2} e^{- 2 x} \right )}} \]
integrate(((exp(ln(4*ln(5)+4)-x)**2-x)*ln(exp(ln(4*ln(5)+4)-x)**2-x)**2+(- 4*x*exp(ln(4*ln(5)+4)-x)**2+4*x**2)*ln(exp(ln(4*ln(5)+4)-x)**2-x)+(4*x**2+ 4)*exp(ln(4*ln(5)+4)-x)**2-4*x**3-2*x+1)/((exp(ln(4*ln(5)+4)-x)**2-x)*ln(e xp(ln(4*ln(5)+4)-x)**2-x)**2+(-4*x*exp(ln(4*ln(5)+4)-x)**2+4*x**2)*ln(exp( ln(4*ln(5)+4)-x)**2-x)+4*x**2*exp(ln(4*ln(5)+4)-x)**2-4*x**3),x)
Time = 0.63 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {4 \, x^{2} - x \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \left (5\right )^{2} + 32 \, \log \left (5\right ) + 16\right ) - 1}{4 \, x - \log \left (-x e^{\left (2 \, x\right )} + 16 \, \log \left (5\right )^{2} + 32 \, \log \left (5\right ) + 16\right )} \]
integrate(((exp(log(4*log(5)+4)-x)^2-x)*log(exp(log(4*log(5)+4)-x)^2-x)^2+ (-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*log(exp(log(4*log(5)+4)-x)^2-x)+(4*x ^2+4)*exp(log(4*log(5)+4)-x)^2-4*x^3-2*x+1)/((exp(log(4*log(5)+4)-x)^2-x)* log(exp(log(4*log(5)+4)-x)^2-x)^2+(-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*lo g(exp(log(4*log(5)+4)-x)^2-x)+4*x^2*exp(log(4*log(5)+4)-x)^2-4*x^3),x, alg orithm=\
(4*x^2 - x*log(-x*e^(2*x) + 16*log(5)^2 + 32*log(5) + 16) - 1)/(4*x - log( -x*e^(2*x) + 16*log(5)^2 + 32*log(5) + 16))
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (31) = 62\).
Time = 1.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.85 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=\frac {2 \, x^{2} - x \log \left (16 \, e^{\left (-2 \, x\right )} \log \left (5\right )^{2} + 32 \, e^{\left (-2 \, x\right )} \log \left (5\right ) - x + 16 \, e^{\left (-2 \, x\right )}\right ) - 1}{2 \, x - \log \left (16 \, e^{\left (-2 \, x\right )} \log \left (5\right )^{2} + 32 \, e^{\left (-2 \, x\right )} \log \left (5\right ) - x + 16 \, e^{\left (-2 \, x\right )}\right )} \]
integrate(((exp(log(4*log(5)+4)-x)^2-x)*log(exp(log(4*log(5)+4)-x)^2-x)^2+ (-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*log(exp(log(4*log(5)+4)-x)^2-x)+(4*x ^2+4)*exp(log(4*log(5)+4)-x)^2-4*x^3-2*x+1)/((exp(log(4*log(5)+4)-x)^2-x)* log(exp(log(4*log(5)+4)-x)^2-x)^2+(-4*x*exp(log(4*log(5)+4)-x)^2+4*x^2)*lo g(exp(log(4*log(5)+4)-x)^2-x)+4*x^2*exp(log(4*log(5)+4)-x)^2-4*x^3),x, alg orithm=\
(2*x^2 - x*log(16*e^(-2*x)*log(5)^2 + 32*e^(-2*x)*log(5) - x + 16*e^(-2*x) ) - 1)/(2*x - log(16*e^(-2*x)*log(5)^2 + 32*e^(-2*x)*log(5) - x + 16*e^(-2 *x)))
Time = 11.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {1-2 x-4 x^3+e^{-2 x} \left (4+4 x^2\right ) (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )}{-4 x^3+4 e^{-2 x} x^2 (4+4 \log (5))^2+\left (4 x^2-4 e^{-2 x} x (4+4 \log (5))^2\right ) \log \left (-x+e^{-2 x} (4+4 \log (5))^2\right )+\left (-x+e^{-2 x} (4+4 \log (5))^2\right ) \log ^2\left (-x+e^{-2 x} (4+4 \log (5))^2\right )} \, dx=x-\frac {1}{2\,x-\ln \left (16\,{\mathrm {e}}^{-2\,x}-x+32\,{\mathrm {e}}^{-2\,x}\,\ln \left (5\right )+16\,{\mathrm {e}}^{-2\,x}\,{\ln \left (5\right )}^2\right )} \]
int((2*x - exp(2*log(4*log(5) + 4) - 2*x)*(4*x^2 + 4) + log(exp(2*log(4*lo g(5) + 4) - 2*x) - x)*(4*x*exp(2*log(4*log(5) + 4) - 2*x) - 4*x^2) + log(e xp(2*log(4*log(5) + 4) - 2*x) - x)^2*(x - exp(2*log(4*log(5) + 4) - 2*x)) + 4*x^3 - 1)/(log(exp(2*log(4*log(5) + 4) - 2*x) - x)*(4*x*exp(2*log(4*log (5) + 4) - 2*x) - 4*x^2) + log(exp(2*log(4*log(5) + 4) - 2*x) - x)^2*(x - exp(2*log(4*log(5) + 4) - 2*x)) - 4*x^2*exp(2*log(4*log(5) + 4) - 2*x) + 4 *x^3),x)