Integrand size = 72, antiderivative size = 23 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {x}{4+2 x-\frac {1}{\log \left (-130 e^{-4-x}\right )}} \] Output:
x/(2*x-1/ln(-130/exp(4+x))+4)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {1-4 \log \left (-130 e^{-4-x}\right )}{-2+4 (2+x) \log \left (-130 e^{-4-x}\right )} \] Input:
Integrate[(x - Log[-130*E^(-4 - x)] + 4*Log[-130*E^(-4 - x)]^2)/(1 + (-8 - 4*x)*Log[-130*E^(-4 - x)] + (16 + 16*x + 4*x^2)*Log[-130*E^(-4 - x)]^2),x ]
Output:
(1 - 4*Log[-130*E^(-4 - x)])/(-2 + 4*(2 + x)*Log[-130*E^(-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.
\(\displaystyle \int \frac {x+4 \log ^2\left (-130 e^{-x-4}\right )-\log \left (-130 e^{-x-4}\right )}{\left (4 x^2+16 x+16\right ) \log ^2\left (-130 e^{-x-4}\right )+(-4 x-8) \log \left (-130 e^{-x-4}\right )+1} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x+4 \log ^2\left (-130 e^{-x-4}\right )-\log \left (-130 e^{-x-4}\right )}{\left (1-2 (x+2) \log \left (-130 e^{-x-4}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (2 x^2+8 x+7\right )}{2 (x+2)^2 \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )^2}+\frac {1}{(x+2)^2}+\frac {2-x}{2 (x+2)^2 \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x}{\left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )^2}dx+\int \frac {1}{(x+2)^2 \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )^2}dx-\frac {1}{2} \int \frac {1}{(x+2) \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )^2}dx+2 \int \frac {1}{(x+2)^2 \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )}dx-\frac {1}{2} \int \frac {1}{(x+2) \left (2 x \log \left (-130 e^{-x-4}\right )+4 \log \left (-130 e^{-x-4}\right )-1\right )}dx-\frac {1}{x+2}\) |
Input:
Int[(x - Log[-130*E^(-4 - x)] + 4*Log[-130*E^(-4 - x)]^2)/(1 + (-8 - 4*x)* Log[-130*E^(-4 - x)] + (16 + 16*x + 4*x^2)*Log[-130*E^(-4 - x)]^2),x]
Output:
$Aborted
Time = 0.68 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83
method | result | size |
norman | \(\frac {\frac {1}{2}-2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )}{2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right ) x +4 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )-1}\) | \(42\) |
parallelrisch | \(\frac {-8 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )+2}{8 \ln \left (-130 \,{\mathrm e}^{-4-x}\right ) x +16 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )-4}\) | \(43\) |
default | \(\frac {\ln \left (-130 \,{\mathrm e}^{-4-x}\right )-\frac {1}{4}}{x^{2}-x \left (\ln \left (-130 \,{\mathrm e}^{-4-x}\right )+\ln \left ({\mathrm e}^{4+x}\right )\right )+x \left (\ln \left ({\mathrm e}^{4+x}\right )-4-x \right )+4 x -2 \ln \left (-130 \,{\mathrm e}^{-4-x}\right )+\frac {1}{2}}\) | \(64\) |
risch | \(-\frac {1}{2+x}-\frac {i x}{2 \left (2+x \right ) \left (-2 x \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{2}+2 x \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{3}-4 \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{2}+4 \pi \operatorname {csgn}\left (i {\mathrm e}^{-4-x}\right )^{3}+2 \pi x +4 \pi -2 i x \ln \left (2\right )-4 i \ln \left (13\right )-2 i x \ln \left (13\right )-4 i \ln \left (5\right )+4 i \ln \left ({\mathrm e}^{4+x}\right )+2 i x \ln \left ({\mathrm e}^{4+x}\right )-4 i \ln \left (2\right )-2 i x \ln \left (5\right )+i\right )}\) | \(142\) |
Input:
int((4*ln(-130/exp(4+x))^2-ln(-130/exp(4+x))+x)/((4*x^2+16*x+16)*ln(-130/e xp(4+x))^2+(-4*x-8)*ln(-130/exp(4+x))+1),x,method=_RETURNVERBOSE)
Output:
(1/2-2*ln(-130/exp(4+x)))/(2*ln(-130/exp(4+x))*x+4*ln(-130/exp(4+x))-1)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {-4 i \, \pi + 4 \, x - 4 \, \log \left (130\right ) + 17}{2 \, {\left (2 \, {\left (i \, \pi + \log \left (130\right )\right )} {\left (x + 2\right )} - 2 \, x^{2} - 12 \, x - 17\right )}} \] Input:
integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*l og(-130/exp(4+x))^2+(-4*x-8)*log(-130/exp(4+x))+1),x, algorithm="fricas")
Output:
1/2*(-4*I*pi + 4*x - 4*log(130) + 17)/(2*(I*pi + log(130))*(x + 2) - 2*x^2 - 12*x - 17)
Result contains complex when optimal does not.
Time = 1.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {- 4 x - 17 + 4 \log {\left (130 \right )} + 4 i \pi }{4 x^{2} + x \left (- 4 \log {\left (130 \right )} + 24 - 4 i \pi \right ) - 8 \log {\left (130 \right )} + 34 - 8 i \pi } \] Input:
integrate((4*ln(-130/exp(4+x))**2-ln(-130/exp(4+x))+x)/((4*x**2+16*x+16)*l n(-130/exp(4+x))**2+(-4*x-8)*ln(-130/exp(4+x))+1),x)
Output:
(-4*x - 17 + 4*log(130) + 4*I*pi)/(4*x**2 + x*(-4*log(130) + 24 - 4*I*pi) - 8*log(130) + 34 - 8*I*pi)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=-\frac {4 \, \log \left (13\right ) + 4 \, \log \left (5\right ) + 4 \, \log \left (2\right ) - 4 \, \log \left (-e^{x}\right ) - 17}{2 \, {\left (2 \, x {\left (\log \left (13\right ) + \log \left (5\right ) + \log \left (2\right ) - 4\right )} - 2 \, {\left (x + 2\right )} \log \left (-e^{x}\right ) + 4 \, \log \left (13\right ) + 4 \, \log \left (5\right ) + 4 \, \log \left (2\right ) - 17\right )}} \] Input:
integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*l og(-130/exp(4+x))^2+(-4*x-8)*log(-130/exp(4+x))+1),x, algorithm="maxima")
Output:
-1/2*(4*log(13) + 4*log(5) + 4*log(2) - 4*log(-e^x) - 17)/(2*x*(log(13) + log(5) + log(2) - 4) - 2*(x + 2)*log(-e^x) + 4*log(13) + 4*log(5) + 4*log( 2) - 17)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {4 i \, \pi - 4 \, x + 4 \, \log \left (130\right ) - 17}{-8 i \, \pi - 4 i \, \pi x + 4 \, x^{2} - 4 \, x \log \left (130\right ) + 24 \, x - 8 \, \log \left (130\right ) + 34} \] Input:
integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*l og(-130/exp(4+x))^2+(-4*x-8)*log(-130/exp(4+x))+1),x, algorithm="giac")
Output:
(4*I*pi - 4*x + 4*log(130) - 17)/(-8*I*pi - 4*I*pi*x + 4*x^2 - 4*x*log(130 ) + 24*x - 8*log(130) + 34)
Time = 3.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {2\,x-2\,\ln \left (130\right )+\frac {17}{2}-\pi \,2{}\mathrm {i}}{-2\,x^2+\left (2\,\ln \left (130\right )-12+\pi \,2{}\mathrm {i}\right )\,x+\pi \,4{}\mathrm {i}+4\,\ln \left (130\right )-17} \] Input:
int((x - log(-130*exp(- x - 4)) + 4*log(-130*exp(- x - 4))^2)/(log(-130*ex p(- x - 4))^2*(16*x + 4*x^2 + 16) - log(-130*exp(- x - 4))*(4*x + 8) + 1), x)
Output:
(2*x - pi*2i - 2*log(130) + 17/2)/(pi*4i + 4*log(130) + x*(pi*2i + 2*log(1 30) - 12) - 2*x^2 - 17)
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx=\frac {\mathrm {log}\left (-\frac {130}{e^{x} e^{4}}\right ) x}{2 \,\mathrm {log}\left (-\frac {130}{e^{x} e^{4}}\right ) x +4 \,\mathrm {log}\left (-\frac {130}{e^{x} e^{4}}\right )-1} \] Input:
int((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*log(-13 0/exp(4+x))^2+(-4*x-8)*log(-130/exp(4+x))+1),x)
Output:
(log(( - 130)/(e**x*e**4))*x)/(2*log(( - 130)/(e**x*e**4))*x + 4*log(( - 1 30)/(e**x*e**4)) - 1)