Integrand size = 78, antiderivative size = 26 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\left (1+e^5-e^{-2-\log ^2((-2-x) x)}\right ) \log (x) \] Output:
(1+exp(5)-1/exp(2+ln((-2-x)*x)^2))*ln(x)
Time = 5.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\left (1+e^5-e^{-2-\log ^2(-x (2+x))}\right ) \log (x) \] Input:
Integrate[(E^(-2 - Log[-2*x - x^2]^2)*(-2 - x + E^(2 + Log[-2*x - x^2]^2)* (2 + x + E^5*(2 + x)) + (4 + 4*x)*Log[x]*Log[-2*x - x^2]))/(2*x + x^2),x]
Output:
(1 + E^5 - E^(-2 - Log[-(x*(2 + x))]^2))*Log[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.
\(\displaystyle \int \frac {e^{-\log ^2\left (-x^2-2 x\right )-2} \left (\left (x+e^5 (x+2)+2\right ) e^{\log ^2\left (-x^2-2 x\right )+2}+(4 x+4) \log (x) \log \left (-x^2-2 x\right )-x-2\right )}{x^2+2 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-\log ^2\left (-x^2-2 x\right )-2} \left (\left (x+e^5 (x+2)+2\right ) e^{\log ^2\left (-x^2-2 x\right )+2}+(4 x+4) \log (x) \log \left (-x^2-2 x\right )-x-2\right )}{x (x+2)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-\log ^2\left (-x^2-2 x\right )-2} (-x+4 x \log (x) \log (-x (x+2))+4 \log (x) \log (-x (x+2))-2)}{x (x+2)}+\frac {\left (1+e^5\right ) e^{\log ^2(-x (x+2))-\log ^2\left (-x^2-2 x\right )}}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{-\log ^2\left (-x^2-2 x\right )-2}}{x}dx+2 \int \frac {e^{-\log ^2\left (-x^2-2 x\right )-2} \log (x) \log ((-x-2) x)}{x}dx+2 \int \frac {e^{-\log ^2\left (-x^2-2 x\right )-2} \log (x) \log ((-x-2) x)}{x+2}dx+\left (1+e^5\right ) \log (x)\) |
Input:
Int[(E^(-2 - Log[-2*x - x^2]^2)*(-2 - x + E^(2 + Log[-2*x - x^2]^2)*(2 + x + E^5*(2 + x)) + (4 + 4*x)*Log[x]*Log[-2*x - x^2]))/(2*x + x^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(61\) vs. \(2(24)=48\).
Time = 0.94 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38
method | result | size |
parallelrisch | \(\left (\ln \left (x \right ) {\mathrm e}^{\ln \left (-x^{2}-2 x \right )^{2}+2} {\mathrm e}^{5}+\ln \left (x \right ) {\mathrm e}^{\ln \left (-x^{2}-2 x \right )^{2}+2}-\ln \left (x \right )\right ) {\mathrm e}^{-\ln \left (-x^{2}-2 x \right )^{2}-2}\) | \(62\) |
risch | \(\text {Expression too large to display}\) | \(632\) |
Input:
int((((2+x)*exp(5)+2+x)*exp(ln(-x^2-2*x)^2+2)+(4+4*x)*ln(-x^2-2*x)*ln(x)-x -2)/(x^2+2*x)/exp(ln(-x^2-2*x)^2+2),x,method=_RETURNVERBOSE)
Output:
(ln(x)*exp(ln(-x^2-2*x)^2+2)*exp(5)+ln(x)*exp(ln(-x^2-2*x)^2+2)-ln(x))/exp (ln(-x^2-2*x)^2+2)
Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx={\left ({\left (e^{5} + 1\right )} e^{\left (\log \left (-x^{2} - 2 \, x\right )^{2} + 2\right )} \log \left (x\right ) - \log \left (x\right )\right )} e^{\left (-\log \left (-x^{2} - 2 \, x\right )^{2} - 2\right )} \] Input:
integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4+4*x)*log(-x^2-2*x) *log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-2*x)^2+2),x, algorithm="fricas")
Output:
((e^5 + 1)*e^(log(-x^2 - 2*x)^2 + 2)*log(x) - log(x))*e^(-log(-x^2 - 2*x)^ 2 - 2)
Timed out. \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\text {Timed out} \] Input:
integrate((((2+x)*exp(5)+2+x)*exp(ln(-x**2-2*x)**2+2)+(4+4*x)*ln(-x**2-2*x )*ln(x)-x-2)/(x**2+2*x)/exp(ln(-x**2-2*x)**2+2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=-{\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} e^{5} + e^{5} \log \left (x + 2\right ) - e^{\left (-\log \left (x\right )^{2} - 2 \, \log \left (x\right ) \log \left (-x - 2\right ) - \log \left (-x - 2\right )^{2} - 2\right )} \log \left (x\right ) + \log \left (x\right ) \] Input:
integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4+4*x)*log(-x^2-2*x) *log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-2*x)^2+2),x, algorithm="maxima")
Output:
-(log(x + 2) - log(x))*e^5 + e^5*log(x + 2) - e^(-log(x)^2 - 2*log(x)*log( -x - 2) - log(-x - 2)^2 - 2)*log(x) + log(x)
\[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\int { \frac {{\left (4 \, {\left (x + 1\right )} \log \left (-x^{2} - 2 \, x\right ) \log \left (x\right ) + {\left ({\left (x + 2\right )} e^{5} + x + 2\right )} e^{\left (\log \left (-x^{2} - 2 \, x\right )^{2} + 2\right )} - x - 2\right )} e^{\left (-\log \left (-x^{2} - 2 \, x\right )^{2} - 2\right )}}{x^{2} + 2 \, x} \,d x } \] Input:
integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4+4*x)*log(-x^2-2*x) *log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-2*x)^2+2),x, algorithm="giac")
Output:
integrate((4*(x + 1)*log(-x^2 - 2*x)*log(x) + ((x + 2)*e^5 + x + 2)*e^(log (-x^2 - 2*x)^2 + 2) - x - 2)*e^(-log(-x^2 - 2*x)^2 - 2)/(x^2 + 2*x), x)
Time = 2.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\ln \left (x\right )\,\left ({\mathrm {e}}^5+1\right )-{\mathrm {e}}^{-{\ln \left (-x^2-2\,x\right )}^2-2}\,\ln \left (x\right ) \] Input:
int(-(exp(- log(- 2*x - x^2)^2 - 2)*(x - exp(log(- 2*x - x^2)^2 + 2)*(x + exp(5)*(x + 2) + 2) - log(- 2*x - x^2)*log(x)*(4*x + 4) + 2))/(2*x + x^2), x)
Output:
log(x)*(exp(5) + 1) - exp(- log(- 2*x - x^2)^2 - 2)*log(x)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{2 x+x^2} \, dx=\frac {\mathrm {log}\left (x \right ) \left (e^{\mathrm {log}\left (-x^{2}-2 x \right )^{2}} e^{7}+e^{\mathrm {log}\left (-x^{2}-2 x \right )^{2}} e^{2}-1\right )}{e^{\mathrm {log}\left (-x^{2}-2 x \right )^{2}} e^{2}} \] Input:
int((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4+4*x)*log(-x^2-2*x)*log(x )-x-2)/(x^2+2*x)/exp(log(-x^2-2*x)^2+2),x)
Output:
(log(x)*(e**(log( - x**2 - 2*x)**2)*e**7 + e**(log( - x**2 - 2*x)**2)*e**2 - 1))/(e**(log( - x**2 - 2*x)**2)*e**2)