Integrand size = 106, antiderivative size = 22 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=-x+\frac {5}{x+\frac {e^{-x} x}{\sqrt {2}}} \] Output:
5/(x+x/exp(1+1/2*ln(2))*exp(1-x))-x
Time = 2.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=-\frac {x^2+\sqrt {2} e^x \left (-5+x^2\right )}{x+\sqrt {2} e^x x} \] Input:
Integrate[(-(E^(2 - 2*x)*x^2) + E^(1 - x + (4 + Log[4])/4)*(-5 + 5*x - 2*x ^2) + E^((4 + Log[4])/2)*(-5 - x^2))/(E^(2 - 2*x)*x^2 + E^((4 + Log[4])/2) *x^2 + 2*E^(1 - x + (4 + Log[4])/4)*x^2),x]
Output:
-((x^2 + Sqrt[2]*E^x*(-5 + x^2))/(x + Sqrt[2]*E^x*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.
\(\displaystyle \int \frac {-e^{2-2 x} x^2+\left (-2 x^2+5 x-5\right ) e^{-x+1+\frac {1}{4} (4+\log (4))}+\left (-x^2-5\right ) e^{\frac {1}{2} (4+\log (4))}}{e^{2-2 x} x^2+2 x^2 e^{-x+1+\frac {1}{4} (4+\log (4))}+x^2 e^{\frac {1}{2} (4+\log (4))}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-x^2+\sqrt {2} e^x \left (-2 x^2+5 x-5\right )-2 e^{2 x} \left (x^2+5\right )}{\left (\sqrt {2} e^x x+x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 (x+1)}{\left (\sqrt {2} e^x+1\right ) x^2}+\frac {-x^2-5}{x^2}-\frac {5}{\left (\sqrt {2} e^x+1\right )^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int \frac {1}{\left (1+\sqrt {2} e^x\right ) x^2}dx-5 \int \frac {1}{\left (1+\sqrt {2} e^x\right )^2 x}dx+5 \int \frac {1}{\left (1+\sqrt {2} e^x\right ) x}dx-x+\frac {5}{x}\) |
Input:
Int[(-(E^(2 - 2*x)*x^2) + E^(1 - x + (4 + Log[4])/4)*(-5 + 5*x - 2*x^2) + E^((4 + Log[4])/2)*(-5 - x^2))/(E^(2 - 2*x)*x^2 + E^((4 + Log[4])/2)*x^2 + 2*E^(1 - x + (4 + Log[4])/4)*x^2),x]
Output:
$Aborted
Time = 1.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
risch | \(-x +\frac {10 \,{\mathrm e}}{x \left (2 \,{\mathrm e}+{\mathrm e}^{1-x} \sqrt {2}\right )}\) | \(29\) |
norman | \(\frac {-x^{2} {\mathrm e}^{1-x}+5 \,{\mathrm e} \sqrt {2}-{\mathrm e} \sqrt {2}\, x^{2}}{x \left ({\mathrm e}^{1+\frac {\ln \left (2\right )}{2}}+{\mathrm e}^{1-x}\right )}\) | \(50\) |
parallelrisch | \(-\frac {{\mathrm e}^{1+\frac {\ln \left (2\right )}{2}} x^{2}+x^{2} {\mathrm e}^{1-x}-5 \,{\mathrm e}^{1+\frac {\ln \left (2\right )}{2}}}{\left ({\mathrm e}^{1+\frac {\ln \left (2\right )}{2}}+{\mathrm e}^{1-x}\right ) x}\) | \(53\) |
Input:
int(((-x^2-5)*exp(1+1/2*ln(2))^2+(-2*x^2+5*x-5)*exp(1-x)*exp(1+1/2*ln(2))- x^2*exp(1-x)^2)/(x^2*exp(1+1/2*ln(2))^2+2*x^2*exp(1-x)*exp(1+1/2*ln(2))+x^ 2*exp(1-x)^2),x,method=_RETURNVERBOSE)
Output:
-x+10*exp(1)/x/(2*exp(1)+exp(1-x)*2^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (21) = 42\).
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.27 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=-\frac {x^{2} e^{\left (-x + \frac {1}{2} \, \log \left (2\right ) + 2\right )} + {\left (x^{2} - 5\right )} e^{\left (\log \left (2\right ) + 2\right )}}{x e^{\left (-x + \frac {1}{2} \, \log \left (2\right ) + 2\right )} + x e^{\left (\log \left (2\right ) + 2\right )}} \] Input:
integrate(((-x^2-5)*exp(1+1/2*log(2))^2+(-2*x^2+5*x-5)*exp(1-x)*exp(1+1/2* log(2))-x^2*exp(1-x)^2)/(x^2*exp(1+1/2*log(2))^2+2*x^2*exp(1-x)*exp(1+1/2* log(2))+x^2*exp(1-x)^2),x, algorithm="fricas")
Output:
-(x^2*e^(-x + 1/2*log(2) + 2) + (x^2 - 5)*e^(log(2) + 2))/(x*e^(-x + 1/2*l og(2) + 2) + x*e^(log(2) + 2))
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=- x + \frac {5 \sqrt {2} e}{x \left (e^{1 - x} + \sqrt {2} e\right )} \] Input:
integrate(((-x**2-5)*exp(1+1/2*ln(2))**2+(-2*x**2+5*x-5)*exp(1-x)*exp(1+1/ 2*ln(2))-x**2*exp(1-x)**2)/(x**2*exp(1+1/2*ln(2))**2+2*x**2*exp(1-x)*exp(1 +1/2*ln(2))+x**2*exp(1-x)**2),x)
Output:
-x + 5*sqrt(2)*E/(x*(exp(1 - x) + sqrt(2)*E))
Exception generated. \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((-x^2-5)*exp(1+1/2*log(2))^2+(-2*x^2+5*x-5)*exp(1-x)*exp(1+1/2* log(2))-x^2*exp(1-x)^2)/(x^2*exp(1+1/2*log(2))^2+2*x^2*exp(1-x)*exp(1+1/2* log(2))+x^2*exp(1-x)^2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (21) = 42\).
Time = 0.15 (sec) , antiderivative size = 530, normalized size of antiderivative = 24.09 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx =\text {Too large to display} \] Input:
integrate(((-x^2-5)*exp(1+1/2*log(2))^2+(-2*x^2+5*x-5)*exp(1-x)*exp(1+1/2* log(2))-x^2*exp(1-x)^2)/(x^2*exp(1+1/2*log(2))^2+2*x^2*exp(1-x)*exp(1+1/2* log(2))+x^2*exp(1-x)^2),x, algorithm="giac")
Output:
-1/2*(2*(2*x - log(2) - 4)^2*e^2 + (2*x - log(2) - 4)^2*e^(-x + 1/2*log(2) + 2) + 2*(2*x - log(2) - 4)*e^2*log(2) + (2*x - log(2) - 4)*e^(-x + 1/2*l og(2) + 2)*log(2) - 4*(2*x - log(2) - 4)*e^2*log(2*e^2 + e^(-x + 1/2*log(2 ) + 2)) - 2*(2*x - log(2) - 4)*e^(-x + 1/2*log(2) + 2)*log(2*e^2 + e^(-x + 1/2*log(2) + 2)) - 4*e^2*log(2)*log(2*e^2 + e^(-x + 1/2*log(2) + 2)) - 2* e^(-x + 1/2*log(2) + 2)*log(2)*log(2*e^2 + e^(-x + 1/2*log(2) + 2)) + 4*(2 *x - log(2) - 4)*e^2*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) + 2*(2*x - log( 2) - 4)*e^(-x + 1/2*log(2) + 2)*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) + 4* e^2*log(2)*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) + 2*e^(-x + 1/2*log(2) + 2)*log(2)*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) + 8*(2*x - log(2) - 4)*e^2 + 4*(2*x - log(2) - 4)*e^(-x + 1/2*log(2) + 2) - 16*e^2*log(2*e^2 + e^(-x + 1/2*log(2) + 2)) - 8*e^(-x + 1/2*log(2) + 2)*log(2*e^2 + e^(-x + 1/2*lo g(2) + 2)) + 16*e^2*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) + 8*e^(-x + 1/2* log(2) + 2)*log(-2*e^2 - e^(-x + 1/2*log(2) + 2)) - 40*e^2)/(2*(2*x - log( 2) - 4)*e^2 + (2*x - log(2) - 4)*e^(-x + 1/2*log(2) + 2) + 2*e^2*log(2) + e^(-x + 1/2*log(2) + 2)*log(2) + 8*e^2 + 4*e^(-x + 1/2*log(2) + 2))
Time = 7.66 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=\frac {5\,\sqrt {2}\,\mathrm {e}}{x\,\left ({\mathrm {e}}^{1-x}+\sqrt {2}\,\mathrm {e}\right )}-x \] Input:
int(-(exp(log(2) + 2)*(x^2 + 5) + x^2*exp(2 - 2*x) + exp(log(2)/2 + 1)*exp (1 - x)*(2*x^2 - 5*x + 5))/(x^2*exp(log(2) + 2) + x^2*exp(2 - 2*x) + 2*x^2 *exp(log(2)/2 + 1)*exp(1 - x)),x)
Output:
(5*2^(1/2)*exp(1))/(x*(exp(1 - x) + 2^(1/2)*exp(1))) - x
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.95 \[ \int \frac {-e^{2-2 x} x^2+e^{1-x+\frac {1}{4} (4+\log (4))} \left (-5+5 x-2 x^2\right )+e^{\frac {1}{2} (4+\log (4))} \left (-5-x^2\right )}{e^{2-2 x} x^2+e^{\frac {1}{2} (4+\log (4))} x^2+2 e^{1-x+\frac {1}{4} (4+\log (4))} x^2} \, dx=\frac {-2 e^{2 x} x^{2}+10 e^{2 x}-5 e^{x} \sqrt {2}+x^{2}}{x \left (2 e^{2 x}-1\right )} \] Input:
int(((-x^2-5)*exp(1+1/2*log(2))^2+(-2*x^2+5*x-5)*exp(1-x)*exp(1+1/2*log(2) )-x^2*exp(1-x)^2)/(x^2*exp(1+1/2*log(2))^2+2*x^2*exp(1-x)*exp(1+1/2*log(2) )+x^2*exp(1-x)^2),x)
Output:
( - 2*e**(2*x)*x**2 + 10*e**(2*x) - 5*e**x*sqrt(2) + x**2)/(x*(2*e**(2*x) - 1))