Integrand size = 67, antiderivative size = 23 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-5+x-x^{1-x+\frac {x^2}{-1-x}} \] Output:
-5-x*exp(ln(x)*(x^2/(-1-x)-x))+x
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=x-x^{\frac {1-2 x^2}{1+x}} \] Input:
Integrate[(1 + 2*x + x^2 + x^((-x - 2*x^2)/(1 + x))*(-1 - x + 2*x^2 + 2*x^ 3 + (x + 4*x^2 + 2*x^3)*Log[x]))/(1 + 2*x + x^2),x]
Output:
x - x^((1 - 2*x^2)/(1 + 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.
\(\displaystyle \int \frac {x^2+x^{\frac {-2 x^2-x}{x+1}} \left (2 x^3+2 x^2+\left (2 x^3+4 x^2+x\right ) \log (x)-x-1\right )+2 x+1}{x^2+2 x+1} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {x^2+x^{\frac {-2 x^2-x}{x+1}} \left (2 x^3+2 x^2+\left (2 x^3+4 x^2+x\right ) \log (x)-x-1\right )+2 x+1}{(x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^{-\frac {x (2 x+1)}{x+1}} \left (2 x^3+2 x^3 \log (x)+2 x^2+4 x^2 \log (x)-x+x \log (x)-1\right )}{(x+1)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {\int x^{\frac {1-2 x^2}{x+1}}dx}{x}dx+2 \log (x) \int x^{\frac {1-2 x^2}{x+1}}dx-\int \frac {x^{\frac {1-2 x^2}{x+1}} \log (x)}{(x+1)^2}dx-x^{1-\frac {x (2 x+1)}{x+1}} \operatorname {Hypergeometric2F1}\left (1,1-\frac {x (2 x+1)}{x+1},2-\frac {x (2 x+1)}{x+1},-x\right )-\frac {(2 x+1) x^{2-\frac {x (2 x+1)}{x+1}} \operatorname {Hypergeometric2F1}\left (1,1-\frac {x (2 x+1)}{x+1},2-\frac {x (2 x+1)}{x+1},-x\right )}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}+\frac {2 \left (1-2 x^2\right ) \left (2-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {1-2 x^2}{x+1}+1} \operatorname {Hypergeometric2F1}\left (1,\frac {1-2 x^2}{x+1}+1,\frac {1-2 x^2}{x+1}+2,-x\right )}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right ) \left (\frac {1-2 x^2}{x+1}+1\right )}+\frac {2 \left (-2 x^2+x+2\right ) \left (3-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {-2 x^2+x+2}{x+1}+1} \operatorname {Hypergeometric2F1}\left (1,\frac {-2 x^2+x+2}{x+1}+1,\frac {-2 x^2+x+2}{x+1}+2,-x\right )}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right ) \left (\frac {-2 x^2+x+2}{x+1}+1\right )}-\frac {\left (1-\frac {x (2 x+1)}{x+1}\right ) x^{-\frac {x (2 x+1)}{x+1}}}{2 x+1}+\frac {x^{-\frac {x (2 x+1)}{x+1}}}{2 x+1}-\frac {x^{1-\frac {x (2 x+1)}{x+1}}}{x+1}+\frac {2 x^{2-\frac {x (2 x+1)}{x+1}}}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}+\frac {2 x^{3-\frac {x (2 x+1)}{x+1}}}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right )}-\frac {2 \left (2-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {1-2 x^2}{x+1}+1}}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}-\frac {2 \left (3-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {-2 x^2+x+2}{x+1}+1}}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right )}+x\) |
Input:
Int[(1 + 2*x + x^2 + x^((-x - 2*x^2)/(1 + x))*(-1 - x + 2*x^2 + 2*x^3 + (x + 4*x^2 + 2*x^3)*Log[x]))/(1 + 2*x + x^2),x]
Output:
$Aborted
Time = 0.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-x \,x^{-\frac {\left (1+2 x \right ) x}{1+x}}+x\) | \(21\) |
parallelrisch | \(-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}+x -\frac {3}{2}\) | \(25\) |
norman | \(\frac {x^{2}-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-1}{1+x}\) | \(56\) |
default | \(x +\frac {-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-2 x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}}{\left (1+x \right )^{2}}\) | \(77\) |
parts | \(x +\frac {-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-2 x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}}{\left (1+x \right )^{2}}\) | \(77\) |
Input:
int((((2*x^3+4*x^2+x)*ln(x)+2*x^3+2*x^2-x-1)*exp((-2*x^2-x)*ln(x)/(1+x))+x ^2+2*x+1)/(x^2+2*x+1),x,method=_RETURNVERBOSE)
Output:
-x*x^(-(1+2*x)*x/(1+x))+x
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=\frac {x x^{\frac {2 \, x^{2} + x}{x + 1}} - x}{x^{\frac {2 \, x^{2} + x}{x + 1}}} \] Input:
integrate((((2*x^3+4*x^2+x)*log(x)+2*x^3+2*x^2-x-1)*exp((-2*x^2-x)*log(x)/ (1+x))+x^2+2*x+1)/(x^2+2*x+1),x, algorithm="fricas")
Output:
(x*x^((2*x^2 + x)/(x + 1)) - x)/x^((2*x^2 + x)/(x + 1))
Time = 54.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=- x e^{\frac {\left (- 2 x^{2} - x\right ) \log {\left (x \right )}}{x + 1}} + x \] Input:
integrate((((2*x**3+4*x**2+x)*ln(x)+2*x**3+2*x**2-x-1)*exp((-2*x**2-x)*ln( x)/(1+x))+x**2+2*x+1)/(x**2+2*x+1),x)
Output:
-x*exp((-2*x**2 - x)*log(x)/(x + 1)) + x
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-x^{2} e^{\left (-2 \, x \log \left (x\right ) - \frac {\log \left (x\right )}{x + 1}\right )} + x \] Input:
integrate((((2*x^3+4*x^2+x)*log(x)+2*x^3+2*x^2-x-1)*exp((-2*x^2-x)*log(x)/ (1+x))+x^2+2*x+1)/(x^2+2*x+1),x, algorithm="maxima")
Output:
-x^2*e^(-2*x*log(x) - log(x)/(x + 1)) + x
Time = 0.75 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=x - \frac {x^{\left (\frac {1}{x + 1}\right )}}{x^{\frac {2 \, x^{2}}{x + 1}}} \] Input:
integrate((((2*x^3+4*x^2+x)*log(x)+2*x^3+2*x^2-x-1)*exp((-2*x^2-x)*log(x)/ (1+x))+x^2+2*x+1)/(x^2+2*x+1),x, algorithm="giac")
Output:
x - x^(1/(x + 1))/x^(2*x^2/(x + 1))
Time = 2.74 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-x\,\left (\frac {1}{x^{\frac {2\,x^2+x}{x+1}}}-1\right ) \] Input:
int((2*x + exp(-(log(x)*(x + 2*x^2))/(x + 1))*(log(x)*(x + 4*x^2 + 2*x^3) - x + 2*x^2 + 2*x^3 - 1) + x^2 + 1)/(2*x + x^2 + 1),x)
Output:
-x*(1/x^((x + 2*x^2)/(x + 1)) - 1)
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=\frac {x \left (x^{2 x} e^{\frac {\mathrm {log}\left (x \right )}{x +1}}-x \right )}{x^{2 x} e^{\frac {\mathrm {log}\left (x \right )}{x +1}}} \] Input:
int((((2*x^3+4*x^2+x)*log(x)+2*x^3+2*x^2-x-1)*exp((-2*x^2-x)*log(x)/(1+x)) +x^2+2*x+1)/(x^2+2*x+1),x)
Output:
(x*(x**(2*x)*e**(log(x)/(x + 1)) - x))/(x**(2*x)*e**(log(x)/(x + 1)))