Integrand size = 67, antiderivative size = 23 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-5+x-x^{1-x+\frac {x^2}{-1-x}} \]
Time = 0.44 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=x-x^{\frac {1-2 x^2}{1+x}} \]
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]
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 |
\(\displaystyle -2 \int \frac {\int x^{\frac {1-2 x^2}{x+1}}dx}{x}dx+2 \log (x) \int x^{\frac {1-2 x^2}{x+1}}dx-\int \frac {x^{\frac {1-2 x^2}{x+1}} \log (x)}{(x+1)^2}dx-x^{1-\frac {x (2 x+1)}{x+1}} \operatorname {Hypergeometric2F1}\left (1,1-\frac {x (2 x+1)}{x+1},2-\frac {x (2 x+1)}{x+1},-x\right )-\frac {(2 x+1) x^{2-\frac {x (2 x+1)}{x+1}} \operatorname {Hypergeometric2F1}\left (1,1-\frac {x (2 x+1)}{x+1},2-\frac {x (2 x+1)}{x+1},-x\right )}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}+\frac {2 \left (1-2 x^2\right ) \left (2-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {1-2 x^2}{x+1}+1} \operatorname {Hypergeometric2F1}\left (1,\frac {1-2 x^2}{x+1}+1,\frac {1-2 x^2}{x+1}+2,-x\right )}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right ) \left (\frac {1-2 x^2}{x+1}+1\right )}+\frac {2 \left (-2 x^2+x+2\right ) \left (3-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {-2 x^2+x+2}{x+1}+1} \operatorname {Hypergeometric2F1}\left (1,\frac {-2 x^2+x+2}{x+1}+1,\frac {-2 x^2+x+2}{x+1}+2,-x\right )}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right ) \left (\frac {-2 x^2+x+2}{x+1}+1\right )}-\frac {\left (1-\frac {x (2 x+1)}{x+1}\right ) x^{-\frac {x (2 x+1)}{x+1}}}{2 x+1}+\frac {x^{-\frac {x (2 x+1)}{x+1}}}{2 x+1}-\frac {x^{1-\frac {x (2 x+1)}{x+1}}}{x+1}+\frac {2 x^{2-\frac {x (2 x+1)}{x+1}}}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}+\frac {2 x^{3-\frac {x (2 x+1)}{x+1}}}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right )}-\frac {2 \left (2-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {1-2 x^2}{x+1}+1}}{(x+1) \left (1-\frac {x (2 x+1)}{x+1}\right )}-\frac {2 \left (3-\frac {x (2 x+1)}{x+1}\right ) x^{\frac {-2 x^2+x+2}{x+1}+1}}{(x+1) \left (2-\frac {x (2 x+1)}{x+1}\right )}+x\) |
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]
3.8.6.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 2.03 (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 | \(x +\frac {-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-2 x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}-x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}-x \right ) \ln \left (x \right )}{1+x}}}{\left (1+x \right )^{2}}\) | \(77\) |
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)
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=\frac {x x^{\frac {2 \, x^{2} + x}{x + 1}} - x}{x^{\frac {2 \, x^{2} + x}{x + 1}}} \]
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=\
Time = 48.54 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=- x e^{\frac {\left (- 2 x^{2} - x\right ) \log {\left (x \right )}}{x + 1}} + x \]
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)
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-x^{2} e^{\left (-2 \, x \log \left (x\right ) - \frac {\log \left (x\right )}{x + 1}\right )} + x \]
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=\
Time = 0.85 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=x - \frac {x^{\left (\frac {1}{x + 1}\right )}}{x^{\frac {2 \, x^{2}}{x + 1}}} \]
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=\
Time = 8.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1+2 x+x^2+x^{\frac {-x-2 x^2}{1+x}} \left (-1-x+2 x^2+2 x^3+\left (x+4 x^2+2 x^3\right ) \log (x)\right )}{1+2 x+x^2} \, dx=-x\,\left (\frac {1}{x^{\frac {2\,x^2+x}{x+1}}}-1\right ) \]