Integrand size = 101, antiderivative size = 30 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {5+e^{5+\frac {x+2 x^2}{1+x}}+x-\log (x)}{x}} \]
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {5+e^{4+2 x+\frac {1}{1+x}}+x}{x}} x^{-1/x} \]
Integrate[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12 *x - 6*x^2 + E^((5 + 6*x + 2*x^2)/(1 + x))*(-1 - x + 3*x^2 + 2*x^3) + (1 + 2*x + x^2)*Log[x]))/(x^2 + 2*x^3 + x^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.
\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^4+2 x^3+x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 \left (x^2+2 x+1\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}-\frac {\log (x)}{x}+1} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(2 x+1) \left (x^2+x-1\right ) \exp \left (\frac {2 x^2+6 x+5}{x+1}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}-\frac {\log (x)}{x}+1\right )}{x^2 (x+1)^2}+\frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}-\frac {\log (x)}{x}+1} (\log (x)-6)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\exp \left (\frac {2 x^2+6 x+5}{x+1}-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1\right )}{x^2}dx+\int \frac {\exp \left (\frac {2 x^2+6 x+5}{x+1}-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1\right )}{x}dx+\int \frac {\exp \left (\frac {2 x^2+6 x+5}{x+1}-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1\right )}{(x+1)^2}dx+\int \frac {\exp \left (\frac {2 x^2+6 x+5}{x+1}-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1\right )}{x+1}dx-6 \int \frac {e^{-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1}}{x^2}dx+\int \frac {e^{-\frac {\log (x)}{x}+\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}+1} \log (x)}{x^2}dx\) |
Int[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12*x - 6 *x^2 + E^((5 + 6*x + 2*x^2)/(1 + x))*(-1 - x + 3*x^2 + 2*x^3) + (1 + 2*x + x^2)*Log[x]))/(x^2 + 2*x^3 + x^4),x]
3.2.14.3.1 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]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 3.71 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \({\mathrm e}^{-\frac {\ln \left (x \right )-{\mathrm e}^{\frac {2 x^{2}+6 x +5}{1+x}}-5-x}{x}}\) | \(33\) |
risch | \(x^{-\frac {1}{x}} {\mathrm e}^{\frac {{\mathrm e}^{\frac {2 x^{2}+6 x +5}{1+x}}+5+x}{x}}\) | \(34\) |
int(((x^2+2*x+1)*ln(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6*x^2-12 *x-6)*exp((-ln(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2),x,metho d=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {x + e^{\left (\frac {2 \, x^{2} + 6 \, x + 5}{x + 1}\right )} - \log \left (x\right ) + 5}{x}\right )} \]
integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 *x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) ,x, algorithm=\
Time = 0.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {x + e^{\frac {2 x^{2} + 6 x + 5}{x + 1}} - \log {\left (x \right )} + 5}{x}} \]
integrate(((x**2+2*x+1)*ln(x)+(2*x**3+3*x**2-x-1)*exp((2*x**2+6*x+5)/(1+x) )-6*x**2-12*x-6)*exp((-ln(x)+exp((2*x**2+6*x+5)/(1+x))+5+x)/x)/(x**4+2*x** 3+x**2),x)
Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (2 \, x + \frac {1}{x + 1} + 4\right )}}{x} - \frac {\log \left (x\right )}{x} + \frac {5}{x} + 1\right )} \]
integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 *x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) ,x, algorithm=\
Time = 0.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (\frac {2 \, x^{2}}{x + 1} + \frac {6 \, x}{x + 1} + \frac {5}{x + 1}\right )}}{x} - \frac {\log \left (x\right )}{x} + \frac {5}{x} + 1\right )} \]
integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 *x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) ,x, algorithm=\
Time = 13.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=\frac {\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {6\,x}{x+1}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+1}}\,{\mathrm {e}}^{\frac {5}{x+1}}}{x}}}{x^{1/x}} \]