Integrand size = 75, antiderivative size = 29 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx=x \left (3+x+x^2 \left (1+\frac {10 x}{x-\frac {x}{e}-x^2}\right )\right ) \]
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx=-\frac {10 (-1+e)^3}{e^2 (1+e (-1+x))}+\frac {(10-7 e) x}{e}-9 x^2+x^3 \]
Integrate[(3 + 2*x + 3*x^2 + E*(-6 + 2*x - 32*x^2 + 6*x^3) + E^2*(3 - 4*x + 32*x^2 - 24*x^3 + 3*x^4))/(1 + E*(-2 + 2*x) + E^2*(1 - 2*x + x^2)),x]
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2452, 2188, 2009}
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.
\(\displaystyle \int \frac {3 x^2+e \left (6 x^3-32 x^2+2 x-6\right )+e^2 \left (3 x^4-24 x^3+32 x^2-4 x+3\right )+2 x+3}{e^2 \left (x^2-2 x+1\right )+e (2 x-2)+1} \, dx\) |
\(\Big \downarrow \) 2452 |
\(\displaystyle \int \frac {3 x^2+e \left (6 x^3-32 x^2+2 x-6\right )+e^2 \left (3 x^4-24 x^3+32 x^2-4 x+3\right )+2 x+3}{e^2 x^2+2 (1-e) e x+(e-1)^2}dx\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle \int \left (3 x^2+\frac {10 \left (-1+3 e-3 e^2+e^3\right )}{e \left (e^2 x^2+2 (1-e) e x+(e-1)^2\right )}-18 x+\frac {10}{e}-7\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^3-9 x^2-\left (7-\frac {10}{e}\right ) x+\frac {10 (1-e)^3}{e^2 (e x-e+1)}\) |
Int[(3 + 2*x + 3*x^2 + E*(-6 + 2*x - 32*x^2 + 6*x^3) + E^2*(3 - 4*x + 32*x ^2 - 24*x^3 + 3*x^4))/(1 + E*(-2 + 2*x) + E^2*(1 - 2*x + x^2)),x]
3.28.49.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[ p, x] && PolyQ[Pq, x] && QuadraticQ[u, x] && !QuadraticMatchQ[u, x]
Time = 0.60 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97
method | result | size |
norman | \(\frac {x^{4} {\mathrm e}+\left (-10 \,{\mathrm e}+1\right ) x^{3}+\left (2 \,{\mathrm e}+1\right ) x^{2}-3 \left ({\mathrm e}^{2}-2 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}}{x \,{\mathrm e}-{\mathrm e}+1}\) | \(57\) |
gosper | \(\frac {\left (x^{4} {\mathrm e}^{2}-10 x^{3} {\mathrm e}^{2}+2 x^{2} {\mathrm e}^{2}+x^{3} {\mathrm e}+x^{2} {\mathrm e}-3 \,{\mathrm e}^{2}+6 \,{\mathrm e}-3\right ) {\mathrm e}^{-1}}{x \,{\mathrm e}-{\mathrm e}+1}\) | \(68\) |
parallelrisch | \(\frac {\left (x^{4} {\mathrm e}^{2}-10 x^{3} {\mathrm e}^{2}+2 x^{2} {\mathrm e}^{2}+x^{3} {\mathrm e}+x^{2} {\mathrm e}-3 \,{\mathrm e}^{2}+6 \,{\mathrm e}-3\right ) {\mathrm e}^{-1}}{x \,{\mathrm e}-{\mathrm e}+1}\) | \(68\) |
risch | \({\mathrm e} \,{\mathrm e}^{-1} x^{3}-9 \,{\mathrm e} \,{\mathrm e}^{-1} x^{2}-7 \,{\mathrm e} \,{\mathrm e}^{-1} x +10 \,{\mathrm e}^{-1} x -\frac {10 \,{\mathrm e}^{-2} {\mathrm e}^{3}}{x \,{\mathrm e}-{\mathrm e}+1}+\frac {30 \,{\mathrm e}^{-2} {\mathrm e}^{2}}{x \,{\mathrm e}-{\mathrm e}+1}-\frac {30 \,{\mathrm e}^{-2} {\mathrm e}}{x \,{\mathrm e}-{\mathrm e}+1}+\frac {10 \,{\mathrm e}^{-2}}{x \,{\mathrm e}-{\mathrm e}+1}\) | \(101\) |
meijerg | \(-3 \,{\mathrm e}^{-3} \left ({\mathrm e}-1\right )^{3} \left (-\frac {x \,{\mathrm e} \left (-\frac {5 x^{3} {\mathrm e}^{3}}{\left ({\mathrm e}-1\right )^{3}}-\frac {10 x^{2} {\mathrm e}^{2}}{\left ({\mathrm e}-1\right )^{2}}-\frac {30 x \,{\mathrm e}}{{\mathrm e}-1}+60\right )}{15 \left ({\mathrm e}-1\right ) \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}-4 \ln \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )\right )+{\mathrm e}^{-4} \left ({\mathrm e}-1\right )^{2} \left (-24 \,{\mathrm e}^{2}+6 \,{\mathrm e}\right ) \left (\frac {x \,{\mathrm e} \left (-\frac {2 x^{2} {\mathrm e}^{2}}{\left ({\mathrm e}-1\right )^{2}}-\frac {6 x \,{\mathrm e}}{{\mathrm e}-1}+12\right )}{4 \left ({\mathrm e}-1\right ) \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}+3 \ln \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )\right )-{\mathrm e}^{-3} \left (32 \,{\mathrm e}^{2}-32 \,{\mathrm e}+3\right ) \left ({\mathrm e}-1\right ) \left (-\frac {x \,{\mathrm e} \left (-\frac {3 x \,{\mathrm e}}{{\mathrm e}-1}+6\right )}{3 \left ({\mathrm e}-1\right ) \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}-2 \ln \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )\right )+{\mathrm e}^{-2} \left (-4 \,{\mathrm e}^{2}+2 \,{\mathrm e}+2\right ) \left (\frac {x \,{\mathrm e}}{\left ({\mathrm e}-1\right ) \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}+\ln \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )\right )+\frac {3 \,{\mathrm e}^{2} x}{\left ({\mathrm e}-1\right )^{2} \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}-\frac {6 \,{\mathrm e} x}{\left ({\mathrm e}-1\right )^{2} \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}+\frac {3 x}{\left ({\mathrm e}-1\right )^{2} \left (1-\frac {x \,{\mathrm e}}{{\mathrm e}-1}\right )}\) | \(384\) |
int(((3*x^4-24*x^3+32*x^2-4*x+3)*exp(1)^2+(6*x^3-32*x^2+2*x-6)*exp(1)+3*x^ 2+2*x+3)/((x^2-2*x+1)*exp(1)^2+(-2+2*x)*exp(1)+1),x,method=_RETURNVERBOSE)
(x^4*exp(1)+(-10*exp(1)+1)*x^3+(2*exp(1)+1)*x^2-3*(exp(1)^2-2*exp(1)+1)/ex p(1))/(x*exp(1)-exp(1)+1)
Time = 0.34 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx=\frac {{\left (x^{4} - 10 \, x^{3} + 2 \, x^{2} + 7 \, x - 10\right )} e^{3} + {\left (x^{3} + x^{2} - 17 \, x + 30\right )} e^{2} + 10 \, {\left (x - 3\right )} e + 10}{{\left (x - 1\right )} e^{3} + e^{2}} \]
integrate(((3*x^4-24*x^3+32*x^2-4*x+3)*exp(1)^2+(6*x^3-32*x^2+2*x-6)*exp(1 )+3*x^2+2*x+3)/((x^2-2*x+1)*exp(1)^2+(-2+2*x)*exp(1)+1),x, algorithm=\
((x^4 - 10*x^3 + 2*x^2 + 7*x - 10)*e^3 + (x^3 + x^2 - 17*x + 30)*e^2 + 10* (x - 3)*e + 10)/((x - 1)*e^3 + e^2)
Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx=x^{3} - 9 x^{2} + x \left (-7 + \frac {10}{e}\right ) + \frac {- 10 e^{3} - 30 e + 10 + 30 e^{2}}{x e^{3} - e^{3} + e^{2}} \]
integrate(((3*x**4-24*x**3+32*x**2-4*x+3)*exp(1)**2+(6*x**3-32*x**2+2*x-6) *exp(1)+3*x**2+2*x+3)/((x**2-2*x+1)*exp(1)**2+(-2+2*x)*exp(1)+1),x)
x**3 - 9*x**2 + x*(-7 + 10*exp(-1)) + (-10*exp(3) - 30*E + 10 + 30*exp(2)) /(x*exp(3) - exp(3) + exp(2))
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx={\left (x^{3} e - 9 \, x^{2} e - x {\left (7 \, e - 10\right )}\right )} e^{\left (-1\right )} - \frac {10 \, {\left (e^{3} - 3 \, e^{2} + 3 \, e - 1\right )}}{x e^{3} - e^{3} + e^{2}} \]
integrate(((3*x^4-24*x^3+32*x^2-4*x+3)*exp(1)^2+(6*x^3-32*x^2+2*x-6)*exp(1 )+3*x^2+2*x+3)/((x^2-2*x+1)*exp(1)^2+(-2+2*x)*exp(1)+1),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx={\left (x^{3} e^{6} - 9 \, x^{2} e^{6} - 7 \, x e^{6} + 10 \, x e^{5}\right )} e^{\left (-6\right )} - \frac {10 \, {\left (e^{3} - 3 \, e^{2} + 3 \, e - 1\right )} e^{\left (-2\right )}}{x e - e + 1} \]
integrate(((3*x^4-24*x^3+32*x^2-4*x+3)*exp(1)^2+(6*x^3-32*x^2+2*x-6)*exp(1 )+3*x^2+2*x+3)/((x^2-2*x+1)*exp(1)^2+(-2+2*x)*exp(1)+1),x, algorithm=\
(x^3*e^6 - 9*x^2*e^6 - 7*x*e^6 + 10*x*e^5)*e^(-6) - 10*(e^3 - 3*e^2 + 3*e - 1)*e^(-2)/(x*e - e + 1)
Time = 13.73 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.76 \[ \int \frac {3+2 x+3 x^2+e \left (-6+2 x-32 x^2+6 x^3\right )+e^2 \left (3-4 x+32 x^2-24 x^3+3 x^4\right )}{1+e (-2+2 x)+e^2 \left (1-2 x+x^2\right )} \, dx=x^2\,\left (3\,{\mathrm {e}}^{-1}\,\left (\mathrm {e}-1\right )-3\,{\mathrm {e}}^{-1}\,\left (4\,\mathrm {e}-1\right )\right )+x\,\left ({\mathrm {e}}^{-2}\,\left (32\,{\mathrm {e}}^2-32\,\mathrm {e}+3\right )-3\,{\mathrm {e}}^{-2}\,{\left (\mathrm {e}-1\right )}^2+2\,{\mathrm {e}}^{-1}\,\left (\mathrm {e}-1\right )\,\left (6\,{\mathrm {e}}^{-1}\,\left (\mathrm {e}-1\right )-6\,{\mathrm {e}}^{-1}\,\left (4\,\mathrm {e}-1\right )\right )\right )+x^3-\frac {10\,{\mathrm {e}}^{-1}\,\left (3\,\mathrm {e}-3\,{\mathrm {e}}^2+{\mathrm {e}}^3-1\right )}{\mathrm {e}-{\mathrm {e}}^2+x\,{\mathrm {e}}^2} \]
int((2*x + exp(1)*(2*x - 32*x^2 + 6*x^3 - 6) + exp(2)*(32*x^2 - 4*x - 24*x ^3 + 3*x^4 + 3) + 3*x^2 + 3)/(exp(2)*(x^2 - 2*x + 1) + exp(1)*(2*x - 2) + 1),x)