\(\int \frac {e^{-\frac {2 (-24 x+55 x^2)}{-20+55 x}} (160 x-1072 x^2+2090 x^3-1210 x^4)}{80-440 x+605 x^2} \, dx\) [7570]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 52, antiderivative size = 22 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=e^{-2 x-\frac {2 x}{5-\frac {55 x}{4}}} x^2 \]

[Out]

x^2/exp(x+x/(-55/4*x+5))^2

Rubi [F]

\[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=\int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx \]

[In]

Int[(160*x - 1072*x^2 + 2090*x^3 - 1210*x^4)/(E^((2*(-24*x + 55*x^2))/(-20 + 55*x))*(80 - 440*x + 605*x^2)),x]

[Out]

(-32*Defer[Int][E^((-2*x*(-24 + 55*x))/(-20 + 55*x)), x])/605 + 2*Defer[Int][x/E^((2*x*(-24 + 55*x))/(-20 + 55
*x)), x] - 2*Defer[Int][x^2/E^((2*x*(-24 + 55*x))/(-20 + 55*x)), x] - (512*Defer[Int][1/(E^((2*x*(-24 + 55*x))
/(-20 + 55*x))*(-4 + 11*x)^2), x])/605 - (256*Defer[Int][1/(E^((2*x*(-24 + 55*x))/(-20 + 55*x))*(-4 + 11*x)),
x])/605

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{5 (-4+11 x)^2} \, dx \\ & = \frac {1}{5} \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{(-4+11 x)^2} \, dx \\ & = \frac {1}{5} \int \frac {e^{-\frac {2 x (-24+55 x)}{-20+55 x}} x \left (160-1072 x+2090 x^2-1210 x^3\right )}{(4-11 x)^2} \, dx \\ & = \frac {1}{5} \int \left (-\frac {32}{121} e^{-\frac {2 x (-24+55 x)}{-20+55 x}}+10 e^{-\frac {2 x (-24+55 x)}{-20+55 x}} x-10 e^{-\frac {2 x (-24+55 x)}{-20+55 x}} x^2-\frac {512 e^{-\frac {2 x (-24+55 x)}{-20+55 x}}}{121 (-4+11 x)^2}-\frac {256 e^{-\frac {2 x (-24+55 x)}{-20+55 x}}}{121 (-4+11 x)}\right ) \, dx \\ & = -\left (\frac {32}{605} \int e^{-\frac {2 x (-24+55 x)}{-20+55 x}} \, dx\right )-\frac {256}{605} \int \frac {e^{-\frac {2 x (-24+55 x)}{-20+55 x}}}{-4+11 x} \, dx-\frac {512}{605} \int \frac {e^{-\frac {2 x (-24+55 x)}{-20+55 x}}}{(-4+11 x)^2} \, dx+2 \int e^{-\frac {2 x (-24+55 x)}{-20+55 x}} x \, dx-2 \int e^{-\frac {2 x (-24+55 x)}{-20+55 x}} x^2 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=e^{\frac {2 (24-55 x) x}{-20+55 x}} x^2 \]

[In]

Integrate[(160*x - 1072*x^2 + 2090*x^3 - 1210*x^4)/(E^((2*(-24*x + 55*x^2))/(-20 + 55*x))*(80 - 440*x + 605*x^
2)),x]

[Out]

E^((2*(24 - 55*x)*x)/(-20 + 55*x))*x^2

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method result size
risch \(x^{2} {\mathrm e}^{-\frac {2 x \left (55 x -24\right )}{5 \left (11 x -4\right )}}\) \(21\)
gosper \(x^{2} {\mathrm e}^{-\frac {2 x \left (55 x -24\right )}{5 \left (11 x -4\right )}}\) \(23\)
norman \(\frac {\left (11 x^{3}-4 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (55 x^{2}-24 x \right )}{55 x -20}}}{11 x -4}\) \(40\)
parallelrisch \(\frac {\left (805255 x^{3}-292820 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (55 x^{2}-24 x \right )}{5 \left (11 x -4\right )}}}{805255 x -292820}\) \(42\)

[In]

int((-1210*x^4+2090*x^3-1072*x^2+160*x)/(605*x^2-440*x+80)/exp((55*x^2-24*x)/(55*x-20))^2,x,method=_RETURNVERB
OSE)

[Out]

x^2*exp(-2/5*x*(55*x-24)/(11*x-4))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=x^{2} e^{\left (-\frac {2 \, {\left (55 \, x^{2} - 24 \, x\right )}}{5 \, {\left (11 \, x - 4\right )}}\right )} \]

[In]

integrate((-1210*x^4+2090*x^3-1072*x^2+160*x)/(605*x^2-440*x+80)/exp((55*x^2-24*x)/(55*x-20))^2,x, algorithm="
fricas")

[Out]

x^2*e^(-2/5*(55*x^2 - 24*x)/(11*x - 4))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=x^{2} e^{- \frac {2 \cdot \left (55 x^{2} - 24 x\right )}{55 x - 20}} \]

[In]

integrate((-1210*x**4+2090*x**3-1072*x**2+160*x)/(605*x**2-440*x+80)/exp((55*x**2-24*x)/(55*x-20))**2,x)

[Out]

x**2*exp(-2*(55*x**2 - 24*x)/(55*x - 20))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=x^{2} e^{\left (-2 \, x + \frac {32}{55 \, {\left (11 \, x - 4\right )}} + \frac {8}{55}\right )} \]

[In]

integrate((-1210*x^4+2090*x^3-1072*x^2+160*x)/(605*x^2-440*x+80)/exp((55*x^2-24*x)/(55*x-20))^2,x, algorithm="
maxima")

[Out]

x^2*e^(-2*x + 32/55/(11*x - 4) + 8/55)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=x^{2} e^{\left (-\frac {2 \, {\left (55 \, x^{2} - 24 \, x\right )}}{5 \, {\left (11 \, x - 4\right )}}\right )} \]

[In]

integrate((-1210*x^4+2090*x^3-1072*x^2+160*x)/(605*x^2-440*x+80)/exp((55*x^2-24*x)/(55*x-20))^2,x, algorithm="
giac")

[Out]

x^2*e^(-2/5*(55*x^2 - 24*x)/(11*x - 4))

Mupad [B] (verification not implemented)

Time = 13.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {2 \left (-24 x+55 x^2\right )}{-20+55 x}} \left (160 x-1072 x^2+2090 x^3-1210 x^4\right )}{80-440 x+605 x^2} \, dx=x^2\,{\mathrm {e}}^{-\frac {22\,x^2}{11\,x-4}}\,{\mathrm {e}}^{\frac {48\,x}{55\,x-20}} \]

[In]

int((exp((2*(24*x - 55*x^2))/(55*x - 20))*(160*x - 1072*x^2 + 2090*x^3 - 1210*x^4))/(605*x^2 - 440*x + 80),x)

[Out]

x^2*exp(-(22*x^2)/(11*x - 4))*exp((48*x)/(55*x - 20))