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

Optimal. Leaf size=22 \[ e^{-2 x-\frac {2 x}{5-\frac {55 x}{4}}} x^2 \]

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Rubi [F]  time = 0.73, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 21, normalized size = 0.95 \begin {gather*} e^{\frac {2 (24-55 x) x}{-20+55 x}} x^2 \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.54, size = 23, normalized size = 1.05 \begin {gather*} x^{2} e^{\left (-\frac {2 \, {\left (55 \, x^{2} - 24 \, x\right )}}{5 \, {\left (11 \, x - 4\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.22, size = 23, normalized size = 1.05 \begin {gather*} x^{2} e^{\left (-\frac {2 \, {\left (55 \, x^{2} - 24 \, x\right )}}{5 \, {\left (11 \, x - 4\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.24, size = 21, normalized size = 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	\(\frac {\left (11 x^{3}-4 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (55 x^{2}-24 x \right )}{55 x -20}}}{11 x -4}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.44, size = 19, normalized size = 0.86 \begin {gather*} x^{2} e^{\left (-2 \, x + \frac {32}{55 \, {\left (11 \, x - 4\right )}} + \frac {8}{55}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.68, size = 28, normalized size = 1.27 \begin {gather*} x^2\,{\mathrm {e}}^{-\frac {22\,x^2}{11\,x-4}}\,{\mathrm {e}}^{\frac {48\,x}{55\,x-20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.20, size = 17, normalized size = 0.77 \begin {gather*} x^{2} e^{- \frac {2 \left (55 x^{2} - 24 x\right )}{55 x - 20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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