3.40.36 \(\int \frac {e^{-x} (e^{5+\frac {4}{-3+x}} (-1440-1120 x+800 x^2-160 x^3)+e^5 (180+60 x-460 x^2+260 x^3-40 x^4))}{9 x^2-6 x^3+x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {20 e^{5-x} \left (-1+8 e^{\frac {4}{-3+x}}+2 x\right )}{x} \]

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Rubi [F]  time = 2.07, 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^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{9 x^2-6 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(5 + 4/(-3 + x))*(-1440 - 1120*x + 800*x^2 - 160*x^3) + E^5*(180 + 60*x - 460*x^2 + 260*x^3 - 40*x^4))/
(E^x*(9*x^2 - 6*x^3 + x^4)),x]

[Out]

40*E^(5 - x) - (20*E^(5 - x))/x - (640*Defer[Int][E^((-11 + 8*x - x^2)/(-3 + x))/(-3 + x)^2, x])/3 + (640*Defe
r[Int][E^((-11 + 8*x - x^2)/(-3 + x))/(-3 + x), x])/9 - 160*Defer[Int][E^((-11 + 8*x - x^2)/(-3 + x))/x^2, x]
- (2080*Defer[Int][E^((-11 + 8*x - x^2)/(-3 + x))/x, x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{x^2 \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (e^{5+\frac {4}{-3+x}} \left (-1440-1120 x+800 x^2-160 x^3\right )+e^5 \left (180+60 x-460 x^2+260 x^3-40 x^4\right )\right )}{(-3+x)^2 x^2} \, dx\\ &=\int \left (-\frac {20 e^{5-x} (-1+x) (1+2 x)}{x^2}+\frac {160 e^{-x+\frac {-11+5 x}{-3+x}} \left (-9-7 x+5 x^2-x^3\right )}{(3-x)^2 x^2}\right ) \, dx\\ &=-\left (20 \int \frac {e^{5-x} (-1+x) (1+2 x)}{x^2} \, dx\right )+160 \int \frac {e^{-x+\frac {-11+5 x}{-3+x}} \left (-9-7 x+5 x^2-x^3\right )}{(3-x)^2 x^2} \, dx\\ &=-\left (20 \int \left (2 e^{5-x}-\frac {e^{5-x}}{x^2}-\frac {e^{5-x}}{x}\right ) \, dx\right )+160 \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}} \left (-9-7 x+5 x^2-x^3\right )}{(3-x)^2 x^2} \, dx\\ &=20 \int \frac {e^{5-x}}{x^2} \, dx+20 \int \frac {e^{5-x}}{x} \, dx-40 \int e^{5-x} \, dx+160 \int \left (-\frac {4 e^{\frac {-11+8 x-x^2}{-3+x}}}{3 (-3+x)^2}+\frac {4 e^{\frac {-11+8 x-x^2}{-3+x}}}{9 (-3+x)}-\frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{x^2}-\frac {13 e^{\frac {-11+8 x-x^2}{-3+x}}}{9 x}\right ) \, dx\\ &=40 e^{5-x}-\frac {20 e^{5-x}}{x}+20 e^5 \text {Ei}(-x)-20 \int \frac {e^{5-x}}{x} \, dx+\frac {640}{9} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{-3+x} \, dx-160 \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{x^2} \, dx-\frac {640}{3} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{(-3+x)^2} \, dx-\frac {2080}{9} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{x} \, dx\\ &=40 e^{5-x}-\frac {20 e^{5-x}}{x}+\frac {640}{9} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{-3+x} \, dx-160 \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{x^2} \, dx-\frac {640}{3} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{(-3+x)^2} \, dx-\frac {2080}{9} \int \frac {e^{\frac {-11+8 x-x^2}{-3+x}}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 28, normalized size = 1.00 \begin {gather*} -\frac {20 e^{5-x} \left (1-8 e^{\frac {4}{-3+x}}-2 x\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + 4/(-3 + x))*(-1440 - 1120*x + 800*x^2 - 160*x^3) + E^5*(180 + 60*x - 460*x^2 + 260*x^3 - 40*
x^4))/(E^x*(9*x^2 - 6*x^3 + x^4)),x]

[Out]

(-20*E^(5 - x)*(1 - 8*E^(4/(-3 + x)) - 2*x))/x

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fricas [A]  time = 1.00, size = 36, normalized size = 1.29 \begin {gather*} \frac {20 \, {\left ({\left (2 \, x - 1\right )} e^{\left (-x + 5\right )} + 8 \, e^{\left (-x + \frac {5 \, x - 11}{x - 3}\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(x-3))+(-40*x^4+260*x^3-460*x^2+60*x+180)*exp(5))/(x^4-
6*x^3+9*x^2)/exp(x),x, algorithm="fricas")

[Out]

20*((2*x - 1)*e^(-x + 5) + 8*e^(-x + (5*x - 11)/(x - 3)))/x

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giac [A]  time = 0.23, size = 44, normalized size = 1.57 \begin {gather*} \frac {20 \, {\left (2 \, x e^{\left (-x + 5\right )} - e^{\left (-x + 5\right )} + 8 \, e^{\left (-\frac {3 \, x^{2} - 13 \, x}{3 \, {\left (x - 3\right )}} + \frac {11}{3}\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(x-3))+(-40*x^4+260*x^3-460*x^2+60*x+180)*exp(5))/(x^4-
6*x^3+9*x^2)/exp(x),x, algorithm="giac")

[Out]

20*(2*x*e^(-x + 5) - e^(-x + 5) + 8*e^(-1/3*(3*x^2 - 13*x)/(x - 3) + 11/3))/x

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maple [A]  time = 0.11, size = 39, normalized size = 1.39

method	result	size
risch	\(\frac {20 \left (2 x -1\right ) {\mathrm e}^{5-x}}{x}+\frac {160 \,{\mathrm e}^{-\frac {x^{2}-8 x +11}{x -3}}}{x}\)	\(39\)
norman	\(\frac {\left (-140 x \,{\mathrm e}^{5}+40 x^{2} {\mathrm e}^{5}-480 \,{\mathrm e}^{5} {\mathrm e}^{\frac {4}{x -3}}+160 x \,{\mathrm e}^{5} {\mathrm e}^{\frac {4}{x -3}}+60 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x \left (x -3\right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(x-3))+(-40*x^4+260*x^3-460*x^2+60*x+180)*exp(5))/(x^4-6*x^3+
9*x^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

20*(2*x-1)/x*exp(5-x)+160/x*exp(-(x^2-8*x+11)/(x-3))

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maxima [A]  time = 0.40, size = 31, normalized size = 1.11 \begin {gather*} \frac {20 \, {\left (2 \, x e^{5} - e^{5} + 8 \, e^{\left (\frac {4}{x - 3} + 5\right )}\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^3+800*x^2-1120*x-1440)*exp(5)*exp(4/(x-3))+(-40*x^4+260*x^3-460*x^2+60*x+180)*exp(5))/(x^4-
6*x^3+9*x^2)/exp(x),x, algorithm="maxima")

[Out]

20*(2*x*e^5 - e^5 + 8*e^(4/(x - 3) + 5))*e^(-x)/x

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mupad [B]  time = 0.47, size = 39, normalized size = 1.39 \begin {gather*} 40\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-\frac {20\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}{x}+\frac {160\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{\frac {4}{x-3}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(5)*(60*x - 460*x^2 + 260*x^3 - 40*x^4 + 180) - exp(5)*exp(4/(x - 3))*(1120*x - 800*x^2 + 160
*x^3 + 1440)))/(9*x^2 - 6*x^3 + x^4),x)

[Out]

40*exp(-x)*exp(5) - (20*exp(-x)*exp(5))/x + (160*exp(-x)*exp(5)*exp(4/(x - 3)))/x

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sympy [A]  time = 0.29, size = 32, normalized size = 1.14 \begin {gather*} \frac {\left (40 x e^{5} - 20 e^{5}\right ) e^{- x}}{x} + \frac {160 e^{5} e^{- x} e^{\frac {4}{x - 3}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x**3+800*x**2-1120*x-1440)*exp(5)*exp(4/(x-3))+(-40*x**4+260*x**3-460*x**2+60*x+180)*exp(5))/
(x**4-6*x**3+9*x**2)/exp(x),x)

[Out]

(40*x*exp(5) - 20*exp(5))*exp(-x)/x + 160*exp(5)*exp(-x)*exp(4/(x - 3))/x

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