3.43.55 \(\int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} (-50-5 x+35 x^2-11 x^3+x^4)}{25 x^3-10 x^4+x^5} \, dx\)

Optimal. Leaf size=29 \[ e^{\frac {e^{-5+x+2 \left (1+\frac {1-\frac {x}{-5+x}}{x}\right )}}{x}} \]

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Rubi [F]  time = 14.75, 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 {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right ) \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(-50 -
5*x + 35*x^2 - 11*x^3 + x^4))/(25*x^3 - 10*x^4 + x^5),x]

[Out]

(2*Defer[Int][E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2)
)/(-5 + x)^2, x])/5 - (2*Defer[Int][E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x
+ 2*x^2)/(-5*x + x^2))/(-5 + x), x])/25 - 2*Defer[Int][E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))
/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x^3, x] - Defer[Int][E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*
x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x^2, x] + (27*Defer[Int][E^(-5 + E^(-5 + x + (-10 - 10*x
+ 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x, x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right ) \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{x^3 \left (25-10 x+x^2\right )} \, dx\\ &=\int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right ) \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{(-5+x)^2 x^3} \, dx\\ &=\int \left (\frac {2 \exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{5 (-5+x)^2}-\frac {2 \exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{25 (-5+x)}-\frac {2 \exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{x^3}-\frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{x^2}+\frac {27 \exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{25 x}\right ) \, dx\\ &=-\left (\frac {2}{25} \int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{-5+x} \, dx\right )+\frac {2}{5} \int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{(-5+x)^2} \, dx+\frac {27}{25} \int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{x} \, dx-2 \int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{x^3} \, dx-\int \frac {\exp \left (-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 3.71, size = 23, normalized size = 0.79 \begin {gather*} e^{\frac {e^{-3-\frac {2}{-5+x}+\frac {2}{x}+x}}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(
-50 - 5*x + 35*x^2 - 11*x^3 + x^4))/(25*x^3 - 10*x^4 + x^5),x]

[Out]

E^(E^(-3 - 2/(-5 + x) + 2/x + x)/x)

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fricas [B]  time = 0.60, size = 77, normalized size = 2.66 \begin {gather*} e^{\left (\frac {x^{3} - 8 \, x^{2} + {\left (x - 5\right )} e^{\left (\frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} + 15 \, x - 10}{x^{2} - 5 \, x} - \frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(x-5)*exp(exp((2*x^2-10*x-10)/(x^2-5*x)
)*exp(x-5)/x)/(x^5-10*x^4+25*x^3),x, algorithm="fricas")

[Out]

e^((x^3 - 8*x^2 + (x - 5)*e^((x^3 - 8*x^2 + 15*x - 10)/(x^2 - 5*x)) + 15*x - 10)/(x^2 - 5*x) - (x^3 - 8*x^2 +
15*x - 10)/(x^2 - 5*x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(x-5)*exp(exp((2*x^2-10*x-10)/(x^2-5*x)
)*exp(x-5)/x)/(x^5-10*x^4+25*x^3),x, algorithm="giac")

[Out]

integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(
x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)

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maple [A]  time = 0.08, size = 29, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{\frac {x^{3}-8 x^{2}+15 x -10}{\left (x -5\right ) x}}}{x}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(x-5)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(
x-5)/x)/(x^5-10*x^4+25*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(1/x*exp((x^3-8*x^2+15*x-10)/(x-5)/x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(x-5)*exp(exp((2*x^2-10*x-10)/(x^2-5*x)
)*exp(x-5)/x)/(x^5-10*x^4+25*x^3),x, algorithm="maxima")

[Out]

integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(
x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)

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mupad [B]  time = 3.41, size = 40, normalized size = 1.38 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{\frac {2\,x}{x-5}}\,{\mathrm {e}}^{\frac {10}{5\,x-x^2}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {10}{x-5}}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 5)*exp((exp(x - 5)*exp((10*x - 2*x^2 + 10)/(5*x - x^2)))/x)*exp((10*x - 2*x^2 + 10)/(5*x - x^2))
*(5*x - 35*x^2 + 11*x^3 - x^4 + 50))/(25*x^3 - 10*x^4 + x^5),x)

[Out]

exp((exp(-5)*exp((2*x)/(x - 5))*exp(10/(5*x - x^2))*exp(x)*exp(-10/(x - 5)))/x)

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sympy [A]  time = 1.07, size = 26, normalized size = 0.90 \begin {gather*} e^{\frac {e^{\frac {2 x^{2} - 10 x - 10}{x^{2} - 5 x}} e^{x - 5}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-11*x**3+35*x**2-5*x-50)*exp((2*x**2-10*x-10)/(x**2-5*x))*exp(x-5)*exp(exp((2*x**2-10*x-10)/(x*
*2-5*x))*exp(x-5)/x)/(x**5-10*x**4+25*x**3),x)

[Out]

exp(exp((2*x**2 - 10*x - 10)/(x**2 - 5*x))*exp(x - 5)/x)

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