3.58.93 \(\int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} (1+e^{\frac {24 e^5}{5 x}} (9 e^x-9 x))-\frac {24 e^5}{5 x}} (8 e^5+e^{\frac {24 e^5}{5 x}} (-15 x^2+15 e^x x^2))}{15 x^2} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x} \]

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Rubi [F]  time = 1.40, 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 (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}\right ) \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((1 + E^((24*E^5)/(5*x))*(9*E^x - 9*x))/(9*E^((24*E^5)/(5*x))) - (24*E^5)/(5*x))*(8*E^5 + E^((24*E^5)/(
5*x))*(-15*x^2 + 15*E^x*x^2)))/(15*x^2),x]

[Out]

Defer[Int][E^(1/(9*E^((24*E^5)/(5*x))) + E^x), x] - Defer[Int][E^(1/(9*E^((24*E^5)/(5*x))) + E^x - x), x] + (8
*Defer[Int][E^(5 + 1/(9*E^((24*E^5)/(5*x))) + E^x - (24*E^5)/(5*x) - x)/x^2, x])/15

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} \int \frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}\right ) \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{x^2} \, dx\\ &=\frac {1}{15} \int \left (15 \exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )+x\right )-\frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}\right ) \left (-8 e^5+15 e^{\frac {24 e^5}{5 x}} x^2\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{15} \int \frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}\right ) \left (-8 e^5+15 e^{\frac {24 e^5}{5 x}} x^2\right )}{x^2} \, dx\right )+\int \exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )+x\right ) \, dx\\ &=-\left (\frac {1}{15} \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-\frac {24 e^5}{5 x}-x} \left (-8 e^5+15 e^{\frac {24 e^5}{5 x}} x^2\right )}{x^2} \, dx\right )+\int e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x} \, dx\\ &=-\left (\frac {1}{15} \int \left (15 e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x}-\frac {8 e^{5+\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-\frac {24 e^5}{5 x}-x}}{x^2}\right ) \, dx\right )+\int e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x} \, dx\\ &=\frac {8}{15} \int \frac {e^{5+\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-\frac {24 e^5}{5 x}-x}}{x^2} \, dx+\int e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x} \, dx-\int e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 1.00 \begin {gather*} e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 + E^((24*E^5)/(5*x))*(9*E^x - 9*x))/(9*E^((24*E^5)/(5*x))) - (24*E^5)/(5*x))*(8*E^5 + E^((24*
E^5)/(5*x))*(-15*x^2 + 15*E^x*x^2)))/(15*x^2),x]

[Out]

E^(1/(9*E^((24*E^5)/(5*x))) + E^x - x)

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fricas [B]  time = 0.73, size = 51, normalized size = 2.04 \begin {gather*} e^{\left (-\frac {{\left (9 \, {\left (5 \, x^{2} - 5 \, x e^{x} + 24 \, e^{5}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 5 \, x\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )}}{45 \, x} + \frac {24 \, e^{5}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)
^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5)/x)^8,x, algorithm="fricas")

[Out]

e^(-1/45*(9*(5*x^2 - 5*x*e^x + 24*e^5)*e^(24/5*e^5/x) - 5*x)*e^(-24/5*e^5/x)/x + 24/5*e^5/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (15 \, {\left (x^{2} e^{x} - x^{2}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} + 8 \, e^{5}\right )} e^{\left (-\frac {1}{9} \, {\left (9 \, {\left (x - e^{x}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 1\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )} - \frac {24 \, e^{5}}{5 \, x}\right )}}{15 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)
^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5)/x)^8,x, algorithm="giac")

[Out]

integrate(1/15*(15*(x^2*e^x - x^2)*e^(24/5*e^5/x) + 8*e^5)*e^(-1/9*(9*(x - e^x)*e^(24/5*e^5/x) - 1)*e^(-24/5*e
^5/x) - 24/5*e^5/x)/x^2, x)

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maple [B]  time = 0.14, size = 43, normalized size = 1.72

method	result	size
risch	\({\mathrm e}^{-\frac {\left (-9 \,{\mathrm e}^{\frac {5 x^{2}+24 \,{\mathrm e}^{5}}{5 x}}+9 \,{\mathrm e}^{\frac {24 \,{\mathrm e}^{5}}{5 x}} x -1\right ) {\mathrm e}^{-\frac {24 \,{\mathrm e}^{5}}{5 x}}}{9}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/
exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5)/x)^8,x,method=_RETURNVERBOSE)

[Out]

exp(-1/9*(-9*exp(1/5*(5*x^2+24*exp(5))/x)+9*exp(24/5*exp(5)/x)*x-1)*exp(-24/5*exp(5)/x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{15} \, \int \frac {{\left (15 \, {\left (x^{2} e^{x} - x^{2}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} + 8 \, e^{5}\right )} e^{\left (-\frac {1}{9} \, {\left (9 \, {\left (x - e^{x}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 1\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )} - \frac {24 \, e^{5}}{5 \, x}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)
^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5)/x)^8,x, algorithm="maxima")

[Out]

1/15*integrate((15*(x^2*e^x - x^2)*e^(24/5*e^5/x) + 8*e^5)*e^(-1/9*(9*(x - e^x)*e^(24/5*e^5/x) - 1)*e^(-24/5*e
^5/x) - 24/5*e^5/x)/x^2, x)

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mupad [B]  time = 4.36, size = 19, normalized size = 0.76 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {24\,{\mathrm {e}}^5}{5\,x}}}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(24*exp(5))/(5*x))*exp(-exp(-(24*exp(5))/(5*x))*((exp((24*exp(5))/(5*x))*(9*x - 9*exp(x)))/9 - 1/9))
*(8*exp(5) + exp((24*exp(5))/(5*x))*(15*x^2*exp(x) - 15*x^2)))/(15*x^2),x)

[Out]

exp(-x)*exp(exp(x))*exp(exp(-(24*exp(5))/(5*x))/9)

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sympy [A]  time = 0.43, size = 34, normalized size = 1.36 \begin {gather*} e^{\left (\frac {\left (- 9 x + 9 e^{x}\right ) e^{\frac {24 e^{5}}{5 x}}}{9} + \frac {1}{9}\right ) e^{- \frac {24 e^{5}}{5 x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((15*exp(x)*x**2-15*x**2)*exp(3/5*exp(5)/x)**8+8*exp(5))*exp(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)
/x)**8+1)/exp(3/5*exp(5)/x)**8)/x**2/exp(3/5*exp(5)/x)**8,x)

[Out]

exp(((-9*x + 9*exp(x))*exp(24*exp(5)/(5*x))/9 + 1/9)*exp(-24*exp(5)/(5*x)))

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