3.31.42 \(\int \frac {e^{.-\frac {3}{2}/x} (e^{.\frac {3}{2}/x} (480 x^3-320 x^4)+e^{e^{.-\frac {3}{2}/x} (-3+e^{.\frac {3}{2}/x} x)} (-81+216 x-144 x^2+e^{.\frac {3}{2}/x} (18 x^2-48 x^3+32 x^4)))}{18 x^2-48 x^3+32 x^4} \, dx\)

Optimal. Leaf size=28 \[ e^{-3 e^{\left .-\frac {3}{2}\right /x}+x}-\frac {10 x^2}{-\frac {3}{4}+x} \]

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Rubi [F]  time = 0.98, 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^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{18 x^2-48 x^3+32 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3/(2*x))*(480*x^3 - 320*x^4) + E^((-3 + E^(3/(2*x))*x)/E^(3/(2*x)))*(-81 + 216*x - 144*x^2 + E^(3/(2*x
))*(18*x^2 - 48*x^3 + 32*x^4)))/(E^(3/(2*x))*(18*x^2 - 48*x^3 + 32*x^4)),x]

[Out]

(40*x^2)/(3 - 4*x) + Defer[Int][E^(-3/E^(3/(2*x)) + x), x] - (9*Defer[Int][E^(-3/E^(3/(2*x)) - 3/(2*x) + x)/x^
2, x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{x^2 \left (18-48 x+32 x^2\right )} \, dx\\ &=\int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{2 x^2 (-3+4 x)^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{\left .-\frac {3}{2}\right /x} \left (e^{\left .\frac {3}{2}\right /x} \left (480 x^3-320 x^4\right )+e^{e^{\left .-\frac {3}{2}\right /x} \left (-3+e^{\left .\frac {3}{2}\right /x} x\right )} \left (-81+216 x-144 x^2+e^{\left .\frac {3}{2}\right /x} \left (18 x^2-48 x^3+32 x^4\right )\right )\right )}{x^2 (-3+4 x)^2} \, dx\\ &=\frac {1}{2} \int \left (2 e^{-3 e^{\left .-\frac {3}{2}\right /x}+x}-\frac {9 e^{-3 e^{\left .-\frac {3}{2}\right /x}-\frac {3}{2 x}+x}}{x^2}+\frac {160 (3-2 x) x}{(3-4 x)^2}\right ) \, dx\\ &=-\left (\frac {9}{2} \int \frac {e^{-3 e^{\left .-\frac {3}{2}\right /x}-\frac {3}{2 x}+x}}{x^2} \, dx\right )+80 \int \frac {(3-2 x) x}{(3-4 x)^2} \, dx+\int e^{-3 e^{\left .-\frac {3}{2}\right /x}+x} \, dx\\ &=\frac {40 x^2}{3-4 x}-\frac {9}{2} \int \frac {e^{-3 e^{\left .-\frac {3}{2}\right /x}-\frac {3}{2 x}+x}}{x^2} \, dx+\int e^{-3 e^{\left .-\frac {3}{2}\right /x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 30, normalized size = 1.07 \begin {gather*} e^{-3 e^{\left .-\frac {3}{2}\right /x}+x}-10 x-\frac {45}{2 (-3+4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3/(2*x))*(480*x^3 - 320*x^4) + E^((-3 + E^(3/(2*x))*x)/E^(3/(2*x)))*(-81 + 216*x - 144*x^2 + E^(
3/(2*x))*(18*x^2 - 48*x^3 + 32*x^4)))/(E^(3/(2*x))*(18*x^2 - 48*x^3 + 32*x^4)),x]

[Out]

E^(-3/E^(3/(2*x)) + x) - 10*x - 45/(2*(-3 + 4*x))

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fricas [A]  time = 0.61, size = 44, normalized size = 1.57 \begin {gather*} -\frac {80 \, x^{2} - 2 \, {\left (4 \, x - 3\right )} e^{\left ({\left (x e^{\left (\frac {3}{2 \, x}\right )} - 3\right )} e^{\left (-\frac {3}{2 \, x}\right )}\right )} - 60 \, x + 45}{2 \, {\left (4 \, x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp(3/2/x)-3)/exp(3/2/x))+(-320*x^4+480
*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^2)/exp(3/2/x),x, algorithm="fricas")

[Out]

-1/2*(80*x^2 - 2*(4*x - 3)*e^((x*e^(3/2/x) - 3)*e^(-3/2/x)) - 60*x + 45)/(4*x - 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (144 \, x^{2} - 2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )} e^{\left (\frac {3}{2 \, x}\right )} - 216 \, x + 81\right )} e^{\left ({\left (x e^{\left (\frac {3}{2 \, x}\right )} - 3\right )} e^{\left (-\frac {3}{2 \, x}\right )}\right )} + 160 \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} e^{\left (\frac {3}{2 \, x}\right )}\right )} e^{\left (-\frac {3}{2 \, x}\right )}}{2 \, {\left (16 \, x^{4} - 24 \, x^{3} + 9 \, x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp(3/2/x)-3)/exp(3/2/x))+(-320*x^4+480
*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^2)/exp(3/2/x),x, algorithm="giac")

[Out]

integrate(-1/2*((144*x^2 - 2*(16*x^4 - 24*x^3 + 9*x^2)*e^(3/2/x) - 216*x + 81)*e^((x*e^(3/2/x) - 3)*e^(-3/2/x)
) + 160*(2*x^4 - 3*x^3)*e^(3/2/x))*e^(-3/2/x)/(16*x^4 - 24*x^3 + 9*x^2), x)

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maple [A]  time = 0.08, size = 30, normalized size = 1.07

method	result	size
risch	\(-10 x -\frac {45}{8 \left (x -\frac {3}{4}\right )}+{\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}}\)	\(30\)
norman	\(\frac {\left (-40 \,{\mathrm e}^{\frac {3}{2 x}} x^{3}+4 x^{2} {\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} {\mathrm e}^{\frac {3}{2 x}}-3 \,{\mathrm e}^{\frac {3}{2 x}} {\mathrm e}^{\left (x \,{\mathrm e}^{\frac {3}{2 x}}-3\right ) {\mathrm e}^{-\frac {3}{2 x}}} x \right ) {\mathrm e}^{-\frac {3}{2 x}}}{x \left (4 x -3\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp(3/2/x)-3)/exp(3/2/x))+(-320*x^4+480*x^3)*
exp(3/2/x))/(32*x^4-48*x^3+18*x^2)/exp(3/2/x),x,method=_RETURNVERBOSE)

[Out]

-10*x-45/8/(x-3/4)+exp((x*exp(3/2/x)-3)*exp(-3/2/x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {5 \, {\left (16 \, x^{2} - 12 \, x - 9\right )}}{2 \, {\left (4 \, x - 3\right )}} - \frac {45}{4 \, x - 3} + \frac {1}{2} \, \int \frac {{\left (2 \, x^{2} e^{\left (x + \frac {3}{2 \, x}\right )} - 9 \, e^{x}\right )} e^{\left (-\frac {3}{2 \, x} - 3 \, e^{\left (-\frac {3}{2 \, x}\right )}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((32*x^4-48*x^3+18*x^2)*exp(3/2/x)-144*x^2+216*x-81)*exp((x*exp(3/2/x)-3)/exp(3/2/x))+(-320*x^4+480
*x^3)*exp(3/2/x))/(32*x^4-48*x^3+18*x^2)/exp(3/2/x),x, algorithm="maxima")

[Out]

-5/2*(16*x^2 - 12*x - 9)/(4*x - 3) - 45/(4*x - 3) + 1/2*integrate((2*x^2*e^(x + 3/2/x) - 9*e^x)*e^(-3/2/x - 3*
e^(-3/2/x))/x^2, x)

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mupad [B]  time = 1.93, size = 25, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{-\frac {3}{{\left ({\mathrm {e}}^{1/x}\right )}^{3/2}}}\,{\mathrm {e}}^x-\frac {45}{8\,\left (x-\frac {3}{4}\right )}-10\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3/(2*x))*(exp(exp(-3/(2*x))*(x*exp(3/(2*x)) - 3))*(216*x - 144*x^2 + exp(3/(2*x))*(18*x^2 - 48*x^3 +
32*x^4) - 81) + exp(3/(2*x))*(480*x^3 - 320*x^4)))/(18*x^2 - 48*x^3 + 32*x^4),x)

[Out]

exp(-3/exp(1/x)^(3/2))*exp(x) - 45/(8*(x - 3/4)) - 10*x

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sympy [A]  time = 0.46, size = 27, normalized size = 0.96 \begin {gather*} - 10 x + e^{\left (x e^{\frac {3}{2 x}} - 3\right ) e^{- \frac {3}{2 x}}} - \frac {45}{8 x - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((32*x**4-48*x**3+18*x**2)*exp(3/2/x)-144*x**2+216*x-81)*exp((x*exp(3/2/x)-3)/exp(3/2/x))+(-320*x**
4+480*x**3)*exp(3/2/x))/(32*x**4-48*x**3+18*x**2)/exp(3/2/x),x)

[Out]

-10*x + exp((x*exp(3/(2*x)) - 3)*exp(-3/(2*x))) - 45/(8*x - 6)

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