3.74.26 \(\int \frac {e^{\frac {9 e^{x/2}}{x}+\frac {x}{2}} (-18+9 x)}{4 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{2} \left (e^{\frac {9 e^{x/2}}{x}}+\log (2)\right ) \]

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Rubi [F]  time = 0.52, 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 {9 e^{x/2}}{x}+\frac {x}{2}} (-18+9 x)}{4 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((9*E^(x/2))/x + x/2)*(-18 + 9*x))/(4*x^2),x]

[Out]

(-9*Defer[Int][E^((18*E^(x/2) + x^2)/(2*x))/x^2, x])/2 + (9*Defer[Int][E^((18*E^(x/2) + x^2)/(2*x))/x, x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{\frac {9 e^{x/2}}{x}+\frac {x}{2}} (-18+9 x)}{x^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{\frac {18 e^{x/2}+x^2}{2 x}} (-18+9 x)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {18 e^{\frac {18 e^{x/2}+x^2}{2 x}}}{x^2}+\frac {9 e^{\frac {18 e^{x/2}+x^2}{2 x}}}{x}\right ) \, dx\\ &=\frac {9}{4} \int \frac {e^{\frac {18 e^{x/2}+x^2}{2 x}}}{x} \, dx-\frac {9}{2} \int \frac {e^{\frac {18 e^{x/2}+x^2}{2 x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 18, normalized size = 0.86 \begin {gather*} \frac {1}{2} e^{\frac {9 e^{x/2}}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((9*E^(x/2))/x + x/2)*(-18 + 9*x))/(4*x^2),x]

[Out]

E^((9*E^(x/2))/x)/2

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fricas [A]  time = 0.65, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{2} \, e^{\left (-\frac {1}{2} \, x + \frac {x^{2} + 18 \, e^{\left (\frac {1}{2} \, x\right )}}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(9*x-18)*exp(1/2*x)*exp(9*exp(1/2*x)/x)/x^2,x, algorithm="fricas")

[Out]

1/2*e^(-1/2*x + 1/2*(x^2 + 18*e^(1/2*x))/x)

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giac [A]  time = 0.13, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{2} \, e^{\left (-\frac {1}{2} \, x + \frac {x^{2} + 18 \, e^{\left (\frac {1}{2} \, x\right )}}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(9*x-18)*exp(1/2*x)*exp(9*exp(1/2*x)/x)/x^2,x, algorithm="giac")

[Out]

1/2*e^(-1/2*x + 1/2*(x^2 + 18*e^(1/2*x))/x)

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maple [A]  time = 0.03, size = 13, normalized size = 0.62

method	result	size
norman	\(\frac {{\mathrm e}^{\frac {9 \,{\mathrm e}^{\frac {x}{2}}}{x}}}{2}\)	\(13\)
risch	\(\frac {{\mathrm e}^{\frac {9 \,{\mathrm e}^{\frac {x}{2}}}{x}}}{2}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(9*x-18)*exp(1/2*x)*exp(9*exp(1/2*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/2*exp(9*exp(1/2*x)/x)

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maxima [A]  time = 0.43, size = 12, normalized size = 0.57 \begin {gather*} \frac {1}{2} \, e^{\left (\frac {9 \, e^{\left (\frac {1}{2} \, x\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(9*x-18)*exp(1/2*x)*exp(9*exp(1/2*x)/x)/x^2,x, algorithm="maxima")

[Out]

1/2*e^(9*e^(1/2*x)/x)

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mupad [B]  time = 4.61, size = 12, normalized size = 0.57 \begin {gather*} \frac {{\mathrm {e}}^{\frac {9\,{\mathrm {e}}^{x/2}}{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/2)*exp((9*exp(x/2))/x)*(9*x - 18))/(4*x^2),x)

[Out]

exp((9*exp(x/2))/x)/2

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sympy [A]  time = 0.18, size = 10, normalized size = 0.48 \begin {gather*} \frac {e^{\frac {9 e^{\frac {x}{2}}}{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(9*x-18)*exp(1/2*x)*exp(9*exp(1/2*x)/x)/x**2,x)

[Out]

exp(9*exp(x/2)/x)/2

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