3.72.22 \(\int \frac {e^{\frac {e^{x/2}+10 x}{2 x}} (-24 x+e^{x/2} (-12+6 x))}{x^3} \, dx\)

Optimal. Leaf size=29 \[ \frac {24 e^{5+\frac {e^{x/2}}{2 x}}}{x}+5 \log (\log (4)) \]

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Rubi [B]  time = 0.08, antiderivative size = 68, normalized size of antiderivative = 2.34, number of steps used = 1, number of rules used = 1, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2288} \begin {gather*} -\frac {24 e^{\frac {x}{2}+\frac {10 x+e^{x/2}}{2 x}} (2-x)}{x^3 \left (\frac {e^{x/2}+20}{x}-\frac {2 \left (10 x+e^{x/2}\right )}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((E^(x/2) + 10*x)/(2*x))*(-24*x + E^(x/2)*(-12 + 6*x)))/x^3,x]

[Out]

(-24*E^(x/2 + (E^(x/2) + 10*x)/(2*x))*(2 - x))/(x^3*((20 + E^(x/2))/x - (2*(E^(x/2) + 10*x))/x^2))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {24 e^{\frac {x}{2}+\frac {e^{x/2}+10 x}{2 x}} (2-x)}{x^3 \left (\frac {20+e^{x/2}}{x}-\frac {2 \left (e^{x/2}+10 x\right )}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.79 \begin {gather*} \frac {24 e^{5+\frac {e^{x/2}}{2 x}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^(x/2) + 10*x)/(2*x))*(-24*x + E^(x/2)*(-12 + 6*x)))/x^3,x]

[Out]

(24*E^(5 + E^(x/2)/(2*x)))/x

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fricas [A]  time = 1.10, size = 19, normalized size = 0.66 \begin {gather*} \frac {24 \, e^{\left (\frac {10 \, x + e^{\left (\frac {1}{2} \, x\right )}}{2 \, x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(1/2*x)-24*x)*exp(1/2*(exp(1/2*x)+10*x)/x)/x^3,x, algorithm="fricas")

[Out]

24*e^(1/2*(10*x + e^(1/2*x))/x)/x

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giac [A]  time = 0.17, size = 19, normalized size = 0.66 \begin {gather*} \frac {24 \, e^{\left (\frac {10 \, x + e^{\left (\frac {1}{2} \, x\right )}}{2 \, x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(1/2*x)-24*x)*exp(1/2*(exp(1/2*x)+10*x)/x)/x^3,x, algorithm="giac")

[Out]

24*e^(1/2*(10*x + e^(1/2*x))/x)/x

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maple [A]  time = 0.05, size = 20, normalized size = 0.69

method	result	size
norman	\(\frac {24 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{2}}+10 x}{2 x}}}{x}\)	\(20\)
risch	\(\frac {24 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{2}}+10 x}{2 x}}}{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x-12)*exp(1/2*x)-24*x)*exp(1/2*(exp(1/2*x)+10*x)/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

24/x*exp(1/2*(exp(1/2*x)+10*x)/x)

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maxima [A]  time = 0.47, size = 17, normalized size = 0.59 \begin {gather*} \frac {24 \, e^{\left (\frac {e^{\left (\frac {1}{2} \, x\right )}}{2 \, x} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(1/2*x)-24*x)*exp(1/2*(exp(1/2*x)+10*x)/x)/x^3,x, algorithm="maxima")

[Out]

24*e^(1/2*e^(1/2*x)/x + 5)/x

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mupad [B]  time = 4.33, size = 17, normalized size = 0.59 \begin {gather*} \frac {24\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x/2}}{2\,x}+5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((5*x + exp(x/2)/2)/x)*(24*x - exp(x/2)*(6*x - 12)))/x^3,x)

[Out]

(24*exp(exp(x/2)/(2*x) + 5))/x

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sympy [A]  time = 0.17, size = 15, normalized size = 0.52 \begin {gather*} \frac {24 e^{\frac {5 x + \frac {e^{\frac {x}{2}}}{2}}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(1/2*x)-24*x)*exp(1/2*(exp(1/2*x)+10*x)/x)/x**3,x)

[Out]

24*exp((5*x + exp(x/2)/2)/x)/x

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