3.66.36 \(\int \frac {e^{e^{\frac {2 (3 e^x+30 x+x^2)}{x^2}}+\frac {2 (3 e^x+30 x+x^2)}{x^2}} (-60 x+e^x (-12+6 x))}{x^3} \, dx\)

Optimal. Leaf size=21 \[ e^{e^{2-\frac {6 \left (-10-\frac {e^x}{x}\right )}{x}}} \]

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Rubi [F]  time = 1.72, 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 (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right ) \left (-60 x+e^x (-12+6 x)\right )}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^((2*(3*E^x + 30*x + x^2))/x^2) + (2*(3*E^x + 30*x + x^2))/x^2)*(-60*x + E^x*(-12 + 6*x)))/x^3,x]

[Out]

-12*Defer[Int][E^(E^((2*(3*E^x + 30*x + x^2))/x^2) + x + (2*(3*E^x + 30*x + x^2))/x^2)/x^3, x] - 60*Defer[Int]
[E^(E^((2*(3*E^x + 30*x + x^2))/x^2) + (2*(3*E^x + 30*x + x^2))/x^2)/x^2, x] + 6*Defer[Int][E^(E^((2*(3*E^x +
30*x + x^2))/x^2) + x + (2*(3*E^x + 30*x + x^2))/x^2)/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {6 \exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right ) (-2+x)}{x^3}-\frac {60 \exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2}\right ) \, dx\\ &=6 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right ) (-2+x)}{x^3} \, dx-60 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2} \, dx\\ &=6 \int \left (-\frac {2 \exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^3}+\frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2}\right ) \, dx-60 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2} \, dx\\ &=6 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2} \, dx-12 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+x+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^3} \, dx-60 \int \frac {\exp \left (e^{\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}}+\frac {2 \left (3 e^x+30 x+x^2\right )}{x^2}\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.58, size = 19, normalized size = 0.90 \begin {gather*} e^{e^{2+\frac {6 e^x}{x^2}+\frac {60}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^((2*(3*E^x + 30*x + x^2))/x^2) + (2*(3*E^x + 30*x + x^2))/x^2)*(-60*x + E^x*(-12 + 6*x)))/x^3,
x]

[Out]

E^E^(2 + (6*E^x)/x^2 + 60/x)

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fricas [B]  time = 0.96, size = 56, normalized size = 2.67 \begin {gather*} e^{\left (\frac {x^{2} e^{\left (\frac {2 \, {\left (x^{2} + 30 \, x + 3 \, e^{x}\right )}}{x^{2}}\right )} + 2 \, x^{2} + 60 \, x + 6 \, e^{x}}{x^{2}} - \frac {2 \, {\left (x^{2} + 30 \, x + 3 \, e^{x}\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(x)-60*x)*exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4*exp(exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4)/x^
3,x, algorithm="fricas")

[Out]

e^((x^2*e^(2*(x^2 + 30*x + 3*e^x)/x^2) + 2*x^2 + 60*x + 6*e^x)/x^2 - 2*(x^2 + 30*x + 3*e^x)/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {6 \, {\left ({\left (x - 2\right )} e^{x} - 10 \, x\right )} e^{\left (\frac {2 \, {\left (x^{2} + 30 \, x + 3 \, e^{x}\right )}}{x^{2}} + e^{\left (\frac {2 \, {\left (x^{2} + 30 \, x + 3 \, e^{x}\right )}}{x^{2}}\right )}\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(x)-60*x)*exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4*exp(exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4)/x^
3,x, algorithm="giac")

[Out]

integrate(6*((x - 2)*e^x - 10*x)*e^(2*(x^2 + 30*x + 3*e^x)/x^2 + e^(2*(x^2 + 30*x + 3*e^x)/x^2))/x^3, x)

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maple [A]  time = 0.39, size = 19, normalized size = 0.90

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{\frac {6 \,{\mathrm e}^{x}+2 x^{2}+60 x}{x^{2}}}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x-12)*exp(x)-60*x)*exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4*exp(exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4)/x^3,x,me
thod=_RETURNVERBOSE)

[Out]

exp(exp(2*(3*exp(x)+x^2+30*x)/x^2))

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maxima [A]  time = 0.59, size = 16, normalized size = 0.76 \begin {gather*} e^{\left (e^{\left (\frac {60}{x} + \frac {6 \, e^{x}}{x^{2}} + 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(x)-60*x)*exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4*exp(exp(1/2*(3*exp(x)+x^2+30*x)/x^2)^4)/x^
3,x, algorithm="maxima")

[Out]

e^(e^(60/x + 6*e^x/x^2 + 2))

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mupad [B]  time = 4.35, size = 18, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {6\,{\mathrm {e}}^x}{x^2}}\,{\mathrm {e}}^{60/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((4*(15*x + (3*exp(x))/2 + x^2/2))/x^2)*exp(exp((4*(15*x + (3*exp(x))/2 + x^2/2))/x^2))*(60*x - exp(x
)*(6*x - 12)))/x^3,x)

[Out]

exp(exp(2)*exp((6*exp(x))/x^2)*exp(60/x))

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sympy [A]  time = 0.65, size = 19, normalized size = 0.90 \begin {gather*} e^{e^{\frac {4 \left (\frac {x^{2}}{2} + 15 x + \frac {3 e^{x}}{2}\right )}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-12)*exp(x)-60*x)*exp(1/2*(3*exp(x)+x**2+30*x)/x**2)**4*exp(exp(1/2*(3*exp(x)+x**2+30*x)/x**2)*
*4)/x**3,x)

[Out]

exp(exp(4*(x**2/2 + 15*x + 3*exp(x)/2)/x**2))

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