3.39.28 \(\int \frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}} (43200-3600 e^4-7200 x)}{-12 x^3+e^4 x^3} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {1800 e^{\frac {x}{3-\frac {e^4}{4}}}}{x^2}} \]

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Rubi [F]  time = 1.20, 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 {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}} \left (43200-3600 e^4-7200 x\right )}{-12 x^3+e^4 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(1800/(E^((4*x)/(-12 + E^4))*x^2) - (4*x)/(-12 + E^4))*(43200 - 3600*E^4 - 7200*x))/(-12*x^3 + E^4*x^3)
,x]

[Out]

-3600*Defer[Int][E^(1800/(E^((4*x)/(-12 + E^4))*x^2) - (4*x)/(-12 + E^4))/x^3, x] + (7200*Defer[Int][E^(1800/(
E^((4*x)/(-12 + E^4))*x^2) - (4*x)/(-12 + E^4))/x^2, x])/(12 - E^4)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}} \left (43200-3600 e^4-7200 x\right )}{\left (-12+e^4\right ) x^3} \, dx\\ &=\frac {\int \frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}} \left (43200-3600 e^4-7200 x\right )}{x^3} \, dx}{-12+e^4}\\ &=\frac {\int \left (-\frac {3600 e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}} \left (-12+e^4\right )}{x^3}-\frac {7200 e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}}}{x^2}\right ) \, dx}{-12+e^4}\\ &=-\left (3600 \int \frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}}}{x^3} \, dx\right )+\frac {7200 \int \frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}-\frac {4 x}{-12+e^4}}}{x^2} \, dx}{12-e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 35, normalized size = 1.59 \begin {gather*} -\frac {e^{\frac {1800 e^{-\frac {4 x}{-12+e^4}}}{x^2}} \left (12-e^4\right )}{-12+e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1800/(E^((4*x)/(-12 + E^4))*x^2) - (4*x)/(-12 + E^4))*(43200 - 3600*E^4 - 7200*x))/(-12*x^3 + E^
4*x^3),x]

[Out]

-((E^(1800/(E^((4*x)/(-12 + E^4))*x^2))*(12 - E^4))/(-12 + E^4))

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fricas [B]  time = 0.60, size = 47, normalized size = 2.14 \begin {gather*} e^{\left (-\frac {4 \, {\left (x^{3} - 450 \, {\left (e^{4} - 12\right )} e^{\left (-\frac {4 \, x}{e^{4} - 12}\right )}\right )}}{x^{2} e^{4} - 12 \, x^{2}} + \frac {4 \, x}{e^{4} - 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3600*exp(2)^2-7200*x+43200)*exp(-2*x/(exp(2)^2-12))^2*exp(900*exp(-2*x/(exp(2)^2-12))^2/x^2)^2/(x^
3*exp(2)^2-12*x^3),x, algorithm="fricas")

[Out]

e^(-4*(x^3 - 450*(e^4 - 12)*e^(-4*x/(e^4 - 12)))/(x^2*e^4 - 12*x^2) + 4*x/(e^4 - 12))

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giac [B]  time = 0.45, size = 57, normalized size = 2.59 \begin {gather*} e^{\left (-\frac {4 \, {\left (x^{3} - 450 \, e^{\left (-\frac {4 \, x}{e^{4} - 12} + 4\right )} + 5400 \, e^{\left (-\frac {4 \, x}{e^{4} - 12}\right )}\right )}}{x^{2} e^{4} - 12 \, x^{2}} + \frac {4 \, x}{e^{4} - 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3600*exp(2)^2-7200*x+43200)*exp(-2*x/(exp(2)^2-12))^2*exp(900*exp(-2*x/(exp(2)^2-12))^2/x^2)^2/(x^
3*exp(2)^2-12*x^3),x, algorithm="giac")

[Out]

e^(-4*(x^3 - 450*e^(-4*x/(e^4 - 12) + 4) + 5400*e^(-4*x/(e^4 - 12)))/(x^2*e^4 - 12*x^2) + 4*x/(e^4 - 12))

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maple [A]  time = 0.21, size = 17, normalized size = 0.77

method	result	size
risch	\({\mathrm e}^{\frac {1800 \,{\mathrm e}^{-\frac {4 x}{{\mathrm e}^{4}-12}}}{x^{2}}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3600*exp(2)^2-7200*x+43200)*exp(-2*x/(exp(2)^2-12))^2*exp(900*exp(-2*x/(exp(2)^2-12))^2/x^2)^2/(x^3*exp(
2)^2-12*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(1800*exp(-4*x/(exp(4)-12))/x^2)

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maxima [A]  time = 0.57, size = 16, normalized size = 0.73 \begin {gather*} e^{\left (\frac {1800 \, e^{\left (-\frac {4 \, x}{e^{4} - 12}\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3600*exp(2)^2-7200*x+43200)*exp(-2*x/(exp(2)^2-12))^2*exp(900*exp(-2*x/(exp(2)^2-12))^2/x^2)^2/(x^
3*exp(2)^2-12*x^3),x, algorithm="maxima")

[Out]

e^(1800*e^(-4*x/(e^4 - 12))/x^2)

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mupad [B]  time = 2.56, size = 16, normalized size = 0.73 \begin {gather*} {\mathrm {e}}^{\frac {1800\,{\mathrm {e}}^{-\frac {4\,x}{{\mathrm {e}}^4-12}}}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(4*x)/(exp(4) - 12))*exp((1800*exp(-(4*x)/(exp(4) - 12)))/x^2)*(7200*x + 3600*exp(4) - 43200))/(x^3
*exp(4) - 12*x^3),x)

[Out]

exp((1800*exp(-(4*x)/(exp(4) - 12)))/x^2)

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sympy [A]  time = 0.30, size = 15, normalized size = 0.68 \begin {gather*} e^{\frac {1800 e^{- \frac {4 x}{-12 + e^{4}}}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3600*exp(2)**2-7200*x+43200)*exp(-2*x/(exp(2)**2-12))**2*exp(900*exp(-2*x/(exp(2)**2-12))**2/x**2)
**2/(x**3*exp(2)**2-12*x**3),x)

[Out]

exp(1800*exp(-4*x/(-12 + exp(4)))/x**2)

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