3.44.43 \(\int \frac {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2)+(-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x) \log (4+e^{e^{-3+e^{e^x}}})}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {4 \left (\frac {5}{x}+\log \left (4+e^{e^{-3+e^{e^x}}}\right )\right )}{e^3 x^2} \]

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Rubi [F]  time = 1.84, 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 {-\frac {240}{e^3}+e^{e^{-3+e^{e^x}}} \left (-\frac {60}{e^3}+4 e^{-6+e^{e^x}+e^x+x} x^2\right )+\left (-\frac {32 x}{e^3}-8 e^{-3+e^{-3+e^{e^x}}} x\right ) \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{4 x^4+e^{e^{-3+e^{e^x}}} x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-240/E^3 + E^E^(-3 + E^E^x)*(-60/E^3 + 4*E^(-6 + E^E^x + E^x + x)*x^2) + ((-32*x)/E^3 - 8*E^(-3 + E^(-3 +
E^E^x))*x)*Log[4 + E^E^(-3 + E^E^x)])/(4*x^4 + E^E^(-3 + E^E^x)*x^4),x]

[Out]

20/(E^3*x^3) + (4*Log[4 + E^E^(-3 + E^E^x)])/(E^3*x^2) + 4*Defer[Int][E^(-6 + E^E^x + E^(-3 + E^E^x) + E^x + x
)/((4 + E^E^(-3 + E^E^x))*x^2), x] - (4*Defer[Int][E^(-3 + E^E^x + E^(-3 + E^E^x) + E^x + x)/((4 + E^E^(-3 + E
^E^x))*x^2), x])/E^3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 e^{-6+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2}-\frac {4 \left (15+2 x \log \left (4+e^{e^{-3+e^{e^x}}}\right )\right )}{e^3 x^4}\right ) \, dx\\ &=4 \int \frac {e^{-6+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2} \, dx-\frac {4 \int \frac {15+2 x \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{x^4} \, dx}{e^3}\\ &=4 \int \frac {e^{-6+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2} \, dx-\frac {4 \int \left (\frac {15}{x^4}+\frac {2 \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{x^3}\right ) \, dx}{e^3}\\ &=\frac {20}{e^3 x^3}+4 \int \frac {e^{-6+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2} \, dx-\frac {8 \int \frac {\log \left (4+e^{e^{-3+e^{e^x}}}\right )}{x^3} \, dx}{e^3}\\ &=\frac {20}{e^3 x^3}+\frac {4 \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{e^3 x^2}+4 \int \frac {e^{-6+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2} \, dx-\frac {4 \int \frac {e^{-3+e^{e^x}+e^{-3+e^{e^x}}+e^x+x}}{\left (4+e^{e^{-3+e^{e^x}}}\right ) x^2} \, dx}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 36, normalized size = 1.29 \begin {gather*} -\frac {4 \left (-\frac {5 e^3}{x^3}-\frac {e^3 \log \left (4+e^{e^{-3+e^{e^x}}}\right )}{x^2}\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240/E^3 + E^E^(-3 + E^E^x)*(-60/E^3 + 4*E^(-6 + E^E^x + E^x + x)*x^2) + ((-32*x)/E^3 - 8*E^(-3 + E
^(-3 + E^E^x))*x)*Log[4 + E^E^(-3 + E^E^x)])/(4*x^4 + E^E^(-3 + E^E^x)*x^4),x]

[Out]

(-4*((-5*E^3)/x^3 - (E^3*Log[4 + E^E^(-3 + E^E^x)])/x^2))/E^6

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fricas [B]  time = 0.54, size = 89, normalized size = 3.18 \begin {gather*} \frac {x e^{\left (2 \, \log \relax (2) - 3\right )} \log \left ({\left (e^{\left ({\left ({\left (2 \, \log \relax (2) - 3\right )} e^{\left (x + e^{x} + 2 \, \log \relax (2) - 3\right )} + e^{\left (x + e^{x} + e^{\left (e^{x}\right )} + 2 \, \log \relax (2) - 6\right )}\right )} e^{\left (-x - e^{x} - 2 \, \log \relax (2) + 3\right )}\right )} + 4 \, e^{\left (2 \, \log \relax (2) - 3\right )}\right )} e^{\left (-2 \, \log \relax (2) + 3\right )}\right ) + 5 \, e^{\left (2 \, \log \relax (2) - 3\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2)-3))*log(exp(exp(-3+exp(exp(x))))+4)
+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*e
xp(2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm="fricas")

[Out]

(x*e^(2*log(2) - 3)*log((e^(((2*log(2) - 3)*e^(x + e^x + 2*log(2) - 3) + e^(x + e^x + e^(e^x) + 2*log(2) - 6))
*e^(-x - e^x - 2*log(2) + 3)) + 4*e^(2*log(2) - 3))*e^(-2*log(2) + 3)) + 5*e^(2*log(2) - 3))/x^3

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giac [A]  time = 0.17, size = 52, normalized size = 1.86 \begin {gather*} \frac {4 \, {\left (x \log \left ({\left (e^{\left (x + e^{x} + e^{\left (e^{\left (e^{x}\right )} - 3\right )} + e^{\left (e^{x}\right )}\right )} + 4 \, e^{\left (x + e^{x} + e^{\left (e^{x}\right )}\right )}\right )} e^{\left (-x - e^{x} - e^{\left (e^{x}\right )}\right )}\right ) + 5\right )} e^{\left (-3\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2)-3))*log(exp(exp(-3+exp(exp(x))))+4)
+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*e
xp(2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm="giac")

[Out]

4*(x*log((e^(x + e^x + e^(e^(e^x) - 3) + e^(e^x)) + 4*e^(x + e^x + e^(e^x)))*e^(-x - e^x - e^(e^x))) + 5)*e^(-
3)/x^3

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maple [A]  time = 0.12, size = 26, normalized size = 0.93

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-3} \ln \left ({\mathrm e}^{{\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}}+4\right )}{x^{2}}+\frac {20 \,{\mathrm e}^{-3}}{x^{3}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(2*ln(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*ln(2)-3))*ln(exp(exp(-3+exp(exp(x))))+4)+(x^2*exp
(x)*exp(2*ln(2)-3)*exp(exp(x))*exp(-3+exp(exp(x)))-15*exp(2*ln(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp(2*ln(2)-
3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x,method=_RETURNVERBOSE)

[Out]

4*exp(-3)/x^2*ln(exp(exp(-3+exp(exp(x))))+4)+20/x^3*exp(-3)

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maxima [A]  time = 0.53, size = 21, normalized size = 0.75 \begin {gather*} \frac {4 \, {\left (x \log \left (e^{\left (e^{\left (e^{\left (e^{x}\right )} - 3\right )}\right )} + 4\right ) + 5\right )} e^{\left (-3\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*log(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*log(2)-3))*log(exp(exp(-3+exp(exp(x))))+4)
+(x^2*exp(x)*exp(2*log(2)-3)*exp(exp(x))*exp(-3+exp(exp(x)))-15*exp(2*log(2)-3))*exp(exp(-3+exp(exp(x))))-60*e
xp(2*log(2)-3))/(x^4*exp(exp(-3+exp(exp(x))))+4*x^4),x, algorithm="maxima")

[Out]

4*(x*log(e^(e^(e^(e^x) - 3)) + 4) + 5)*e^(-3)/x^3

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mupad [B]  time = 0.26, size = 22, normalized size = 0.79 \begin {gather*} \frac {4\,{\mathrm {e}}^{-3}\,\left (x\,\ln \left ({\mathrm {e}}^{{\mathrm {e}}^{-3}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}+4\right )+5\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*exp(2*log(2) - 3) + exp(exp(exp(exp(x)) - 3))*(15*exp(2*log(2) - 3) - x^2*exp(exp(exp(x)) - 3)*exp(ex
p(x))*exp(2*log(2) - 3)*exp(x)) + log(exp(exp(exp(exp(x)) - 3)) + 4)*(8*x*exp(2*log(2) - 3) + 2*x*exp(exp(exp(
exp(x)) - 3))*exp(2*log(2) - 3)))/(x^4*exp(exp(exp(exp(x)) - 3)) + 4*x^4),x)

[Out]

(4*exp(-3)*(x*log(exp(exp(-3)*exp(exp(exp(x)))) + 4) + 5))/x^3

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sympy [A]  time = 5.11, size = 29, normalized size = 1.04 \begin {gather*} \frac {4 \log {\left (e^{e^{e^{e^{x}} - 3}} + 4 \right )}}{x^{2} e^{3}} + \frac {20}{x^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*ln(2)-3)*exp(exp(-3+exp(exp(x))))-8*x*exp(2*ln(2)-3))*ln(exp(exp(-3+exp(exp(x))))+4)+(x
**2*exp(x)*exp(2*ln(2)-3)*exp(exp(x))*exp(-3+exp(exp(x)))-15*exp(2*ln(2)-3))*exp(exp(-3+exp(exp(x))))-60*exp(2
*ln(2)-3))/(x**4*exp(exp(-3+exp(exp(x))))+4*x**4),x)

[Out]

4*exp(-3)*log(exp(exp(exp(exp(x)) - 3)) + 4)/x**2 + 20*exp(-3)/x**3

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