3.53.2 \(\int \frac {-3 e^{\frac {2 (2 e-2 e^4+2 x-2 x \log (2))}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} (-2 e+2 e^4+x)}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 (2 e-2 e^4+2 x-2 x \log (2))}{x}} x} \, dx\)

Optimal. Leaf size=27 \[ \frac {x}{-3+e^{-2 \left (\frac {e-e^4+x}{x}-\log (2)\right )}} \]

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Rubi [F]  time = 3.85, 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 {-3 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x+e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} \left (-2 e+2 e^4+x\right )}{x-6 e^{\frac {2 e-2 e^4+2 x-2 x \log (2)}{x}} x+9 e^{\frac {2 \left (2 e-2 e^4+2 x-2 x \log (2)\right )}{x}} x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x + E^((2*E - 2*E^4 + 2*x - 2*x*Log[2])/x)*(-2*E + 2*E^4 +
x))/(x - 6*E^((2*E - 2*E^4 + 2*x - 2*x*Log[2])/x)*x + 9*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x),x]

[Out]

Defer[Int][E^(2 + (2*E)/x)/(-3*E^(2 + (2*E)/x) + 4*E^((2*E^4)/x)), x] + 6*(1 - E^3)*Defer[Subst][Defer[Int][E^
(5 + 4*E*(1 - E^3)*x)/((4 - 3*E^(2 + 2*E*(1 - E^3)*x))^2*x), x], x, x^(-1)] + 2*(1 - E^3)*Defer[Subst][Defer[I
nt][E^(3 + 2*E*x)/(4*E^(2*E^4*x)*x - 3*E^(2 + 2*E*x)*x), x], x, x^(-1)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2+\frac {2 e}{x}} \left (-8 e^{1+\frac {2 e^4}{x}} \left (1-e^3\right )-3 e^{2+\frac {2 e}{x}} x+4 e^{\frac {2 e^4}{x}} x\right )}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x} \, dx\\ &=\int \left (\frac {6 e^{5+\frac {4 e}{x}} \left (-1+e^3\right )}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x}+\frac {e^{2+\frac {2 e}{x}} \left (-2 e+2 e^4+x\right )}{\left (-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}\right ) x}\right ) \, dx\\ &=-\left (\left (6 \left (1-e^3\right )\right ) \int \frac {e^{5+\frac {4 e}{x}}}{\left (3 e^{2+\frac {2 e}{x}}-4 e^{\frac {2 e^4}{x}}\right )^2 x} \, dx\right )+\int \frac {e^{2+\frac {2 e}{x}} \left (-2 e+2 e^4+x\right )}{\left (-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}\right ) x} \, dx\\ &=\left (6 \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{5+4 e x}}{\left (4 e^{2 e^4 x}-3 e^{2+2 e x}\right )^2 x} \, dx,x,\frac {1}{x}\right )+\int \left (\frac {e^{2+\frac {2 e}{x}}}{-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}}+\frac {2 e^{3+\frac {2 e}{x}} \left (-1+e^3\right )}{\left (-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}\right ) x}\right ) \, dx\\ &=-\left (\left (2 \left (1-e^3\right )\right ) \int \frac {e^{3+\frac {2 e}{x}}}{\left (-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}\right ) x} \, dx\right )+\left (6 \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{5+4 e x-4 e^4 x}}{\left (4-3 e^{2+2 e \left (1-e^3\right ) x}\right )^2 x} \, dx,x,\frac {1}{x}\right )+\int \frac {e^{2+\frac {2 e}{x}}}{-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}} \, dx\\ &=\left (2 \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{3+2 e x}}{4 e^{2 e^4 x} x-3 e^{2+2 e x} x} \, dx,x,\frac {1}{x}\right )+\left (6 \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{5+4 e \left (1-e^3\right ) x}}{\left (4-3 e^{2+2 e \left (1-e^3\right ) x}\right )^2 x} \, dx,x,\frac {1}{x}\right )+\int \frac {e^{2+\frac {2 e}{x}}}{-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 39, normalized size = 1.44 \begin {gather*} \frac {e^{2+\frac {2 e}{x}} x}{-3 e^{2+\frac {2 e}{x}}+4 e^{\frac {2 e^4}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x + E^((2*E - 2*E^4 + 2*x - 2*x*Log[2])/x)*(-2*E + 2*
E^4 + x))/(x - 6*E^((2*E - 2*E^4 + 2*x - 2*x*Log[2])/x)*x + 9*E^((2*(2*E - 2*E^4 + 2*x - 2*x*Log[2]))/x)*x),x]

[Out]

(E^(2 + (2*E)/x)*x)/(-3*E^(2 + (2*E)/x) + 4*E^((2*E^4)/x))

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fricas [A]  time = 0.65, size = 49, normalized size = 1.81 \begin {gather*} -\frac {x e^{\left (-\frac {2 \, {\left (x \log \relax (2) - x + e^{4} - e\right )}}{x}\right )}}{3 \, e^{\left (-\frac {2 \, {\left (x \log \relax (2) - x + e^{4} - e\right )}}{x}\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*exp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2
*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)
/x)+x),x, algorithm="fricas")

[Out]

-x*e^(-2*(x*log(2) - x + e^4 - e)/x)/(3*e^(-2*(x*log(2) - x + e^4 - e)/x) - 1)

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giac [A]  time = 0.22, size = 37, normalized size = 1.37 \begin {gather*} -\frac {x e^{\left (\frac {2 \, {\left (x - e^{4} + e\right )}}{x}\right )}}{3 \, e^{\left (\frac {2 \, {\left (x - e^{4} + e\right )}}{x}\right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*exp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2
*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)
/x)+x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.51, size = 28, normalized size = 1.04

method	result	size
risch	\(-\frac {x}{3}-\frac {x}{3 \left (\frac {3 \,{\mathrm e}^{\frac {2 \,{\mathrm e}+2 x -2 \,{\mathrm e}^{4}}{x}}}{4}-1\right )}\)	\(28\)
norman	\(-\frac {x \,{\mathrm e}^{\frac {-2 x \ln \relax (2)-2 \,{\mathrm e}^{4}+2 \,{\mathrm e}+2 x}{x}}}{3 \,{\mathrm e}^{\frac {-2 x \ln \relax (2)-2 \,{\mathrm e}^{4}+2 \,{\mathrm e}+2 x}{x}}-1}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*exp(1)+x)*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+
2*x)/x))/(9*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)+x),x,m
ethod=_RETURNVERBOSE)

[Out]

-1/3*x-1/3*x/(3/4*exp(2*(exp(1)+x-exp(4))/x)-1)

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maxima [A]  time = 0.55, size = 37, normalized size = 1.37 \begin {gather*} \frac {x e^{\left (\frac {2 \, e}{x} + 2\right )}}{4 \, e^{\left (\frac {2 \, e^{4}}{x}\right )} - 3 \, e^{\left (\frac {2 \, e}{x} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2+(2*exp(4)-2*exp(1)+x)*exp((-2*x*log(2)-2*exp(4)+2
*exp(1)+2*x)/x))/(9*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)/x)^2-6*x*exp((-2*x*log(2)-2*exp(4)+2*exp(1)+2*x)
/x)+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.00, size = 31, normalized size = 1.15 \begin {gather*} -\frac {x}{3}-\frac {x}{3\,\left (\frac {3\,{\mathrm {e}}^{\frac {2\,\mathrm {e}}{x}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^2}{4}-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x + 2*exp(1) - 2*exp(4) - 2*x*log(2))/x)*(x - 2*exp(1) + 2*exp(4)) - 3*x*exp((2*(2*x + 2*exp(1) -
2*exp(4) - 2*x*log(2)))/x))/(x - 6*x*exp((2*x + 2*exp(1) - 2*exp(4) - 2*x*log(2))/x) + 9*x*exp((2*(2*x + 2*exp
(1) - 2*exp(4) - 2*x*log(2)))/x)),x)

[Out]

- x/3 - x/(3*((3*exp((2*exp(1))/x)*exp(-(2*exp(4))/x)*exp(2))/4 - 1))

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sympy [A]  time = 0.18, size = 32, normalized size = 1.19 \begin {gather*} - \frac {x}{3} - \frac {x}{9 e^{\frac {- 2 x \log {\relax (2 )} + 2 x - 2 e^{4} + 2 e}{x}} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)**2+(2*exp(4)-2*exp(1)+x)*exp((-2*x*ln(2)-2*exp(4)+2*
exp(1)+2*x)/x))/(9*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x)**2-6*x*exp((-2*x*ln(2)-2*exp(4)+2*exp(1)+2*x)/x
)+x),x)

[Out]

-x/3 - x/(9*exp((-2*x*log(2) + 2*x - 2*exp(4) + 2*E)/x) - 3)

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