3.58.96 \(\int \frac {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} (3 x-2 e x+4 x^2) \log ^2(x)+(e^2 x+6 x^2+4 x^3+e (-3 x-4 x^2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (4 \left (-1-\frac {3}{-e+e^{\frac {4}{\log (x)}}+2 x}\right )\right ) \]

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Rubi [F]  time = 3.76, 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 {12 e^{\frac {4}{\log (x)}}-6 x \log ^2(x)}{e^{\frac {8}{\log (x)}} x \log ^2(x)+e^{\frac {4}{\log (x)}} \left (3 x-2 e x+4 x^2\right ) \log ^2(x)+\left (e^2 x+6 x^2+4 x^3+e \left (-3 x-4 x^2\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12*E^(4/Log[x]) - 6*x*Log[x]^2)/(E^(8/Log[x])*x*Log[x]^2 + E^(4/Log[x])*(3*x - 2*E*x + 4*x^2)*Log[x]^2 +
(E^2*x + 6*x^2 + 4*x^3 + E*(-3*x - 4*x^2))*Log[x]^2),x]

[Out]

2*Defer[Int][(3*(1 - E/3) + E^(4/Log[x]) + 2*x)^(-1), x] - 2*Defer[Int][(-E + E^(4/Log[x]) + 2*x)^(-1), x] - 4
*E*Defer[Int][1/((E - E^(4/Log[x]) - 2*x)*x*Log[x]^2), x] + 8*Defer[Int][1/((3*(1 - E/3) + E^(4/Log[x]) + 2*x)
*Log[x]^2), x] + 4*(3 - E)*Defer[Int][1/(x*(3*(1 - E/3) + E^(4/Log[x]) + 2*x)*Log[x]^2), x] - 8*Defer[Int][1/(
(-E + E^(4/Log[x]) + 2*x)*Log[x]^2), x]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 36, normalized size = 1.50 \begin {gather*} -\log \left (-e+e^{\frac {4}{\log (x)}}+2 x\right )+\log \left (3-e+e^{\frac {4}{\log (x)}}+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^(4/Log[x]) - 6*x*Log[x]^2)/(E^(8/Log[x])*x*Log[x]^2 + E^(4/Log[x])*(3*x - 2*E*x + 4*x^2)*Log[x
]^2 + (E^2*x + 6*x^2 + 4*x^3 + E*(-3*x - 4*x^2))*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.64, size = 36, normalized size = 1.50 \begin {gather*} \log \left (2 \, x - e + e^{\frac {4}{\log \relax (x)}} + 3\right ) - \log \left (2 \, x - e + e^{\frac {4}{\log \relax (x)}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.06, size = 37, normalized size = 1.54

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {4}{\ln \relax (x )}}+3+2 x -{\mathrm e}\right )-\ln \left (2 x -{\mathrm e}+{\mathrm e}^{\frac {4}{\ln \relax (x )}}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(4/ln(x))-6*x*ln(x)^2)/(x*ln(x)^2*exp(4/ln(x))^2+(-2*x*exp(1)+4*x^2+3*x)*ln(x)^2*exp(4/ln(x))+(x*ex
p(1)^2+(-4*x^2-3*x)*exp(1)+4*x^3+6*x^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 36, normalized size = 1.50 \begin {gather*} \log \left (2 \, x - e + e^{\frac {4}{\log \relax (x)}} + 3\right ) - \log \left (2 \, x - e + e^{\frac {4}{\log \relax (x)}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {12\,{\mathrm {e}}^{\frac {4}{\ln \relax (x)}}-6\,x\,{\ln \relax (x)}^2}{{\ln \relax (x)}^2\,\left (x\,{\mathrm {e}}^2-\mathrm {e}\,\left (4\,x^2+3\,x\right )+6\,x^2+4\,x^3\right )+x\,{\mathrm {e}}^{\frac {8}{\ln \relax (x)}}\,{\ln \relax (x)}^2+{\mathrm {e}}^{\frac {4}{\ln \relax (x)}}\,{\ln \relax (x)}^2\,\left (3\,x-2\,x\,\mathrm {e}+4\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(4/log(x)) - 6*x*log(x)^2)/(log(x)^2*(x*exp(2) - exp(1)*(3*x + 4*x^2) + 6*x^2 + 4*x^3) + x*exp(8/lo
g(x))*log(x)^2 + exp(4/log(x))*log(x)^2*(3*x - 2*x*exp(1) + 4*x^2)),x)

[Out]

int((12*exp(4/log(x)) - 6*x*log(x)^2)/(log(x)^2*(x*exp(2) - exp(1)*(3*x + 4*x^2) + 6*x^2 + 4*x^3) + x*exp(8/lo
g(x))*log(x)^2 + exp(4/log(x))*log(x)^2*(3*x - 2*x*exp(1) + 4*x^2)), x)

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sympy [A]  time = 0.62, size = 31, normalized size = 1.29 \begin {gather*} - \log {\left (2 x + e^{\frac {4}{\log {\relax (x )}}} - e \right )} + \log {\left (2 x + e^{\frac {4}{\log {\relax (x )}}} - e + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*exp(4/ln(x))-6*x*ln(x)**2)/(x*ln(x)**2*exp(4/ln(x))**2+(-2*x*exp(1)+4*x**2+3*x)*ln(x)**2*exp(4/l
n(x))+(x*exp(1)**2+(-4*x**2-3*x)*exp(1)+4*x**3+6*x**2)*ln(x)**2),x)

[Out]

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

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