3.36.59 \(\int \frac {e^{e^5} (4 x^2-4 x^4)+(e^{e^5} (-8 x^2+2 x^4+e^5 (-8+2 x^2))+e^{e^5} (-24+6 x^2) \log (\frac {1}{2} (-4+x^2))) \log (e^5+x^2+3 \log (\frac {1}{2} (-4+x^2)))}{(-4 x^2+x^4+e^5 (-4+x^2)+(-12+3 x^2) \log (\frac {1}{2} (-4+x^2))) \log ^2(e^5+x^2+3 \log (\frac {1}{2} (-4+x^2)))} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 e^{e^5} x}{\log \left (e^5+x^2+3 \log \left (-2+\frac {x^2}{2}\right )\right )} \]

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Rubi [F]  time = 1.95, 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^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^5*(4*x^2 - 4*x^4) + (E^E^5*(-8*x^2 + 2*x^4 + E^5*(-8 + 2*x^2)) + E^E^5*(-24 + 6*x^2)*Log[(-4 + x^2)/2
])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]])/((-4*x^2 + x^4 + E^5*(-4 + x^2) + (-12 + 3*x^2)*Log[(-4 + x^2)/2])*Lo
g[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2),x]

[Out]

-12*E^E^5*Defer[Int][1/((E^5 + x^2 + 3*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2), x] - 12*E^E
^5*Defer[Int][1/((-2 + x)*(E^5 + x^2 + 3*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2), x] - 4*E^
E^5*Defer[Int][x^2/((E^5 + x^2 + 3*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2), x] + 12*E^E^5*D
efer[Int][1/((2 + x)*(E^5 + x^2 + 3*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2), x] + 2*E^E^5*D
efer[Int][Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{e^5} \left (-\frac {2 x^2 \left (-1+x^2\right )}{\left (-4+x^2\right ) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}+\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )\right )}{\log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {-\frac {2 x^2 \left (-1+x^2\right )}{\left (-4+x^2\right ) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}+\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \int \left (-\frac {2 x^2 \left (-1+x^2\right )}{(-2+x) (2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}+\frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}\right ) \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \frac {x^2 \left (-1+x^2\right )}{(-2+x) (2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \frac {x^2 \left (-1+x^2\right )}{\left (-4+x^2\right ) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \left (\frac {3}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}+\frac {x^2}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}+\frac {12}{\left (-4+x^2\right ) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}\right ) \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \frac {x^2}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (12 e^{e^5}\right ) \int \frac {1}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (48 e^{e^5}\right ) \int \frac {1}{\left (-4+x^2\right ) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \frac {x^2}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (12 e^{e^5}\right ) \int \frac {1}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (48 e^{e^5}\right ) \int \left (\frac {1}{4 (-2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}-\frac {1}{4 (2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}\right ) \, dx\\ &=\left (2 e^{e^5}\right ) \int \frac {1}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (4 e^{e^5}\right ) \int \frac {x^2}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (12 e^{e^5}\right ) \int \frac {1}{\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx-\left (12 e^{e^5}\right ) \int \frac {1}{(-2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx+\left (12 e^{e^5}\right ) \int \frac {1}{(2+x) \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 30, normalized size = 1.00 \begin {gather*} \frac {2 e^{e^5} x}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^5*(4*x^2 - 4*x^4) + (E^E^5*(-8*x^2 + 2*x^4 + E^5*(-8 + 2*x^2)) + E^E^5*(-24 + 6*x^2)*Log[(-4 +
x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]])/((-4*x^2 + x^4 + E^5*(-4 + x^2) + (-12 + 3*x^2)*Log[(-4 + x^2)/
2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2),x]

[Out]

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

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fricas [A]  time = 1.14, size = 25, normalized size = 0.83 \begin {gather*} \frac {2 \, x e^{\left (e^{5}\right )}}{\log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4-8*x^2)*exp(exp(5)))*log(3*log(1/2*x^
2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(exp(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/
2*x^2-2)+x^2+exp(5))^2,x, algorithm="fricas")

[Out]

2*x*e^(e^5)/log(x^2 + e^5 + 3*log(1/2*x^2 - 2))

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giac [B]  time = 0.61, size = 121, normalized size = 4.03 \begin {gather*} \frac {2 \, {\left (x^{3} e^{\left (e^{5}\right )} + 3 \, x e^{\left (e^{5}\right )} \log \left (\frac {1}{2} \, x^{2} - 2\right ) + x e^{\left (e^{5} + 5\right )}\right )}}{x^{2} \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) + e^{5} \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) - 3 \, \log \relax (2) \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) + 3 \, \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) \log \left (x^{2} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4-8*x^2)*exp(exp(5)))*log(3*log(1/2*x^
2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(exp(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/
2*x^2-2)+x^2+exp(5))^2,x, algorithm="giac")

[Out]

2*(x^3*e^(e^5) + 3*x*e^(e^5)*log(1/2*x^2 - 2) + x*e^(e^5 + 5))/(x^2*log(x^2 + e^5 + 3*log(1/2*x^2 - 2)) + e^5*
log(x^2 + e^5 + 3*log(1/2*x^2 - 2)) - 3*log(2)*log(x^2 + e^5 + 3*log(1/2*x^2 - 2)) + 3*log(x^2 + e^5 + 3*log(1
/2*x^2 - 2))*log(x^2 - 4))

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maple [A]  time = 0.06, size = 26, normalized size = 0.87

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{{\mathrm e}^{5}} x}{\ln \left (3 \ln \left (\frac {x^{2}}{2}-2\right )+x^{2}+{\mathrm e}^{5}\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x^2-24)*exp(exp(5))*ln(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4-8*x^2)*exp(exp(5)))*ln(3*ln(1/2*x^2-2)+x^2+
exp(5))+(-4*x^4+4*x^2)*exp(exp(5)))/((3*x^2-12)*ln(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/ln(3*ln(1/2*x^2-2)+x^2
+exp(5))^2,x,method=_RETURNVERBOSE)

[Out]

2*exp(exp(5))/ln(3*ln(1/2*x^2-2)+x^2+exp(5))*x

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maxima [A]  time = 0.71, size = 31, normalized size = 1.03 \begin {gather*} \frac {2 \, x e^{\left (e^{5}\right )}}{\log \left (x^{2} + e^{5} - 3 \, \log \relax (2) + 3 \, \log \left (x + 2\right ) + 3 \, \log \left (x - 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4-8*x^2)*exp(exp(5)))*log(3*log(1/2*x^
2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(exp(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/
2*x^2-2)+x^2+exp(5))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.97, size = 64, normalized size = 2.13 \begin {gather*} x\,{\mathrm {e}}^{{\mathrm {e}}^5}+\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^5}}{\ln \left (3\,\ln \left (\frac {x^2}{2}-2\right )+{\mathrm {e}}^5+x^2\right )}-\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x-x^3}+\frac {x^4\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3*log(x^2/2 - 2) + exp(5) + x^2)*(exp(exp(5))*(exp(5)*(2*x^2 - 8) - 8*x^2 + 2*x^4) + log(x^2/2 - 2)*e
xp(exp(5))*(6*x^2 - 24)) + exp(exp(5))*(4*x^2 - 4*x^4))/(log(3*log(x^2/2 - 2) + exp(5) + x^2)^2*(x^4 - 4*x^2 +
log(x^2/2 - 2)*(3*x^2 - 12) + exp(5)*(x^2 - 4))),x)

[Out]

x*exp(exp(5)) + (2*x*exp(exp(5)))/log(3*log(x^2/2 - 2) + exp(5) + x^2) - (x^2*exp(exp(5)))/(x - x^3) + (x^4*ex
p(exp(5)))/(x - x^3)

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sympy [A]  time = 0.43, size = 26, normalized size = 0.87 \begin {gather*} \frac {2 x e^{e^{5}}}{\log {\left (x^{2} + 3 \log {\left (\frac {x^{2}}{2} - 2 \right )} + e^{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x**2-24)*exp(exp(5))*ln(1/2*x**2-2)+((2*x**2-8)*exp(5)+2*x**4-8*x**2)*exp(exp(5)))*ln(3*ln(1/2*
x**2-2)+x**2+exp(5))+(-4*x**4+4*x**2)*exp(exp(5)))/((3*x**2-12)*ln(1/2*x**2-2)+(x**2-4)*exp(5)+x**4-4*x**2)/ln
(3*ln(1/2*x**2-2)+x**2+exp(5))**2,x)

[Out]

2*x*exp(exp(5))/log(x**2 + 3*log(x**2/2 - 2) + exp(5))

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