3.17.38 \(\int \frac {-1-2 x^2-2 e^2 x^2}{(-x+2 x^3+2 e^2 x^3) \log (\frac {-1+2 x^2+2 e^2 x^2}{2 x+2 e^2 x})} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {6}{\log \left (\frac {1}{2 \left (-1-e^2\right ) x}+x\right )}\right ) \]

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Rubi [F]  time = 0.82, 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 {-1-2 x^2-2 e^2 x^2}{\left (-x+2 x^3+2 e^2 x^3\right ) \log \left (\frac {-1+2 x^2+2 e^2 x^2}{2 x+2 e^2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 20, normalized size = 0.83 \begin {gather*} -\log \left (\log \left (x-\frac {1}{2 x+2 e^2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 2*x^2 - 2*E^2*x^2)/((-x + 2*x^3 + 2*E^2*x^3)*Log[(-1 + 2*x^2 + 2*E^2*x^2)/(2*x + 2*E^2*x)]),x]

[Out]

-Log[Log[x - (2*x + 2*E^2*x)^(-1)]]

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fricas [A]  time = 0.79, size = 28, normalized size = 1.17 \begin {gather*} -\log \left (\log \left (\frac {2 \, x^{2} e^{2} + 2 \, x^{2} - 1}{2 \, {\left (x e^{2} + x\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(2)-2*x^2-1)/(2*x^3*exp(2)+2*x^3-x)/log((2*x^2*exp(2)+2*x^2-1)/(2*exp(2)*x+2*x)),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 28, normalized size = 1.17 \begin {gather*} -\log \left (\log \left (\frac {2 \, x^{2} e^{2} + 2 \, x^{2} - 1}{2 \, {\left (x e^{2} + x\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(2)-2*x^2-1)/(2*x^3*exp(2)+2*x^3-x)/log((2*x^2*exp(2)+2*x^2-1)/(2*exp(2)*x+2*x)),x, algor
ithm="giac")

[Out]

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

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maple [A]  time = 0.31, size = 31, normalized size = 1.29

method	result	size
norman	\(-\ln \left (\ln \left (\frac {2 x^{2} {\mathrm e}^{2}+2 x^{2}-1}{2 \,{\mathrm e}^{2} x +2 x}\right )\right )\)	\(31\)
risch	\(-\ln \left (\ln \left (\frac {2 x^{2} {\mathrm e}^{2}+2 x^{2}-1}{2 \,{\mathrm e}^{2} x +2 x}\right )\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^2*exp(2)-2*x^2-1)/(2*x^3*exp(2)+2*x^3-x)/ln((2*x^2*exp(2)+2*x^2-1)/(2*exp(2)*x+2*x)),x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [A]  time = 0.53, size = 31, normalized size = 1.29 \begin {gather*} -\log \left (-\log \relax (2) + \log \left (2 \, x^{2} {\left (e^{2} + 1\right )} - 1\right ) - \log \relax (x) - \log \left (e^{2} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(2)-2*x^2-1)/(2*x^3*exp(2)+2*x^3-x)/log((2*x^2*exp(2)+2*x^2-1)/(2*exp(2)*x+2*x)),x, algor
ithm="maxima")

[Out]

-log(-log(2) + log(2*x^2*(e^2 + 1) - 1) - log(x) - log(e^2 + 1))

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mupad [B]  time = 5.52, size = 30, normalized size = 1.25 \begin {gather*} -\ln \left (\ln \left (\frac {2\,x^2\,{\mathrm {e}}^2+2\,x^2-1}{2\,x+2\,x\,{\mathrm {e}}^2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2*exp(2) + 2*x^2 + 1)/(log((2*x^2*exp(2) + 2*x^2 - 1)/(2*x + 2*x*exp(2)))*(2*x^3*exp(2) - x + 2*x^3)
),x)

[Out]

-log(log((2*x^2*exp(2) + 2*x^2 - 1)/(2*x + 2*x*exp(2))))

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sympy [A]  time = 0.29, size = 29, normalized size = 1.21 \begin {gather*} - \log {\left (\log {\left (\frac {2 x^{2} + 2 x^{2} e^{2} - 1}{2 x + 2 x e^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log((2*x**2 + 2*x**2*exp(2) - 1)/(2*x + 2*x*exp(2))))

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