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

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

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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 = {} $$\begin{array}{c}\int \frac{-1-2x^2-2e^2{x}^2}{(-x+2x^3+2e^2{x}^3)\log \left(\frac{-1+2x^2+2e^2{x}^2}{2x+2e^{2}x}\right)}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-1+(-2-2e^2)x^2}{(-x+2x^3+2e^2{x}^3)\log \left(\frac{-1+2x^2+2e^2{x}^2}{2x+2e^{2}x}\right)}dx\\ & =\int \frac{-1+(-2-2e^2)x^2}{(-x+(2+2e^2)x^3)\log \left(\frac{-1+2x^2+2e^2{x}^2}{2x+2e^{2}x}\right)}dx\\ & =\int \frac{-1+(-2-2e^2)x^2}{x(-1+(2+2e^2)x^2)\log \left(\frac{-1+2x^2+2e^2{x}^2}{2x+2e^{2}x}\right)}dx\\ & =\int (\frac{1}{x\log (x-\frac{1}{2x+2e^{2}x})}+\frac{4(1+e^2)x}{(1-2(1+e^2)x^2)\log (x-\frac{1}{2x+2e^{2}x})})dx\\ & =\left(4(1+e^2)\right)\int \frac{x}{(1-2(1+e^2)x^2)\log (x-\frac{1}{2x+2e^{2}x})}dx+\int \frac{1}{x\log (x-\frac{1}{2x+2e^{2}x})}dx\\ & =\left(4(1+e^2)\right)\int (\frac{1}{2\sqrt{2(1+e^2)}(1-\sqrt{2(1+e^2)}x)\log (x-\frac{1}{2x+2e^{2}x})}-\frac{1}{2\sqrt{2(1+e^2)}(1+\sqrt{2(1+e^2)}x)\log (x-\frac{1}{2x+2e^{2}x})})dx+\int \frac{1}{x\log (x-\frac{1}{2x+2e^{2}x})}dx\\ & =\sqrt{2(1+e^2)}\int \frac{1}{(1-\sqrt{2(1+e^2)}x)\log (x-\frac{1}{2x+2e^{2}x})}dx-\sqrt{2(1+e^2)}\int \frac{1}{(1+\sqrt{2(1+e^2)}x)\log (x-\frac{1}{2x+2e^{2}x})}dx+\int \frac{1}{x\log (x-\frac{1}{2x+2e^{2}x})}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}-\log (\log \left(\frac{2x^2{e}^2+2x^2-1}{2(x{e}^2+x)}\right))\end{array}$$

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 $$\begin{array}{c}-\log (\log \left(\frac{2x^2{e}^2+2x^2-1}{2(x{e}^2+x)}\right))\end{array}$$

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	$-\ln (\ln \left(\frac{2x^2{\mathrm{e}}^2+2x^2-1}{2{\mathrm{e}}^{2}x+2x}\right))$	$31$
risch	$-\ln (\ln \left(\frac{2x^2{\mathrm{e}}^2+2x^2-1}{2{\mathrm{e}}^{2}x+2x}\right))$	$31$

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 $$\begin{array}{c}-\log (-\log (2)+\log (2x^2(e^2+1)-1)-\log (x)-\log (e^2+1))\end{array}$$

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 $$\begin{array}{c}-\ln (\ln \left(\frac{2x^2{\mathrm{e}}^2+2x^2-1}{2x+2x{\mathrm{e}}^2}\right))\end{array}$$

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 $$\begin{array}{c}-\log (\log \left(\frac{2x^2+2x^2{e}^2-1}{2x+2x{e}^2}\right))\end{array}$$

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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