\(\int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 (-2+2 x^2)+(2-2 e^2-2 x^2) \log (x)+\log ^2(x)} \, dx\) [7684]

Optimal result

Integrand size = 61, antiderivative size = 17 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 x}{-1+e^2+x^2-\log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 x}{-1+e^2+x^2-\log (x)} \]

[In]

Integrate[(4*E^2 - 4*x^2 - 4*Log[x])/(1 + E^4 - 2*x^2 + x^4 + E^2*(-2 + 2*x^2) + (2 - 2*E^2 - 2*x^2)*Log[x] +
Log[x]^2),x]

[Out]

(4*x)/(-1 + E^2 + x^2 - Log[x])

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {4 x}{x^{2}+{\mathrm e}^{2}-\ln \left (x \right )-1}\)	\(17\)
norman	\(\frac {4 x}{x^{2}+{\mathrm e}^{2}-\ln \left (x \right )-1}\)	\(17\)
risch	\(\frac {4 x}{x^{2}+{\mathrm e}^{2}-\ln \left (x \right )-1}\)	\(17\)
parallelrisch	\(\frac {4 x}{x^{2}+{\mathrm e}^{2}-\ln \left (x \right )-1}\)	\(17\)

[In]

int((-4*ln(x)+4*exp(2)-4*x^2)/(ln(x)^2+(-2*exp(2)-2*x^2+2)*ln(x)+exp(2)^2+(2*x^2-2)*exp(2)+x^4-2*x^2+1),x,meth
od=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 \, x}{x^{2} + e^{2} - \log \left (x\right ) - 1} \]

[In]

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

[Out]

4*x/(x^2 + e^2 - log(x) - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=- \frac {4 x}{- x^{2} + \log {\left (x \right )} - e^{2} + 1} \]

[In]

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

[Out]

-4*x/(-x**2 + log(x) - exp(2) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 \, x}{x^{2} + e^{2} - \log \left (x\right ) - 1} \]

[In]

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

[Out]

4*x/(x^2 + e^2 - log(x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4 \, x}{x^{2} + e^{2} - \log \left (x\right ) - 1} \]

[In]

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

[Out]

4*x/(x^2 + e^2 - log(x) - 1)

Mupad [B] (verification not implemented)

Time = 12.91 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {4 e^2-4 x^2-4 \log (x)}{1+e^4-2 x^2+x^4+e^2 \left (-2+2 x^2\right )+\left (2-2 e^2-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {4\,x}{{\mathrm {e}}^2-\ln \left (x\right )+x^2-1} \]

[In]

int(-(4*log(x) - 4*exp(2) + 4*x^2)/(exp(4) + log(x)^2 + exp(2)*(2*x^2 - 2) - log(x)*(2*exp(2) + 2*x^2 - 2) - 2
*x^2 + x^4 + 1),x)

[Out]

(4*x)/(exp(2) - log(x) + x^2 - 1)