3.1.10 \(\int \frac {e^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+(-32 x+8 x^2) \log (x)+(32-32 x+2 x^2) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx\)

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

________________________________________________________________________________________

Rubi [F]  time = 0.45, 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^2 (-4+x)+2 e^2 \log ^2(x)}{8 x^2+\left (-32 x+8 x^2\right ) \log (x)+\left (32-32 x+2 x^2\right ) \log ^2(x)+(32-8 x) \log ^3(x)+8 \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^2*(-4 + x) + 2*E^2*Log[x]^2)/(8*x^2 + (-32*x + 8*x^2)*Log[x] + (32 - 32*x + 2*x^2)*Log[x]^2 + (32 - 8*x
)*Log[x]^3 + 8*Log[x]^4),x]

[Out]

E^2*Defer[Int][(x - 2*Log[x])^(-2), x] - 6*E^2*Defer[Int][1/((4 + x)*(x - 2*Log[x])^2), x] - 4*E^2*Defer[Int][
1/((4 + x)^2*(x - 2*Log[x])), x] + (E^2*Defer[Int][1/((4 + x)*(2 + Log[x])^2), x])/2 - 2*E^2*Defer[Int][1/((4
+ x)^2*(2 + Log[x])), x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 22, normalized size = 0.92 \begin {gather*} -\frac {e^2 x}{2 (x-2 \log (x)) (2+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-4 + x) + 2*E^2*Log[x]^2)/(8*x^2 + (-32*x + 8*x^2)*Log[x] + (32 - 32*x + 2*x^2)*Log[x]^2 + (32
- 8*x)*Log[x]^3 + 8*Log[x]^4),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 23, normalized size = 0.96 \begin {gather*} -\frac {x e^{2}}{2 \, {\left ({\left (x - 4\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} + 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*log(x)^2+(x-4)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3+(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x
)*log(x)+8*x^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.26, size = 25, normalized size = 1.04 \begin {gather*} -\frac {x e^{2}}{2 \, {\left (x \log \relax (x) - 2 \, \log \relax (x)^{2} + 2 \, x - 4 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*log(x)^2+(x-4)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3+(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x
)*log(x)+8*x^2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.10, size = 20, normalized size = 0.83

method	result	size
norman	\(-\frac {{\mathrm e}^{2} x}{2 \left (\ln \relax (x )+2\right ) \left (x -2 \ln \relax (x )\right )}\)	\(20\)
risch	\(-\frac {{\mathrm e}^{2} x}{2 \left (x \ln \relax (x )-2 \ln \relax (x )^{2}+2 x -4 \ln \relax (x )\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2)*ln(x)^2+(x-4)*exp(2))/(8*ln(x)^4+(-8*x+32)*ln(x)^3+(2*x^2-32*x+32)*ln(x)^2+(8*x^2-32*x)*ln(x)+8*
x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.63, size = 23, normalized size = 0.96 \begin {gather*} -\frac {x e^{2}}{2 \, {\left ({\left (x - 4\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} + 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*log(x)^2+(x-4)*exp(2))/(8*log(x)^4+(-8*x+32)*log(x)^3+(2*x^2-32*x+32)*log(x)^2+(8*x^2-32*x
)*log(x)+8*x^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.69, size = 19, normalized size = 0.79 \begin {gather*} -\frac {x\,{\mathrm {e}}^2}{2\,\left (x-2\,\ln \relax (x)\right )\,\left (\ln \relax (x)+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2)*log(x)^2 + exp(2)*(x - 4))/(log(x)^2*(2*x^2 - 32*x + 32) + 8*log(x)^4 - log(x)*(32*x - 8*x^2) +
8*x^2 - log(x)^3*(8*x - 32)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 22, normalized size = 0.92 \begin {gather*} \frac {x e^{2}}{- 4 x + \left (8 - 2 x\right ) \log {\relax (x )} + 4 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*ln(x)**2+(x-4)*exp(2))/(8*ln(x)**4+(-8*x+32)*ln(x)**3+(2*x**2-32*x+32)*ln(x)**2+(8*x**2-32
*x)*ln(x)+8*x**2),x)

[Out]

x*exp(2)/(-4*x + (8 - 2*x)*log(x) + 4*log(x)**2)

________________________________________________________________________________________