\(\int \frac {e^4 (-8+2 x)+(16 x-8 x^2+x^3+e^4 (-16 x+8 x^2-x^3)) \log (x^2)-e^4 x \log (x^2) \log (\log (x^2))}{e^4 (16 x-8 x^2+x^3) \log (x^2)} \, dx\) [4314]

Optimal result

Integrand size = 84, antiderivative size = 20 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-x+\frac {x}{e^4}+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx \]

[In]

Int[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3))*Log[x^2] - E^4*x*Log[x^2]*Log[Log[x^2]]
)/(E^4*(16*x - 8*x^2 + x^3)*Log[x^2]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx}{e^4} \\ & = \frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \left (16-8 x+x^2\right ) \log \left (x^2\right )} \, dx}{e^4} \\ & = \frac {\int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(-4+x)^2 x \log \left (x^2\right )} \, dx}{e^4} \\ & = \frac {\int \left (1-e^4+\frac {2 e^4}{(-4+x) x \log \left (x^2\right )}-\frac {e^4 \log \left (\log \left (x^2\right )\right )}{(-4+x)^2}\right ) \, dx}{e^4} \\ & = -\left (\left (1-\frac {1}{e^4}\right ) x\right )+2 \int \frac {1}{(-4+x) x \log \left (x^2\right )} \, dx-\int \frac {\log \left (\log \left (x^2\right )\right )}{(-4+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\left (-1+\frac {1}{e^4}\right ) x+\frac {\log \left (\log \left (x^2\right )\right )}{-4+x} \]

[In]

Integrate[(E^4*(-8 + 2*x) + (16*x - 8*x^2 + x^3 + E^4*(-16*x + 8*x^2 - x^3))*Log[x^2] - E^4*x*Log[x^2]*Log[Log
[x^2]])/(E^4*(16*x - 8*x^2 + x^3)*Log[x^2]),x]

[Out]

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{-4} \left (2 x^{2} {\mathrm e}^{4}+32-2 \ln \left (\ln \left (x^{2}\right )\right ) {\mathrm e}^{4}-2 x^{2}-32 \,{\mathrm e}^{4}\right )}{2 \left (x -4\right )}\)	\(39\)

[In]

int((-x*exp(4)*ln(x^2)*ln(ln(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*ln(x^2)+(2*x-8)*exp(4))/(x^3-8*x^
2+16*x)/exp(4)/ln(x^2),x,method=_RETURNVERBOSE)

[Out]

-1/2/exp(4)*(2*x^2*exp(4)+32-2*ln(ln(x^2))*exp(4)-2*x^2-32*exp(4))/(x-4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {{\left (x^{2} - {\left (x^{2} - 4 \, x\right )} e^{4} + e^{4} \log \left (\log \left (x^{2}\right )\right ) - 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \]

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {x \left (1 - e^{4}\right )}{e^{4}} + \frac {\log {\left (\log {\left (x^{2} \right )} \right )}}{x - 4} \]

[In]

integrate((-x*exp(4)*ln(x**2)*ln(ln(x**2))+((-x**3+8*x**2-16*x)*exp(4)+x**3-8*x**2+16*x)*ln(x**2)+(2*x-8)*exp(
4))/(x**3-8*x**2+16*x)/exp(4)/ln(x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-\frac {{\left (x^{2} {\left (e^{4} - 1\right )} - 4 \, x {\left (e^{4} - 1\right )} - e^{4} \log \left (2\right ) - e^{4} \log \left (\log \left (x\right )\right )\right )} e^{\left (-4\right )}}{x - 4} \]

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=-\frac {{\left (x^{2} e^{4} - x^{2} - 4 \, x e^{4} - e^{4} \log \left (\log \left (x^{2}\right )\right ) + 4 \, x\right )} e^{\left (-4\right )}}{x - 4} \]

[In]

integrate((-x*exp(4)*log(x^2)*log(log(x^2))+((-x^3+8*x^2-16*x)*exp(4)+x^3-8*x^2+16*x)*log(x^2)+(2*x-8)*exp(4))
/(x^3-8*x^2+16*x)/exp(4)/log(x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.84 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^4 (-8+2 x)+\left (16 x-8 x^2+x^3+e^4 \left (-16 x+8 x^2-x^3\right )\right ) \log \left (x^2\right )-e^4 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{e^4 \left (16 x-8 x^2+x^3\right ) \log \left (x^2\right )} \, dx=\frac {\ln \left (\ln \left (x^2\right )\right )}{x-4}+x\,\left ({\mathrm {e}}^{-4}-1\right ) \]

[In]

int((exp(-4)*(log(x^2)*(16*x - exp(4)*(16*x - 8*x^2 + x^3) - 8*x^2 + x^3) + exp(4)*(2*x - 8) - x*log(x^2)*exp(
4)*log(log(x^2))))/(log(x^2)*(16*x - 8*x^2 + x^3)),x)

[Out]

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