\(\int \frac {16-4 e^x+(-8+e^x (2+4 x)) \log (x)+(2+3 x) \log ^2(x)}{(-8 x+2 e^x x) \log (x)+(2 x+x^2) \log ^2(x)} \, dx\) [771]

Optimal result

Integrand size = 60, antiderivative size = 25 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log \left (\frac {\left (x^2+\frac {2 x \left (-4+e^x+\log (x)\right )}{\log (x)}\right )^2}{x}\right ) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

2*x + Log[x] - 2*Log[Log[x]] + 18*Defer[Int][(-8 + 2*E^x + 2*Log[x] + x*Log[x])^(-1), x] + 4*Defer[Int][1/(x*(
-8 + 2*E^x + 2*Log[x] + x*Log[x])), x] - 2*Defer[Int][Log[x]/(-8 + 2*E^x + 2*Log[x] + x*Log[x]), x] - 2*Defer[
Int][(x*Log[x])/(-8 + 2*E^x + 2*Log[x] + x*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-16+4 e^x-\left (-8+e^x (2+4 x)\right ) \log (x)-(2+3 x) \log ^2(x)}{x \log (x) \left (8-2 e^x-2 \log (x)-x \log (x)\right )} \, dx \\ & = \int \left (\frac {-2+\log (x)+2 x \log (x)}{x \log (x)}-\frac {2 \left (-2-9 x+x \log (x)+x^2 \log (x)\right )}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )}\right ) \, dx \\ & = -\left (2 \int \frac {-2-9 x+x \log (x)+x^2 \log (x)}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )} \, dx\right )+\int \frac {-2+\log (x)+2 x \log (x)}{x \log (x)} \, dx \\ & = -\left (2 \int \left (-\frac {9}{-8+2 e^x+2 \log (x)+x \log (x)}-\frac {2}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )}+\frac {\log (x)}{-8+2 e^x+2 \log (x)+x \log (x)}+\frac {x \log (x)}{-8+2 e^x+2 \log (x)+x \log (x)}\right ) \, dx\right )+\int \left (\frac {1+2 x}{x}-\frac {2}{x \log (x)}\right ) \, dx \\ & = -\left (2 \int \frac {1}{x \log (x)} \, dx\right )-2 \int \frac {\log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx-2 \int \frac {x \log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx+4 \int \frac {1}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )} \, dx+18 \int \frac {1}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx+\int \frac {1+2 x}{x} \, dx \\ & = -\left (2 \int \frac {\log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx\right )-2 \int \frac {x \log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx-2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+4 \int \frac {1}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )} \, dx+18 \int \frac {1}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx+\int \left (2+\frac {1}{x}\right ) \, dx \\ & = 2 x+\log (x)-2 \log (\log (x))-2 \int \frac {\log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx-2 \int \frac {x \log (x)}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx+4 \int \frac {1}{x \left (-8+2 e^x+2 \log (x)+x \log (x)\right )} \, dx+18 \int \frac {1}{-8+2 e^x+2 \log (x)+x \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log (x)-2 \log (\log (x))+2 \log \left (8-2 e^x-2 \log (x)-x \log (x)\right ) \]

[In]

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

[Out]

Log[x] - 2*Log[Log[x]] + 2*Log[8 - 2*E^x - 2*Log[x] - x*Log[x]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
norman	\(\ln \left (x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \ln \left (x \right )+2 \ln \left (x \right )+2 \,{\mathrm e}^{x}-8\right )\)	\(26\)
parallelrisch	\(\ln \left (x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \ln \left (x \right )+2 \ln \left (x \right )+2 \,{\mathrm e}^{x}-8\right )\)	\(26\)
risch	\(\ln \left (x \right )+2 \ln \left (2+x \right )-2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (\ln \left (x \right )+\frac {2 \,{\mathrm e}^{x}-8}{2+x}\right )\)	\(32\)

[In]

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

[Out]

ln(x)-2*ln(ln(x))+2*ln(x*ln(x)+2*ln(x)+2*exp(x)-8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (x + 2\right ) + \log \left (x\right ) + 2 \, \log \left (\frac {{\left (x + 2\right )} \log \left (x\right ) + 2 \, e^{x} - 8}{x + 2}\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

2*log(x + 2) + log(x) + 2*log(((x + 2)*log(x) + 2*e^x - 8)/(x + 2)) - 2*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\log {\left (x \right )} + 2 \log {\left (\frac {x \log {\left (x \right )}}{2} + e^{x} + \log {\left (x \right )} - 4 \right )} - 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (\frac {1}{2} \, {\left (x + 2\right )} \log \left (x\right ) + e^{x} - 4\right ) + \log \left (x\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2 \, \log \left (x \log \left (x\right ) + 2 \, e^{x} + 2 \, \log \left (x\right ) - 8\right ) + \log \left (x\right ) - 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

2*log(x*log(x) + 2*e^x + 2*log(x) - 8) + log(x) - 2*log(log(x))

Mupad [B] (verification not implemented)

Time = 8.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {16-4 e^x+\left (-8+e^x (2+4 x)\right ) \log (x)+(2+3 x) \log ^2(x)}{\left (-8 x+2 e^x x\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=2\,\ln \left (2\,{\mathrm {e}}^x+2\,\ln \left (x\right )+x\,\ln \left (x\right )-8\right )-2\,\ln \left (\ln \left (x\right )\right )+\ln \left (x\right ) \]

[In]

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

[Out]

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