\(\int \frac {-4 x+8 x^2+e^4 (x+2 x^2)+e^2 (4 x+8 x^2)+(-4+4 x+4 e^2 x+e^4 x) \log (-4+4 x+4 e^2 x+e^4 x)}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+(8-8 x-8 e^2 x-2 e^4 x) \log (5)+(-4 x+4 x^2+4 e^2 x^2+e^4 x^2) \log (-4+4 x+4 e^2 x+e^4 x)} \, dx\) [5548]

Optimal result

Integrand size = 158, antiderivative size = 24 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (2-\frac {x \left (x+\log \left (-4+\left (2+e^2\right )^2 x\right )\right )}{\log (5)}\right ) \]

[Out]

ln(2-x/ln(5)*(ln((exp(2)+2)^2*x-4)+x))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6, 6820, 6816} \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x^2+x \log \left (\left (2+e^2\right )^2 x-4\right )-2 \log (5)\right ) \]

[In]

Int[(-4*x + 8*x^2 + E^4*(x + 2*x^2) + E^2*(4*x + 8*x^2) + (-4 + 4*x + 4*E^2*x + E^4*x)*Log[-4 + 4*x + 4*E^2*x
+ E^4*x])/(-4*x^2 + 4*x^3 + 4*E^2*x^3 + E^4*x^3 + (8 - 8*x - 8*E^2*x - 2*E^4*x)*Log[5] + (-4*x + 4*x^2 + 4*E^2
*x^2 + E^4*x^2)*Log[-4 + 4*x + 4*E^2*x + E^4*x]),x]

[Out]

Log[x^2 - 2*Log[5] + x*Log[-4 + (2 + E^2)^2*x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+e^4 x^3+\left (4+4 e^2\right ) x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx \\ & = \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+\left (4+4 e^2+e^4\right ) x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx \\ & = \int \frac {-x \left (-4+8 x+e^4 (1+2 x)+e^2 (4+8 x)\right )-\left (-4+\left (2+e^2\right )^2 x\right ) \log \left (-4+\left (2+e^2\right )^2 x\right )}{\left (4-\left (2+e^2\right )^2 x\right ) \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right )} \, dx \\ & = \log \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x^2-2 \log (5)+x \log \left (-4+\left (2+e^2\right )^2 x\right )\right ) \]

[In]

Integrate[(-4*x + 8*x^2 + E^4*(x + 2*x^2) + E^2*(4*x + 8*x^2) + (-4 + 4*x + 4*E^2*x + E^4*x)*Log[-4 + 4*x + 4*
E^2*x + E^4*x])/(-4*x^2 + 4*x^3 + 4*E^2*x^3 + E^4*x^3 + (8 - 8*x - 8*E^2*x - 2*E^4*x)*Log[5] + (-4*x + 4*x^2 +
4*E^2*x^2 + E^4*x^2)*Log[-4 + 4*x + 4*E^2*x + E^4*x]),x]

[Out]

Log[x^2 - 2*Log[5] + x*Log[-4 + (2 + E^2)^2*x]]

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(\ln \left (\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) x +x^{2}-2 \ln \left (5\right )\right )\)	\(29\)
norman	\(\ln \left (-\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) x -x^{2}+2 \ln \left (5\right )\right )\)	\(32\)
risch	\(\ln \left (x \right )+\ln \left (\ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-\frac {-x^{2}+2 \ln \left (5\right )}{x}\right )\)	\(36\)
default	\(\ln \left (-2 \,{\mathrm e}^{8} \ln \left (5\right )+\left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-16 \,{\mathrm e}^{6} \ln \left (5\right )+4 \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-48 \,{\mathrm e}^{4} \ln \left (5\right )+4 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{4}+\left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )^{2}+4 \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )-64 \,{\mathrm e}^{2} \ln \left (5\right )+16 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right ) {\mathrm e}^{2}+8 x \,{\mathrm e}^{4}+32 \,{\mathrm e}^{2} x +32 x -16-32 \ln \left (5\right )+16 \ln \left (x \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +4 x -4\right )\right )\)	\(244\)

[In]

int(((x*exp(2)^2+4*exp(2)*x+4*x-4)*ln(x*exp(2)^2+4*exp(2)*x+4*x-4)+(2*x^2+x)*exp(2)^2+(8*x^2+4*x)*exp(2)+8*x^2
-4*x)/((x^2*exp(2)^2+4*x^2*exp(2)+4*x^2-4*x)*ln(x*exp(2)^2+4*exp(2)*x+4*x-4)+(-2*x*exp(2)^2-8*exp(2)*x-8*x+8)*
ln(5)+x^3*exp(2)^2+4*x^3*exp(2)+4*x^3-4*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x*exp(2)^2+4*exp(2)*x+4*x-4)*x+x^2-2*ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x^{2} + x \log \left (x e^{4} + 4 \, x e^{2} + 4 \, x - 4\right ) - 2 \, \log \left (5\right )}{x}\right ) \]

[In]

integrate(((x*exp(2)^2+4*exp(2)*x+4*x-4)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(2*x^2+x)*exp(2)^2+(8*x^2+4*x)*exp(2
)+8*x^2-4*x)/((x^2*exp(2)^2+4*x^2*exp(2)+4*x^2-4*x)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(-2*x*exp(2)^2-8*exp(2)*x
-8*x+8)*log(5)+x^3*exp(2)^2+4*x^3*exp(2)+4*x^3-4*x^2),x, algorithm="fricas")

[Out]

log(x) + log((x^2 + x*log(x*e^4 + 4*x*e^2 + 4*x - 4) - 2*log(5))/x)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (4 x + 4 x e^{2} + x e^{4} - 4 \right )} + \frac {x^{2} - 2 \log {\left (5 \right )}}{x} \right )} \]

[In]

integrate(((x*exp(2)**2+4*exp(2)*x+4*x-4)*ln(x*exp(2)**2+4*exp(2)*x+4*x-4)+(2*x**2+x)*exp(2)**2+(8*x**2+4*x)*e
xp(2)+8*x**2-4*x)/((x**2*exp(2)**2+4*x**2*exp(2)+4*x**2-4*x)*ln(x*exp(2)**2+4*exp(2)*x+4*x-4)+(-2*x*exp(2)**2-
8*exp(2)*x-8*x+8)*ln(5)+x**3*exp(2)**2+4*x**3*exp(2)+4*x**3-4*x**2),x)

[Out]

log(x) + log(log(4*x + 4*x*exp(2) + x*exp(4) - 4) + (x**2 - 2*log(5))/x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x^{2} + x \log \left (x {\left (e^{4} + 4 \, e^{2} + 4\right )} - 4\right ) - 2 \, \log \left (5\right )}{x}\right ) \]

[In]

integrate(((x*exp(2)^2+4*exp(2)*x+4*x-4)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(2*x^2+x)*exp(2)^2+(8*x^2+4*x)*exp(2
)+8*x^2-4*x)/((x^2*exp(2)^2+4*x^2*exp(2)+4*x^2-4*x)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(-2*x*exp(2)^2-8*exp(2)*x
-8*x+8)*log(5)+x^3*exp(2)^2+4*x^3*exp(2)+4*x^3-4*x^2),x, algorithm="maxima")

[Out]

log(x) + log((x^2 + x*log(x*(e^4 + 4*e^2 + 4) - 4) - 2*log(5))/x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\log \left (-x^{2} - x \log \left (x e^{4} + 4 \, x e^{2} + 4 \, x - 4\right ) + 2 \, \log \left (5\right )\right ) \]

[In]

integrate(((x*exp(2)^2+4*exp(2)*x+4*x-4)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(2*x^2+x)*exp(2)^2+(8*x^2+4*x)*exp(2
)+8*x^2-4*x)/((x^2*exp(2)^2+4*x^2*exp(2)+4*x^2-4*x)*log(x*exp(2)^2+4*exp(2)*x+4*x-4)+(-2*x*exp(2)^2-8*exp(2)*x
-8*x+8)*log(5)+x^3*exp(2)^2+4*x^3*exp(2)+4*x^3-4*x^2),x, algorithm="giac")

[Out]

log(-x^2 - x*log(x*e^4 + 4*x*e^2 + 4*x - 4) + 2*log(5))

Mupad [B] (verification not implemented)

Time = 12.57 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-4 x+8 x^2+e^4 \left (x+2 x^2\right )+e^2 \left (4 x+8 x^2\right )+\left (-4+4 x+4 e^2 x+e^4 x\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )}{-4 x^2+4 x^3+4 e^2 x^3+e^4 x^3+\left (8-8 x-8 e^2 x-2 e^4 x\right ) \log (5)+\left (-4 x+4 x^2+4 e^2 x^2+e^4 x^2\right ) \log \left (-4+4 x+4 e^2 x+e^4 x\right )} \, dx=\ln \left (\frac {x^2-\ln \left (25\right )+x\,\ln \left (4\,x+4\,x\,{\mathrm {e}}^2+x\,{\mathrm {e}}^4-4\right )}{x}\right )+\ln \left (x\right ) \]

[In]

int((log(4*x + 4*x*exp(2) + x*exp(4) - 4)*(4*x + 4*x*exp(2) + x*exp(4) - 4) - 4*x + exp(2)*(4*x + 8*x^2) + exp
(4)*(x + 2*x^2) + 8*x^2)/(log(4*x + 4*x*exp(2) + x*exp(4) - 4)*(4*x^2*exp(2) - 4*x + x^2*exp(4) + 4*x^2) - log
(5)*(8*x + 8*x*exp(2) + 2*x*exp(4) - 8) + 4*x^3*exp(2) + x^3*exp(4) - 4*x^2 + 4*x^3),x)

[Out]

log((x^2 - log(25) + x*log(4*x + 4*x*exp(2) + x*exp(4) - 4))/x) + log(x)