\(\int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log (e^2+2 e x+x^2)}{12 e x^3+12 x^4+(e x+x^2) \log (e^2+2 e x+x^2)} \, dx\) [60]

Optimal result

Integrand size = 70, antiderivative size = 24 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=-4+\log \left (\frac {\log (5) \left (3+\frac {\log \left ((e+x)^2\right )}{4 x^2}\right )}{x}\right ) \]

[Out]

ln((3+1/4*ln((x+exp(1))^2)/x^2)/x*ln(5))-4

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6820, 6874, 6816} \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=\log \left (12 x^2+\log \left ((x+e)^2\right )\right )-3 \log (x) \]

[In]

Int[(2*x - 12*E*x^2 - 12*x^3 + (-3*E - 3*x)*Log[E^2 + 2*E*x + x^2])/(12*E*x^3 + 12*x^4 + (E*x + x^2)*Log[E^2 +
2*E*x + x^2]),x]

[Out]

-3*Log[x] + Log[12*x^2 + Log[(E + x)^2]]

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]]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x \left (-1+6 e x+6 x^2\right )-3 (e+x) \log \left ((e+x)^2\right )}{x (e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )} \, dx \\ & = \int \left (-\frac {3}{x}+\frac {2 \left (1+12 e x+12 x^2\right )}{(e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )}\right ) \, dx \\ & = -3 \log (x)+2 \int \frac {1+12 e x+12 x^2}{(e+x) \left (12 x^2+\log \left ((e+x)^2\right )\right )} \, dx \\ & = -3 \log (x)+\log \left (12 x^2+\log \left ((e+x)^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=-3 \log (x)+\log \left (12 x^2+\log \left ((e+x)^2\right )\right ) \]

[In]

Integrate[(2*x - 12*E*x^2 - 12*x^3 + (-3*E - 3*x)*Log[E^2 + 2*E*x + x^2])/(12*E*x^3 + 12*x^4 + (E*x + x^2)*Log
[E^2 + 2*E*x + x^2]),x]

[Out]

-3*Log[x] + Log[12*x^2 + Log[(E + x)^2]]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
risch	\(-3 \ln \left (x \right )+\ln \left (12 x^{2}+\ln \left ({\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}\right )\right )\)	\(25\)
norman	\(-3 \ln \left (x \right )+\ln \left (12 x^{2}+\ln \left ({\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}\right )\right )\)	\(27\)
parallelrisch	\(-3 \ln \left (x \right )+\ln \left (x^{2}+\frac {\ln \left ({\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}\right )}{12}\right )\)	\(27\)

[In]

int(((-3*exp(1)-3*x)*ln(exp(1)^2+2*x*exp(1)+x^2)-12*x^2*exp(1)-12*x^3+2*x)/((x*exp(1)+x^2)*ln(exp(1)^2+2*x*exp
(1)+x^2)+12*x^3*exp(1)+12*x^4),x,method=_RETURNVERBOSE)

[Out]

-3*ln(x)+ln(12*x^2+ln(exp(2)+2*x*exp(1)+x^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=\log \left (12 \, x^{2} + \log \left (x^{2} + 2 \, x e + e^{2}\right )\right ) - 3 \, \log \left (x\right ) \]

[In]

integrate(((-3*exp(1)-3*x)*log(exp(1)^2+2*x*exp(1)+x^2)-12*x^2*exp(1)-12*x^3+2*x)/((x*exp(1)+x^2)*log(exp(1)^2
+2*x*exp(1)+x^2)+12*x^3*exp(1)+12*x^4),x, algorithm="fricas")

[Out]

log(12*x^2 + log(x^2 + 2*x*e + e^2)) - 3*log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=- 3 \log {\left (x \right )} + \log {\left (12 x^{2} + \log {\left (x^{2} + 2 e x + e^{2} \right )} \right )} \]

[In]

integrate(((-3*exp(1)-3*x)*ln(exp(1)**2+2*x*exp(1)+x**2)-12*x**2*exp(1)-12*x**3+2*x)/((x*exp(1)+x**2)*ln(exp(1
)**2+2*x*exp(1)+x**2)+12*x**3*exp(1)+12*x**4),x)

[Out]

-3*log(x) + log(12*x**2 + log(x**2 + 2*E*x + exp(2)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=\log \left (6 \, x^{2} + \log \left (x + e\right )\right ) - 3 \, \log \left (x\right ) \]

[In]

integrate(((-3*exp(1)-3*x)*log(exp(1)^2+2*x*exp(1)+x^2)-12*x^2*exp(1)-12*x^3+2*x)/((x*exp(1)+x^2)*log(exp(1)^2
+2*x*exp(1)+x^2)+12*x^3*exp(1)+12*x^4),x, algorithm="maxima")

[Out]

log(6*x^2 + log(x + e)) - 3*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=\log \left (-12 \, x^{2} - \log \left (x^{2} + 2 \, x e + e^{2}\right )\right ) - 3 \, \log \left (x\right ) \]

[In]

integrate(((-3*exp(1)-3*x)*log(exp(1)^2+2*x*exp(1)+x^2)-12*x^2*exp(1)-12*x^3+2*x)/((x*exp(1)+x^2)*log(exp(1)^2
+2*x*exp(1)+x^2)+12*x^3*exp(1)+12*x^4),x, algorithm="giac")

[Out]

log(-12*x^2 - log(x^2 + 2*x*e + e^2)) - 3*log(x)

Mupad [B] (verification not implemented)

Time = 8.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {2 x-12 e x^2-12 x^3+(-3 e-3 x) \log \left (e^2+2 e x+x^2\right )}{12 e x^3+12 x^4+\left (e x+x^2\right ) \log \left (e^2+2 e x+x^2\right )} \, dx=\ln \left (\frac {\ln \left ({\left (x+\mathrm {e}\right )}^2\right )}{12}+x^2\right )-3\,\ln \left (x\right ) \]

[In]

int(-(12*x^2*exp(1) - 2*x + log(exp(2) + 2*x*exp(1) + x^2)*(3*x + 3*exp(1)) + 12*x^3)/(log(exp(2) + 2*x*exp(1)
+ x^2)*(x*exp(1) + x^2) + 12*x^3*exp(1) + 12*x^4),x)

[Out]

log(log((x + exp(1))^2)/12 + x^2) - 3*log(x)