\(\int \frac {(4 x-4 x^3+e (2-4 x^2)) \log (6-x^2+\log (e x+x^2))}{6 x^2-x^4+e (6 x-x^3)+(e x+x^2) \log (e x+x^2)} \, dx\) [7877]

Optimal result

Integrand size = 75, antiderivative size = 16 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log ^2\left (6-x^2+\log (x (e+x))\right ) \]

[Out]

ln(ln(x*(x+exp(1)))-x^2+6)^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6818} \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log ^2\left (-x^2+\log \left (x^2+e x\right )+6\right ) \]

[In]

Int[((4*x - 4*x^3 + E*(2 - 4*x^2))*Log[6 - x^2 + Log[E*x + x^2]])/(6*x^2 - x^4 + E*(6*x - x^3) + (E*x + x^2)*L
og[E*x + x^2]),x]

[Out]

Log[6 - x^2 + Log[E*x + x^2]]^2

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \log ^2\left (6-x^2+\log \left (e x+x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log ^2\left (6-x^2+\log (x (e+x))\right ) \]

[In]

Integrate[((4*x - 4*x^3 + E*(2 - 4*x^2))*Log[6 - x^2 + Log[E*x + x^2]])/(6*x^2 - x^4 + E*(6*x - x^3) + (E*x +
x^2)*Log[E*x + x^2]),x]

[Out]

Log[6 - x^2 + Log[x*(E + x)]]^2

Maple [F]

\[\int \frac {\left (\left (-4 x^{2}+2\right ) {\mathrm e}-4 x^{3}+4 x \right ) \ln \left (\ln \left (x \,{\mathrm e}+x^{2}\right )-x^{2}+6\right )}{\left (x \,{\mathrm e}+x^{2}\right ) \ln \left (x \,{\mathrm e}+x^{2}\right )+\left (-x^{3}+6 x \right ) {\mathrm e}-x^{4}+6 x^{2}}d x\]

[In]

int(((-4*x^2+2)*exp(1)-4*x^3+4*x)*ln(ln(x*exp(1)+x^2)-x^2+6)/((x*exp(1)+x^2)*ln(x*exp(1)+x^2)+(-x^3+6*x)*exp(1
)-x^4+6*x^2),x)

[Out]

int(((-4*x^2+2)*exp(1)-4*x^3+4*x)*ln(ln(x*exp(1)+x^2)-x^2+6)/((x*exp(1)+x^2)*ln(x*exp(1)+x^2)+(-x^3+6*x)*exp(1
)-x^4+6*x^2),x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log \left (-x^{2} + \log \left (x^{2} + x e\right ) + 6\right )^{2} \]

[In]

integrate(((-4*x^2+2)*exp(1)-4*x^3+4*x)*log(log(x*exp(1)+x^2)-x^2+6)/((x*exp(1)+x^2)*log(x*exp(1)+x^2)+(-x^3+6
*x)*exp(1)-x^4+6*x^2),x, algorithm="fricas")

[Out]

log(-x^2 + log(x^2 + x*e) + 6)^2

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log {\left (- x^{2} + \log {\left (x^{2} + e x \right )} + 6 \right )}^{2} \]

[In]

integrate(((-4*x**2+2)*exp(1)-4*x**3+4*x)*ln(ln(x*exp(1)+x**2)-x**2+6)/((x*exp(1)+x**2)*ln(x*exp(1)+x**2)+(-x*
*3+6*x)*exp(1)-x**4+6*x**2),x)

[Out]

log(-x**2 + log(x**2 + E*x) + 6)**2

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log \left (-x^{2} + \log \left (x + e\right ) + \log \left (x\right ) + 6\right )^{2} \]

[In]

integrate(((-4*x^2+2)*exp(1)-4*x^3+4*x)*log(log(x*exp(1)+x^2)-x^2+6)/((x*exp(1)+x^2)*log(x*exp(1)+x^2)+(-x^3+6
*x)*exp(1)-x^4+6*x^2),x, algorithm="maxima")

[Out]

log(-x^2 + log(x + e) + log(x) + 6)^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx=\log \left (-x^{2} + \log \left (x^{2} + x e\right ) + 6\right )^{2} \]

[In]

integrate(((-4*x^2+2)*exp(1)-4*x^3+4*x)*log(log(x*exp(1)+x^2)-x^2+6)/((x*exp(1)+x^2)*log(x*exp(1)+x^2)+(-x^3+6
*x)*exp(1)-x^4+6*x^2),x, algorithm="giac")

[Out]

log(-x^2 + log(x^2 + x*e) + 6)^2

Mupad [B] (verification not implemented)

Time = 13.65 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\left (4 x-4 x^3+e \left (2-4 x^2\right )\right ) \log \left (6-x^2+\log \left (e x+x^2\right )\right )}{6 x^2-x^4+e \left (6 x-x^3\right )+\left (e x+x^2\right ) \log \left (e x+x^2\right )} \, dx={\ln \left (\ln \left (x^2+\mathrm {e}\,x\right )-x^2+6\right )}^2 \]

[In]

int(-(log(log(x*exp(1) + x^2) - x^2 + 6)*(exp(1)*(4*x^2 - 2) - 4*x + 4*x^3))/(exp(1)*(6*x - x^3) + log(x*exp(1
) + x^2)*(x*exp(1) + x^2) + 6*x^2 - x^4),x)

[Out]

log(log(x*exp(1) + x^2) - x^2 + 6)^2