\(\int \frac {-6-3 x^2-24 e^{\log ^2(4+4 x^2+x^4)} x \log (4+4 x^2+x^4)}{(10+10 x+5 x^2+5 x^3+e^{\log ^2(4+4 x^2+x^4)} (10+5 x^2)) \log ^2(1+e^{\log ^2(4+4 x^2+x^4)}+x)} \, dx\) [3312]

Optimal result

Integrand size = 98, antiderivative size = 22 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \log \left (1+e^{\log ^2\left (\left (2+x^2\right )^2\right )}+x\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6818} \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \log \left (e^{\log ^2\left (x^4+4 x^2+4\right )}+x+1\right )} \]

[In]

Int[(-6 - 3*x^2 - 24*E^Log[4 + 4*x^2 + x^4]^2*x*Log[4 + 4*x^2 + x^4])/((10 + 10*x + 5*x^2 + 5*x^3 + E^Log[4 +
4*x^2 + x^4]^2*(10 + 5*x^2))*Log[1 + E^Log[4 + 4*x^2 + x^4]^2 + x]^2),x]

[Out]

3/(5*Log[1 + E^Log[4 + 4*x^2 + x^4]^2 + x])

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}& = \frac {3}{5 \log \left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \log \left (1+e^{\log ^2\left (\left (2+x^2\right )^2\right )}+x\right )} \]

[In]

Integrate[(-6 - 3*x^2 - 24*E^Log[4 + 4*x^2 + x^4]^2*x*Log[4 + 4*x^2 + x^4])/((10 + 10*x + 5*x^2 + 5*x^3 + E^Lo
g[4 + 4*x^2 + x^4]^2*(10 + 5*x^2))*Log[1 + E^Log[4 + 4*x^2 + x^4]^2 + x]^2),x]

[Out]

3/(5*Log[1 + E^Log[(2 + x^2)^2]^2 + x])

Maple [A] (verified)

Time = 7.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
parallelrisch	\(\frac {3}{5 \ln \left ({\mathrm e}^{\ln \left (x^{4}+4 x^{2}+4\right )^{2}}+x +1\right )}\)	\(23\)

[In]

int((-24*x*ln(x^4+4*x^2+4)*exp(ln(x^4+4*x^2+4)^2)-3*x^2-6)/((5*x^2+10)*exp(ln(x^4+4*x^2+4)^2)+5*x^3+5*x^2+10*x
+10)/ln(exp(ln(x^4+4*x^2+4)^2)+x+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \, \log \left (x + e^{\left (\log \left (x^{4} + 4 \, x^{2} + 4\right )^{2}\right )} + 1\right )} \]

[In]

integrate((-24*x*log(x^4+4*x^2+4)*exp(log(x^4+4*x^2+4)^2)-3*x^2-6)/((5*x^2+10)*exp(log(x^4+4*x^2+4)^2)+5*x^3+5
*x^2+10*x+10)/log(exp(log(x^4+4*x^2+4)^2)+x+1)^2,x, algorithm="fricas")

[Out]

3/5/log(x + e^(log(x^4 + 4*x^2 + 4)^2) + 1)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \log {\left (x + e^{\log {\left (x^{4} + 4 x^{2} + 4 \right )}^{2}} + 1 \right )}} \]

[In]

integrate((-24*x*ln(x**4+4*x**2+4)*exp(ln(x**4+4*x**2+4)**2)-3*x**2-6)/((5*x**2+10)*exp(ln(x**4+4*x**2+4)**2)+
5*x**3+5*x**2+10*x+10)/ln(exp(ln(x**4+4*x**2+4)**2)+x+1)**2,x)

[Out]

3/(5*log(x + exp(log(x**4 + 4*x**2 + 4)**2) + 1))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \, \log \left (x + e^{\left (4 \, \log \left (x^{2} + 2\right )^{2}\right )} + 1\right )} \]

[In]

integrate((-24*x*log(x^4+4*x^2+4)*exp(log(x^4+4*x^2+4)^2)-3*x^2-6)/((5*x^2+10)*exp(log(x^4+4*x^2+4)^2)+5*x^3+5
*x^2+10*x+10)/log(exp(log(x^4+4*x^2+4)^2)+x+1)^2,x, algorithm="maxima")

[Out]

3/5/log(x + e^(4*log(x^2 + 2)^2) + 1)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5 \, \log \left (x + e^{\left (\log \left (x^{4} + 4 \, x^{2} + 4\right )^{2}\right )} + 1\right )} \]

[In]

integrate((-24*x*log(x^4+4*x^2+4)*exp(log(x^4+4*x^2+4)^2)-3*x^2-6)/((5*x^2+10)*exp(log(x^4+4*x^2+4)^2)+5*x^3+5
*x^2+10*x+10)/log(exp(log(x^4+4*x^2+4)^2)+x+1)^2,x, algorithm="giac")

[Out]

3/5/log(x + e^(log(x^4 + 4*x^2 + 4)^2) + 1)

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6-3 x^2-24 e^{\log ^2\left (4+4 x^2+x^4\right )} x \log \left (4+4 x^2+x^4\right )}{\left (10+10 x+5 x^2+5 x^3+e^{\log ^2\left (4+4 x^2+x^4\right )} \left (10+5 x^2\right )\right ) \log ^2\left (1+e^{\log ^2\left (4+4 x^2+x^4\right )}+x\right )} \, dx=\frac {3}{5\,\ln \left (x+{\mathrm {e}}^{{\ln \left (x^4+4\,x^2+4\right )}^2}+1\right )} \]

[In]

int(-(3*x^2 + 24*x*exp(log(4*x^2 + x^4 + 4)^2)*log(4*x^2 + x^4 + 4) + 6)/(log(x + exp(log(4*x^2 + x^4 + 4)^2)
+ 1)^2*(10*x + exp(log(4*x^2 + x^4 + 4)^2)*(5*x^2 + 10) + 5*x^2 + 5*x^3 + 10)),x)

[Out]

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