\(\int \frac {4 x^2+(2+2 x^2) \log (\frac {1}{1+x^2})}{(2+2 x^2+(1+x^2) \log (6)) \log ^2(\frac {1}{1+x^2})} \, dx\) [1579]

Optimal result

Integrand size = 50, antiderivative size = 19 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )} \]

[Out]

2*x/(ln(6)+2)/ln(1/(x^2+1))

Rubi [F]

\[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx \]

[In]

Int[(4*x^2 + (2 + 2*x^2)*Log[(1 + x^2)^(-1)])/((2 + 2*x^2 + (1 + x^2)*Log[6])*Log[(1 + x^2)^(-1)]^2),x]

[Out]

(4*Defer[Int][x^2/((1 + x^2)*Log[(1 + x^2)^(-1)]^2), x])/(2 + Log[6]) + (2*Defer[Int][Log[(1 + x^2)^(-1)]^(-1)
, x])/(2 + Log[6])

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+\log (6)+x^2 (2+\log (6))\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx \\ & = \int \left (\frac {4 x^2}{\left (1+x^2\right ) (2+\log (6)) \log ^2\left (\frac {1}{1+x^2}\right )}+\frac {2}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )}\right ) \, dx \\ & = \frac {2 \int \frac {1}{\log \left (\frac {1}{1+x^2}\right )} \, dx}{2+\log (6)}+\frac {4 \int \frac {x^2}{\left (1+x^2\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx}{2+\log (6)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )} \]

[In]

Integrate[(4*x^2 + (2 + 2*x^2)*Log[(1 + x^2)^(-1)])/((2 + 2*x^2 + (1 + x^2)*Log[6])*Log[(1 + x^2)^(-1)]^2),x]

[Out]

(2*x)/((2 + Log[6])*Log[(1 + x^2)^(-1)])

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
norman	\(\frac {2 x}{\left (\ln \left (6\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\)	\(20\)
parallelrisch	\(\frac {2 x}{\left (\ln \left (6\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\)	\(20\)
risch	\(\frac {2 x}{\left (\ln \left (2\right )+\ln \left (3\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\)	\(22\)

[In]

int(((2*x^2+2)*ln(1/(x^2+1))+4*x^2)/((x^2+1)*ln(6)+2*x^2+2)/ln(1/(x^2+1))^2,x,method=_RETURNVERBOSE)

[Out]

2*x/(ln(6)+2)/ln(1/(x^2+1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 \, x}{{\left (\log \left (6\right ) + 2\right )} \log \left (\frac {1}{x^{2} + 1}\right )} \]

[In]

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

[Out]

2*x/((log(6) + 2)*log(1/(x^2 + 1)))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{\left (\log {\left (6 \right )} + 2\right ) \log {\left (\frac {1}{x^{2} + 1} \right )}} \]

[In]

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

[Out]

2*x/((log(6) + 2)*log(1/(x**2 + 1)))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2 \, x}{{\left (\log \left (3\right ) + \log \left (2\right ) + 2\right )} \log \left (x^{2} + 1\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2 \, x}{\log \left (6\right ) \log \left (x^{2} + 1\right ) + 2 \, \log \left (x^{2} + 1\right )} \]

[In]

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

[Out]

-2*x/(log(6)*log(x^2 + 1) + 2*log(x^2 + 1))

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2\,x}{\ln \left (x^2+1\right )\,\left (\ln \left (6\right )+2\right )} \]

[In]

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

[Out]

-(2*x)/(log(x^2 + 1)*(log(6) + 2))