\(\int \frac {x+(4+8 x) \log (\frac {2+4 x}{x})}{(16+32 x) \log (\frac {2+4 x}{x})+(8 x+16 x^2) \log (\frac {2+4 x}{x}) \log (\log (\frac {2+4 x}{x}))+(x^2+2 x^3) \log (\frac {2+4 x}{x}) \log ^2(\log (\frac {2+4 x}{x}))} \, dx\) [6758]

Optimal result

Integrand size = 102, antiderivative size = 19 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=-1+\frac {x}{4+x \log \left (\log \left (4+\frac {2}{x}\right )\right )} \]

[Out]

-1+x/(4+ln(ln(2/x+4))*x)

Rubi [F]

\[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx \]

[In]

Int[(x + (4 + 8*x)*Log[(2 + 4*x)/x])/((16 + 32*x)*Log[(2 + 4*x)/x] + (8*x + 16*x^2)*Log[(2 + 4*x)/x]*Log[Log[(
2 + 4*x)/x]] + (x^2 + 2*x^3)*Log[(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]]^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{4+x \log \left (\log \left (4+\frac {2}{x}\right )\right )} \]

[In]

Integrate[(x + (4 + 8*x)*Log[(2 + 4*x)/x])/((16 + 32*x)*Log[(2 + 4*x)/x] + (8*x + 16*x^2)*Log[(2 + 4*x)/x]*Log
[Log[(2 + 4*x)/x]] + (x^2 + 2*x^3)*Log[(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]]^2),x]

[Out]

x/(4 + x*Log[Log[4 + 2/x]])

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {x}{x \ln \left (\ln \left (\frac {4 x +2}{x}\right )\right )+4}\)	\(21\)
default	\(\frac {1}{-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )}\)	\(29\)
parts	\(\frac {4 \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right ) \left (1+2 x \right )}{\left (\frac {4 \left (1+2 x \right ) \ln \left (2\right )}{x}+\frac {4 \left (1+2 x \right ) \ln \left (\frac {1+2 x}{x}\right )}{x}+1\right ) x \left (-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )\right )}+\frac {1}{\left (\frac {4 \left (1+2 x \right ) \ln \left (2\right )}{x}+\frac {4 \left (1+2 x \right ) \ln \left (\frac {1+2 x}{x}\right )}{x}+1\right ) \left (-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )\right )}\)	\(154\)

[In]

int(((8*x+4)*ln((4*x+2)/x)+x)/((2*x^3+x^2)*ln((4*x+2)/x)*ln(ln((4*x+2)/x))^2+(16*x^2+8*x)*ln((4*x+2)/x)*ln(ln(
(4*x+2)/x))+(32*x+16)*ln((4*x+2)/x)),x,method=_RETURNVERBOSE)

[Out]

x/(x*ln(ln(2*(1+2*x)/x))+4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 4} \]

[In]

integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log((4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2
)/x)*log(log((4*x+2)/x))+(32*x+16)*log((4*x+2)/x)),x, algorithm="fricas")

[Out]

x/(x*log(log(2*(2*x + 1)/x)) + 4)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log {\left (\log {\left (\frac {4 x + 2}{x} \right )} \right )} + 4} \]

[In]

integrate(((8*x+4)*ln((4*x+2)/x)+x)/((2*x**3+x**2)*ln((4*x+2)/x)*ln(ln((4*x+2)/x))**2+(16*x**2+8*x)*ln((4*x+2)
/x)*ln(ln((4*x+2)/x))+(32*x+16)*ln((4*x+2)/x)),x)

[Out]

x/(x*log(log((4*x + 2)/x)) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log \left (\log \left (2\right ) + \log \left (2 \, x + 1\right ) - \log \left (x\right )\right ) + 4} \]

[In]

integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log((4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2
)/x)*log(log((4*x+2)/x))+(32*x+16)*log((4*x+2)/x)),x, algorithm="maxima")

[Out]

x/(x*log(log(2) + log(2*x + 1) - log(x)) + 4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (19) = 38\).

Time = 0.43 (sec) , antiderivative size = 345, normalized size of antiderivative = 18.16 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {8 \, x^{2} \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 8 \, x^{2} \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + x^{2} \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + 4 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 4 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )}{8 \, x^{2} \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - 8 \, x^{2} \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + x^{2} \log \left (4 \, x + 2\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - x^{2} \log \left (x\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 4 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - 4 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 32 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 32 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + 4 \, x \log \left (4 \, x + 2\right ) - 4 \, x \log \left (x\right ) + 16 \, \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 16 \, \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )} \]

[In]

integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log((4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2
)/x)*log(log((4*x+2)/x))+(32*x+16)*log((4*x+2)/x)),x, algorithm="giac")

[Out]

(8*x^2*log(4*x + 2)*log(2*(2*x + 1)/x) - 8*x^2*log(x)*log(2*(2*x + 1)/x) + x^2*log(2*(2*x + 1)/x) + 4*x*log(4*
x + 2)*log(2*(2*x + 1)/x) - 4*x*log(x)*log(2*(2*x + 1)/x))/(8*x^2*log(4*x + 2)*log(2*(2*x + 1)/x)*log(log(2*(2
*x + 1)/x)) - 8*x^2*log(x)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) + x^2*log(4*x + 2)*log(log(2*(2*x + 1)/x
)) - x^2*log(x)*log(log(2*(2*x + 1)/x)) + 4*x*log(4*x + 2)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) - 4*x*lo
g(x)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) + 32*x*log(4*x + 2)*log(2*(2*x + 1)/x) - 32*x*log(x)*log(2*(2*
x + 1)/x) + 4*x*log(4*x + 2) - 4*x*log(x) + 16*log(4*x + 2)*log(2*(2*x + 1)/x) - 16*log(x)*log(2*(2*x + 1)/x))

Mupad [B] (verification not implemented)

Time = 13.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 8.79 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x\,{\left (\ln \left (\frac {4\,x+2}{x}\right )+2\,x\,\ln \left (\frac {4\,x+2}{x}\right )\right )}^2\,\left (x+4\,\ln \left (\frac {4\,x+2}{x}\right )+8\,x\,\ln \left (\frac {4\,x+2}{x}\right )\right )}{\ln \left (\frac {4\,x+2}{x}\right )\,\left (2\,x+1\right )\,\left (x\,\ln \left (\ln \left (\frac {4\,x+2}{x}\right )\right )+4\right )\,\left (16\,x^2\,{\ln \left (\frac {4\,x+2}{x}\right )}^2+2\,x^2\,\ln \left (\frac {4\,x+2}{x}\right )+16\,x\,{\ln \left (\frac {4\,x+2}{x}\right )}^2+x\,\ln \left (\frac {4\,x+2}{x}\right )+4\,{\ln \left (\frac {4\,x+2}{x}\right )}^2\right )} \]

[In]

int((x + log((4*x + 2)/x)*(8*x + 4))/(log((4*x + 2)/x)*(32*x + 16) + log(log((4*x + 2)/x))^2*log((4*x + 2)/x)*
(x^2 + 2*x^3) + log(log((4*x + 2)/x))*log((4*x + 2)/x)*(8*x + 16*x^2)),x)

[Out]

(x*(log((4*x + 2)/x) + 2*x*log((4*x + 2)/x))^2*(x + 4*log((4*x + 2)/x) + 8*x*log((4*x + 2)/x)))/(log((4*x + 2)
/x)*(2*x + 1)*(x*log(log((4*x + 2)/x)) + 4)*(16*x*log((4*x + 2)/x)^2 + 2*x^2*log((4*x + 2)/x) + 4*log((4*x + 2
)/x)^2 + 16*x^2*log((4*x + 2)/x)^2 + x*log((4*x + 2)/x)))