\(\int \frac {(4+5 x+x^2) \log (\frac {4+x}{2+2 x})+\log (2 x) (3 x+(-4-5 x-x^2) \log (\frac {4+x}{2+2 x}))}{(4 x^2+5 x^3+x^4) \log ^2(\frac {4+x}{2+2 x})} \, dx\) [9879]

Optimal result

Integrand size = 85, antiderivative size = 25 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-7+\frac {\log (2 x)}{x \log \left (\frac {4+x}{2 (1+x)}\right )} \]

[Out]

ln(2*x)/ln((4+x)/(2+2*x))/x-7

Rubi [F]

\[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \]

[In]

Int[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^
2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x)]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{x^2 \left (4+5 x+x^2\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = \int \frac {\log (2 x) \left (\frac {3 x}{4+5 x+x^2}-\log \left (\frac {4+x}{2+2 x}\right )\right )+\log \left (\frac {4+x}{2+2 x}\right )}{x^2 \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = \int \left (\frac {3 \log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx \\ & = 3 \int \frac {\log (2 x)}{x (1+x) (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1-\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx \\ & = 3 \int \left (\frac {\log (2 x)}{4 x \log ^2\left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{3 (1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}+\frac {\log (2 x)}{12 (4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )}\right ) \, dx+\int \left (\frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}-\frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {\log (2 x)}{(4+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\frac {3}{4} \int \frac {\log (2 x)}{x \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{(1+x) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx+\int \frac {1}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx-\int \frac {\log (2 x)}{x^2 \log \left (\frac {4+x}{2+2 x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log (2 x)}{x \log \left (\frac {4+x}{2+2 x}\right )} \]

[In]

Integrate[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/
((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x)]^2),x]

[Out]

Log[2*x]/(x*Log[(4 + x)/(2 + 2*x)])

Maple [A] (verified)

Time = 11.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {\ln \left (2 x \right )}{\ln \left (\frac {4+x}{2+2 x}\right ) x}\)	\(22\)
risch	\(\frac {2 i \ln \left (2 x \right )}{x \left (\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )-\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{3}-2 i \ln \left (2\right )+2 i \ln \left (4+x \right )-2 i \ln \left (1+x \right )\right )}\)	\(129\)

[In]

int((((-x^2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x^2+5*x+4)*ln((4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/ln((4+x)/(2
+2*x))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log \left (2 \, x\right )}{x \log \left (\frac {x + 4}{2 \, {\left (x + 1\right )}}\right )} \]

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2+2*x))^2,x, algorithm="fricas")

[Out]

log(2*x)/(x*log(1/2*(x + 4)/(x + 1)))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log {\left (2 x \right )}}{x \log {\left (\frac {x + 4}{2 x + 2} \right )}} \]

[In]

integrate((((-x**2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x**2+5*x+4)*ln((4+x)/(2+2*x)))/(x**4+5*x**3+4*x**2)/
ln((4+x)/(2+2*x))**2,x)

[Out]

log(2*x)/(x*log((x + 4)/(2*x + 2)))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \]

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2+2*x))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \]

[In]

integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log((4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/lo
g((4+x)/(2+2*x))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\ln \left (2\,x\right )}{x\,\ln \left (\frac {x+4}{2\,x+2}\right )} \]

[In]

int((log(2*x)*(3*x - log((x + 4)/(2*x + 2))*(5*x + x^2 + 4)) + log((x + 4)/(2*x + 2))*(5*x + x^2 + 4))/(log((x
+ 4)/(2*x + 2))^2*(4*x^2 + 5*x^3 + x^4)),x)

[Out]

log(2*x)/(x*log((x + 4)/(2*x + 2)))