\(\int \frac {8 x+13 x^2+4 x^3+(-x^2-x^3) \log (\frac {x}{1+x})}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+(-8 x-16 x^2-10 x^3-2 x^4) \log (\frac {x}{1+x})+(x^2+x^3) \log ^2(\frac {x}{1+x})} \, dx\) [7865]

Optimal result

Integrand size = 106, antiderivative size = 24 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^2}{4+x+x \left (3+x-\log \left (\frac {x}{1+x}\right )\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx \]

[In]

Int[(8*x + 13*x^2 + 4*x^3 + (-x^2 - x^3)*Log[x/(1 + x)])/(16 + 48*x + 56*x^2 + 32*x^3 + 9*x^4 + x^5 + (-8*x -
16*x^2 - 10*x^3 - 2*x^4)*Log[x/(1 + x)] + (x^2 + x^3)*Log[x/(1 + x)]^2),x]

[Out]

-Defer[Int][(4 + 4*x + x^2 - x*Log[x/(1 + x)])^(-2), x] + 5*Defer[Int][x/(4 + 4*x + x^2 - x*Log[x/(1 + x)])^2,
x] - Defer[Int][x^3/(4 + 4*x + x^2 - x*Log[x/(1 + x)])^2, x] + Defer[Int][1/((1 + x)*(4 + 4*x + x^2 - x*Log[x
/(1 + x)])^2), x] + Defer[Int][x/(4 + 4*x + x^2 - x*Log[x/(1 + x)]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (8+13 x+4 x^2-x (1+x) \log \left (\frac {x}{1+x}\right )\right )}{(1+x) \left ((2+x)^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx \\ & = \int \left (-\frac {x \left (-4-5 x+x^2+x^3\right )}{(1+x) \left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2}+\frac {x}{4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )}\right ) \, dx \\ & = -\int \frac {x \left (-4-5 x+x^2+x^3\right )}{(1+x) \left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx+\int \frac {x}{4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )} \, dx \\ & = \int \frac {x}{4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )} \, dx-\int \left (\frac {1}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2}-\frac {5 x}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2}+\frac {x^3}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2}-\frac {1}{(1+x) \left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2}\right ) \, dx \\ & = 5 \int \frac {x}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx-\int \frac {1}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx-\int \frac {x^3}{\left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx+\int \frac {1}{(1+x) \left (4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )\right )^2} \, dx+\int \frac {x}{4+4 x+x^2-x \log \left (\frac {x}{1+x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^2}{(2+x)^2-x \log \left (\frac {x}{1+x}\right )} \]

[In]

Integrate[(8*x + 13*x^2 + 4*x^3 + (-x^2 - x^3)*Log[x/(1 + x)])/(16 + 48*x + 56*x^2 + 32*x^3 + 9*x^4 + x^5 + (-
8*x - 16*x^2 - 10*x^3 - 2*x^4)*Log[x/(1 + x)] + (x^2 + x^3)*Log[x/(1 + x)]^2),x]

[Out]

x^2/((2 + x)^2 - x*Log[x/(1 + x)])

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {x^{2}}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\)	\(26\)
parallelrisch	\(\frac {x^{2}}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\)	\(26\)
norman	\(\frac {-4 x +x \ln \left (\frac {x}{1+x}\right )-4}{x^{2}-x \ln \left (\frac {x}{1+x}\right )+4 x +4}\)	\(38\)
derivativedivides	\(\frac {\left (1-\frac {1}{1+x}\right )^{2}}{\left (1-\frac {1}{1+x}\right )^{2} \ln \left (1-\frac {1}{1+x}\right )+\left (1-\frac {1}{1+x}\right )^{2}-\left (1-\frac {1}{1+x}\right ) \ln \left (1-\frac {1}{1+x}\right )+\frac {4}{1+x}}\)	\(77\)
default	\(\frac {\left (1-\frac {1}{1+x}\right )^{2}}{\left (1-\frac {1}{1+x}\right )^{2} \ln \left (1-\frac {1}{1+x}\right )+\left (1-\frac {1}{1+x}\right )^{2}-\left (1-\frac {1}{1+x}\right ) \ln \left (1-\frac {1}{1+x}\right )+\frac {4}{1+x}}\)	\(77\)

[In]

int(((-x^3-x^2)*ln(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*ln(x/(1+x))^2+(-2*x^4-10*x^3-16*x^2-8*x)*ln(x/(1+x))+
x^5+9*x^4+32*x^3+56*x^2+48*x+16),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^{2}}{x^{2} - x \log \left (\frac {x}{x + 1}\right ) + 4 \, x + 4} \]

[In]

integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(
x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+16),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=- \frac {x^{2}}{- x^{2} + x \log {\left (\frac {x}{x + 1} \right )} - 4 x - 4} \]

[In]

integrate(((-x**3-x**2)*ln(x/(1+x))+4*x**3+13*x**2+8*x)/((x**3+x**2)*ln(x/(1+x))**2+(-2*x**4-10*x**3-16*x**2-8
*x)*ln(x/(1+x))+x**5+9*x**4+32*x**3+56*x**2+48*x+16),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\frac {x^{2}}{x^{2} + x \log \left (x + 1\right ) - x \log \left (x\right ) + 4 \, x + 4} \]

[In]

integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(
x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+16),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=-\frac {x^{2}}{{\left (x + 1\right )}^{2} {\left (\frac {x \log \left (\frac {x}{x + 1}\right )}{x + 1} - \frac {x^{2} \log \left (\frac {x}{x + 1}\right )}{{\left (x + 1\right )}^{2}} + \frac {4 \, x}{x + 1} - \frac {x^{2}}{{\left (x + 1\right )}^{2}} - 4\right )}} \]

[In]

integrate(((-x^3-x^2)*log(x/(1+x))+4*x^3+13*x^2+8*x)/((x^3+x^2)*log(x/(1+x))^2+(-2*x^4-10*x^3-16*x^2-8*x)*log(
x/(1+x))+x^5+9*x^4+32*x^3+56*x^2+48*x+16),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {8 x+13 x^2+4 x^3+\left (-x^2-x^3\right ) \log \left (\frac {x}{1+x}\right )}{16+48 x+56 x^2+32 x^3+9 x^4+x^5+\left (-8 x-16 x^2-10 x^3-2 x^4\right ) \log \left (\frac {x}{1+x}\right )+\left (x^2+x^3\right ) \log ^2\left (\frac {x}{1+x}\right )} \, dx=\int \frac {8\,x+13\,x^2+4\,x^3-\ln \left (\frac {x}{x+1}\right )\,\left (x^3+x^2\right )}{48\,x+{\ln \left (\frac {x}{x+1}\right )}^2\,\left (x^3+x^2\right )-\ln \left (\frac {x}{x+1}\right )\,\left (2\,x^4+10\,x^3+16\,x^2+8\,x\right )+56\,x^2+32\,x^3+9\,x^4+x^5+16} \,d x \]

[In]

int((8*x + 13*x^2 + 4*x^3 - log(x/(x + 1))*(x^2 + x^3))/(48*x + log(x/(x + 1))^2*(x^2 + x^3) - log(x/(x + 1))*
(8*x + 16*x^2 + 10*x^3 + 2*x^4) + 56*x^2 + 32*x^3 + 9*x^4 + x^5 + 16),x)

[Out]

int((8*x + 13*x^2 + 4*x^3 - log(x/(x + 1))*(x^2 + x^3))/(48*x + log(x/(x + 1))^2*(x^2 + x^3) - log(x/(x + 1))*
(8*x + 16*x^2 + 10*x^3 + 2*x^4) + 56*x^2 + 32*x^3 + 9*x^4 + x^5 + 16), x)