\(\int \frac {56+32 x-2 x^2+(20+12 x) \log (2 x+x^2)+(2+x) \log ^2(2 x+x^2)}{10+45 x+18 x^2-x^3+(16 x+8 x^2) \log (2 x+x^2)+(2 x+x^2) \log ^2(2 x+x^2)} \, dx\) [1516]

Optimal result

Integrand size = 92, antiderivative size = 28 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (\frac {-5+x \left (-4+x-(4+\log (x (2+x)))^2\right )}{e^3 \log (25)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6873, 6816} \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (-x^2+20 x+x \log ^2(x (x+2))+8 x \log (x (x+2))+5\right ) \]

[In]

Int[(56 + 32*x - 2*x^2 + (20 + 12*x)*Log[2*x + x^2] + (2 + x)*Log[2*x + x^2]^2)/(10 + 45*x + 18*x^2 - x^3 + (1
6*x + 8*x^2)*Log[2*x + x^2] + (2*x + x^2)*Log[2*x + x^2]^2),x]

[Out]

Log[5 + 20*x - x^2 + 8*x*Log[x*(2 + x)] + x*Log[x*(2 + x)]^2]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{(2+x) \left (5+20 x-x^2+8 x \log (x (2+x))+x \log ^2(x (2+x))\right )} \, dx \\ & = \log \left (5+20 x-x^2+8 x \log (x (2+x))+x \log ^2(x (2+x))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (5+20 x-x^2+8 x \log (x (2+x))+x \log ^2(x (2+x))\right ) \]

[In]

Integrate[(56 + 32*x - 2*x^2 + (20 + 12*x)*Log[2*x + x^2] + (2 + x)*Log[2*x + x^2]^2)/(10 + 45*x + 18*x^2 - x^
3 + (16*x + 8*x^2)*Log[2*x + x^2] + (2*x + x^2)*Log[2*x + x^2]^2),x]

[Out]

Log[5 + 20*x - x^2 + 8*x*Log[x*(2 + x)] + x*Log[x*(2 + x)]^2]

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21

method	result	size
norman	\(\ln \left (-\ln \left (x^{2}+2 x \right )^{2} x +x^{2}-8 \ln \left (x^{2}+2 x \right ) x -20 x -5\right )\)	\(34\)
parallelrisch	\(\ln \left (-\ln \left (x^{2}+2 x \right )^{2} x +x^{2}-8 \ln \left (x^{2}+2 x \right ) x -20 x -5\right )\)	\(34\)
risch	\(\ln \left (x \right )+\ln \left (\ln \left (x^{2}+2 x \right )^{2}+8 \ln \left (x^{2}+2 x \right )-\frac {x^{2}-20 x -5}{x}\right )\)	\(39\)

[In]

int(((2+x)*ln(x^2+2*x)^2+(12*x+20)*ln(x^2+2*x)-2*x^2+32*x+56)/((x^2+2*x)*ln(x^2+2*x)^2+(8*x^2+16*x)*ln(x^2+2*x
)-x^3+18*x^2+45*x+10),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(x^2+2*x)^2*x+x^2-8*ln(x^2+2*x)*x-20*x-5)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x \log \left (x^{2} + 2 \, x\right )^{2} - x^{2} + 8 \, x \log \left (x^{2} + 2 \, x\right ) + 20 \, x + 5}{x}\right ) \]

[In]

integrate(((2+x)*log(x^2+2*x)^2+(12*x+20)*log(x^2+2*x)-2*x^2+32*x+56)/((x^2+2*x)*log(x^2+2*x)^2+(8*x^2+16*x)*l
og(x^2+2*x)-x^3+18*x^2+45*x+10),x, algorithm="fricas")

[Out]

log(x) + log((x*log(x^2 + 2*x)^2 - x^2 + 8*x*log(x^2 + 2*x) + 20*x + 5)/x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x^{2} + 2 x \right )}^{2} + 8 \log {\left (x^{2} + 2 x \right )} + \frac {- x^{2} + 20 x + 5}{x} \right )} \]

[In]

integrate(((2+x)*ln(x**2+2*x)**2+(12*x+20)*ln(x**2+2*x)-2*x**2+32*x+56)/((x**2+2*x)*ln(x**2+2*x)**2+(8*x**2+16
*x)*ln(x**2+2*x)-x**3+18*x**2+45*x+10),x)

[Out]

log(x) + log(log(x**2 + 2*x)**2 + 8*log(x**2 + 2*x) + (-x**2 + 20*x + 5)/x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x \log \left (x + 2\right )^{2} + x \log \left (x\right )^{2} - x^{2} + 2 \, {\left (x \log \left (x\right ) + 4 \, x\right )} \log \left (x + 2\right ) + 8 \, x \log \left (x\right ) + 20 \, x + 5}{x}\right ) \]

[In]

integrate(((2+x)*log(x^2+2*x)^2+(12*x+20)*log(x^2+2*x)-2*x^2+32*x+56)/((x^2+2*x)*log(x^2+2*x)^2+(8*x^2+16*x)*l
og(x^2+2*x)-x^3+18*x^2+45*x+10),x, algorithm="maxima")

[Out]

log(x) + log((x*log(x + 2)^2 + x*log(x)^2 - x^2 + 2*(x*log(x) + 4*x)*log(x + 2) + 8*x*log(x) + 20*x + 5)/x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\log \left (-x \log \left (x^{2} + 2 \, x\right )^{2} + x^{2} - 8 \, x \log \left (x^{2} + 2 \, x\right ) - 20 \, x - 5\right ) \]

[In]

integrate(((2+x)*log(x^2+2*x)^2+(12*x+20)*log(x^2+2*x)-2*x^2+32*x+56)/((x^2+2*x)*log(x^2+2*x)^2+(8*x^2+16*x)*l
og(x^2+2*x)-x^3+18*x^2+45*x+10),x, algorithm="giac")

[Out]

log(-x*log(x^2 + 2*x)^2 + x^2 - 8*x*log(x^2 + 2*x) - 20*x - 5)

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {56+32 x-2 x^2+(20+12 x) \log \left (2 x+x^2\right )+(2+x) \log ^2\left (2 x+x^2\right )}{10+45 x+18 x^2-x^3+\left (16 x+8 x^2\right ) \log \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log ^2\left (2 x+x^2\right )} \, dx=\ln \left (8\,\ln \left (x^2+2\,x\right )-x+\frac {5}{x}+{\ln \left (x^2+2\,x\right )}^2+20\right )+\ln \left (x\right ) \]

[In]

int((32*x + log(2*x + x^2)*(12*x + 20) + log(2*x + x^2)^2*(x + 2) - 2*x^2 + 56)/(45*x + log(2*x + x^2)*(16*x +
8*x^2) + log(2*x + x^2)^2*(2*x + x^2) + 18*x^2 - x^3 + 10),x)

[Out]

log(8*log(2*x + x^2) - x + 5/x + log(2*x + x^2)^2 + 20) + log(x)