\(\int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx\) [7845]

Optimal result

Integrand size = 33, antiderivative size = 17 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (3-e^3+\frac {x}{2}+\log (2+x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6873, 6816} \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (-x-2 \log (x+2)-2 \left (3-e^3\right )\right ) \]

[In]

Int[(4 + x)/(12 + E^3*(-4 - 2*x) + 8*x + x^2 + (4 + 2*x)*Log[2 + x]),x]

[Out]

Log[-2*(3 - E^3) - x - 2*Log[2 + x]]

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 {4+x}{(2+x) \left (6 \left (1-\frac {e^3}{3}\right )+x+2 \log (2+x)\right )} \, dx \\ & = \log \left (-2 \left (3-e^3\right )-x-2 \log (2+x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (6-2 e^3+x+2 \log (2+x)\right ) \]

[In]

Integrate[(4 + x)/(12 + E^3*(-4 - 2*x) + 8*x + x^2 + (4 + 2*x)*Log[2 + x]),x]

[Out]

Log[6 - 2*E^3 + x + 2*Log[2 + x]]

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\ln \left (-2 \,{\mathrm e}^{3}+2 \ln \left (2+x \right )+6+x \right )\)	\(15\)
default	\(\ln \left (-2 \,{\mathrm e}^{3}+2 \ln \left (2+x \right )+6+x \right )\)	\(15\)
risch	\(\ln \left (\ln \left (2+x \right )+\frac {x}{2}-{\mathrm e}^{3}+3\right )\)	\(15\)
parallelrisch	\(\ln \left (-2 \,{\mathrm e}^{3}+2 \ln \left (2+x \right )+6+x \right )\)	\(15\)
norman	\(\ln \left (-x +2 \,{\mathrm e}^{3}-2 \ln \left (2+x \right )-6\right )\)	\(17\)

[In]

int((4+x)/((4+2*x)*ln(2+x)+(-2*x-4)*exp(3)+x^2+8*x+12),x,method=_RETURNVERBOSE)

[Out]

ln(-2*exp(3)+2*ln(2+x)+6+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (x - 2 \, e^{3} + 2 \, \log \left (x + 2\right ) + 6\right ) \]

[In]

integrate((4+x)/((4+2*x)*log(2+x)+(-2*x-4)*exp(3)+x^2+8*x+12),x, algorithm="fricas")

[Out]

log(x - 2*e^3 + 2*log(x + 2) + 6)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log {\left (\frac {x}{2} + \log {\left (x + 2 \right )} - e^{3} + 3 \right )} \]

[In]

integrate((4+x)/((4+2*x)*ln(2+x)+(-2*x-4)*exp(3)+x**2+8*x+12),x)

[Out]

log(x/2 + log(x + 2) - exp(3) + 3)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (\frac {1}{2} \, x - e^{3} + \log \left (x + 2\right ) + 3\right ) \]

[In]

integrate((4+x)/((4+2*x)*log(2+x)+(-2*x-4)*exp(3)+x^2+8*x+12),x, algorithm="maxima")

[Out]

log(1/2*x - e^3 + log(x + 2) + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\log \left (x - 2 \, e^{3} + 2 \, \log \left (x + 2\right ) + 6\right ) \]

[In]

integrate((4+x)/((4+2*x)*log(2+x)+(-2*x-4)*exp(3)+x^2+8*x+12),x, algorithm="giac")

[Out]

log(x - 2*e^3 + 2*log(x + 2) + 6)

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {4+x}{12+e^3 (-4-2 x)+8 x+x^2+(4+2 x) \log (2+x)} \, dx=\ln \left (x+2\,\ln \left (x+2\right )-2\,{\mathrm {e}}^3+6\right ) \]

[In]

int((x + 4)/(8*x + x^2 + log(x + 2)*(2*x + 4) - exp(3)*(2*x + 4) + 12),x)

[Out]

log(x + 2*log(x + 2) - 2*exp(3) + 6)