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

Optimal. Leaf size=17 \[ \log \left (3-e^3+\frac {x}{2}+\log (2+x)\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6741, 6684} \begin {gather*} \log \left (-x-2 \log (x+2)-2 \left (3-e^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 15, normalized size = 0.88 \begin {gather*} \log \left (6-2 e^3+x+2 \log (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.68, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x - 2 \, e^{3} + 2 \, \log \left (x + 2\right ) + 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+x)/((2*x+4)*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)

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giac [A]  time = 0.23, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x - 2 \, e^{3} + 2 \, \log \left (x + 2\right ) + 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+x)/((2*x+4)*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)

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maple [A]  time = 0.06, size = 15, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 14, normalized size = 0.82 \begin {gather*} \log \left (\frac {1}{2} \, x - e^{3} + \log \left (x + 2\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4+x)/((2*x+4)*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)

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mupad [B]  time = 6.03, size = 14, normalized size = 0.82 \begin {gather*} \ln \left (x+2\,\ln \left (x+2\right )-2\,{\mathrm {e}}^3+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.20, size = 14, normalized size = 0.82 \begin {gather*} \log {\left (\frac {x}{2} + \log {\left (x + 2 \right )} - e^{3} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x/2 + log(x + 2) - exp(3) + 3)

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