3.280 \(\int \frac{x+x^2}{(4+x) (-4+x^2)} \, dx\)

Optimal. Leaf size=15 \[ \log (x+4)-\frac{1}{2} \tanh ^{-1}\left (\frac{x}{2}\right ) \]

[Out]

-ArcTanh[x/2]/2 + Log[4 + x]

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Rubi [A]  time = 0.0604497, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1593, 1629, 207} \[ \log (x+4)-\frac{1}{2} \tanh ^{-1}\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x + x^2)/((4 + x)*(-4 + x^2)),x]

[Out]

-ArcTanh[x/2]/2 + Log[4 + x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x+x^2}{(4+x) \left (-4+x^2\right )} \, dx &=\int \frac{x (1+x)}{(4+x) \left (-4+x^2\right )} \, dx\\ &=\int \left (\frac{1}{4+x}+\frac{1}{-4+x^2}\right ) \, dx\\ &=\log (4+x)+\int \frac{1}{-4+x^2} \, dx\\ &=-\frac{1}{2} \tanh ^{-1}\left (\frac{x}{2}\right )+\log (4+x)\\ \end{align*}

Mathematica [A]  time = 0.0057477, size = 23, normalized size = 1.53 \[ \frac{1}{4} \log (2-x)-\frac{1}{4} \log (x+2)+\log (x+4) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[2 - x]/4 - Log[2 + x]/4 + Log[4 + x]

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Maple [A]  time = 0.008, size = 18, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( 2+x \right ) }{4}}+\ln \left ( 4+x \right ) +{\frac{\ln \left ( -2+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+x)/(4+x)/(x^2-4),x)

[Out]

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

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Maxima [A]  time = 1.48995, size = 23, normalized size = 1.53 \begin{align*} \log \left (x + 4\right ) - \frac{1}{4} \, \log \left (x + 2\right ) + \frac{1}{4} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x)/(4+x)/(x^2-4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.66527, size = 62, normalized size = 4.13 \begin{align*} \log \left (x + 4\right ) - \frac{1}{4} \, \log \left (x + 2\right ) + \frac{1}{4} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x)/(4+x)/(x^2-4),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.122952, size = 17, normalized size = 1.13 \begin{align*} \frac{\log{\left (x - 2 \right )}}{4} - \frac{\log{\left (x + 2 \right )}}{4} + \log{\left (x + 4 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26201, size = 27, normalized size = 1.8 \begin{align*} \log \left ({\left | x + 4 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | x + 2 \right |}\right ) + \frac{1}{4} \, \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+x)/(4+x)/(x^2-4),x, algorithm="giac")

[Out]

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