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]

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

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Rubi [A]  time = 0.06, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.96, size = 17, normalized size = 1.13 \[ \log \left (x + 4\right ) - \frac {1}{4} \, \log \left (x + 2\right ) + \frac {1}{4} \, \log \left (x - 2\right ) \]

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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giac [A]  time = 0.32, size = 20, normalized size = 1.33 \[ \log \left ({\left | x + 4 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 2 \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 18, normalized size = 1.20 \[ \frac {\ln \left (x -2\right )}{4}-\frac {\ln \left (x +2\right )}{4}+\ln \left (x +4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.31, size = 17, normalized size = 1.13 \[ \log \left (x + 4\right ) - \frac {1}{4} \, \log \left (x + 2\right ) + \frac {1}{4} \, \log \left (x - 2\right ) \]

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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mupad [B]  time = 0.06, size = 19, normalized size = 1.27 \[ \ln \left (x+4\right )+\frac {\mathrm {atanh}\left (\frac {90}{7\,\left (21\,x+48\right )}-\frac {8}{7}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^2)/((x^2 - 4)*(x + 4)),x)

[Out]

log(x + 4) + atanh(90/(7*(21*x + 48)) - 8/7)/2

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sympy [A]  time = 0.14, size = 17, normalized size = 1.13 \[ \frac {\log {\left (x - 2 \right )}}{4} - \frac {\log {\left (x + 2 \right )}}{4} + \log {\left (x + 4 \right )} \]

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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