3.278 \(\int \frac{1}{(1+x^2) (4+x^2)} \, dx\)

Optimal. Leaf size=17 \[ \frac{1}{3} \tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\left (\frac{x}{2}\right ) \]

[Out]

-ArcTan[x/2]/6 + ArcTan[x]/3

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Rubi [A]  time = 0.0065376, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {391, 203} \[ \frac{1}{3} \tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + x^2)*(4 + x^2)),x]

[Out]

-ArcTan[x/2]/6 + ArcTan[x]/3

Rule 391

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.0061688, size = 17, normalized size = 1. \[ \frac{1}{6} \tan ^{-1}\left (\frac{2}{x}\right )+\frac{1}{3} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 + x^2)*(4 + x^2)),x]

[Out]

ArcTan[2/x]/6 + ArcTan[x]/3

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Maple [A]  time = 0.006, size = 12, normalized size = 0.7 \begin{align*} -{\frac{1}{6}\arctan \left ({\frac{x}{2}} \right ) }+{\frac{\arctan \left ( x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2+1)/(x^2+4),x)

[Out]

-1/6*arctan(1/2*x)+1/3*arctan(x)

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Maxima [A]  time = 1.60655, size = 15, normalized size = 0.88 \begin{align*} -\frac{1}{6} \, \arctan \left (\frac{1}{2} \, x\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2+1)/(x^2+4),x, algorithm="maxima")

[Out]

-1/6*arctan(1/2*x) + 1/3*arctan(x)

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Fricas [A]  time = 1.46146, size = 49, normalized size = 2.88 \begin{align*} -\frac{1}{6} \, \arctan \left (\frac{1}{2} \, x\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2+1)/(x^2+4),x, algorithm="fricas")

[Out]

-1/6*arctan(1/2*x) + 1/3*arctan(x)

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Sympy [A]  time = 0.123822, size = 10, normalized size = 0.59 \begin{align*} - \frac{\operatorname{atan}{\left (\frac{x}{2} \right )}}{6} + \frac{\operatorname{atan}{\left (x \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**2+1)/(x**2+4),x)

[Out]

-atan(x/2)/6 + atan(x)/3

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Giac [A]  time = 1.12569, size = 15, normalized size = 0.88 \begin{align*} -\frac{1}{6} \, \arctan \left (\frac{1}{2} \, x\right ) + \frac{1}{3} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2+1)/(x^2+4),x, algorithm="giac")

[Out]

-1/6*arctan(1/2*x) + 1/3*arctan(x)