3.238 \(\int \frac{1}{\sqrt{1-x^2} (4+x^2)} \, dx\)

Optimal. Leaf size=31 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{1-x^2}}\right )}{2 \sqrt{5}} \]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[1 - x^2])]/(2*Sqrt[5])

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Rubi [A]  time = 0.0075575, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{1-x^2}}\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - x^2]*(4 + x^2)),x]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[1 - x^2])]/(2*Sqrt[5])

Rule 377

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

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}{\sqrt{1-x^2} \left (4+x^2\right )} \, dx &=\operatorname{Subst}\left (\int \frac{1}{4+5 x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{1-x^2}}\right )}{2 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.0076511, size = 31, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{1-x^2}}\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - x^2]*(4 + x^2)),x]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[1 - x^2])]/(2*Sqrt[5])

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Maple [A]  time = 0.01, size = 29, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{5}}{10}\arctan \left ({\frac{x\sqrt{5}}{2\,{x}^{2}-2}\sqrt{-{x}^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*5^(1/2)*arctan(1/2*5^(1/2)*(-x^2+1)^(1/2)/(x^2-1)*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} + 4\right )} \sqrt{-x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^2 + 4)*sqrt(-x^2 + 1)), x)

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Fricas [A]  time = 2.07916, size = 70, normalized size = 2.26 \begin{align*} -\frac{1}{10} \, \sqrt{5} \arctan \left (\frac{2 \, \sqrt{5} \sqrt{-x^{2} + 1}}{5 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*sqrt(5)*arctan(2/5*sqrt(5)*sqrt(-x^2 + 1)/x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 4\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-(x - 1)*(x + 1))*(x**2 + 4)), x)

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Giac [B]  time = 1.08879, size = 69, normalized size = 2.23 \begin{align*} \frac{1}{20} \, \sqrt{5}{\left (\pi \mathrm{sgn}\left (x\right ) + 2 \, \arctan \left (-\frac{\sqrt{5} x{\left (\frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{5 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(5)*(pi*sgn(x) + 2*arctan(-1/5*sqrt(5)*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1)))