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

Optimal. Leaf size=31 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{15} x}{2 \sqrt{4 x^2+1}}\right )}{2 \sqrt{15}} \]

[Out]

ArcTanh[(Sqrt[15]*x)/(2*Sqrt[1 + 4*x^2])]/(2*Sqrt[15])

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Rubi [A]  time = 0.0099887, 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, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{15} x}{2 \sqrt{4 x^2+1}}\right )}{2 \sqrt{15}} \]

Antiderivative was successfully verified.

[In]

Int[1/((4 + x^2)*Sqrt[1 + 4*x^2]),x]

[Out]

ArcTanh[(Sqrt[15]*x)/(2*Sqrt[1 + 4*x^2])]/(2*Sqrt[15])

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 206

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

Rubi steps

\begin{align*} \int \frac{1}{\left (4+x^2\right ) \sqrt{1+4 x^2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{4-15 x^2} \, dx,x,\frac{x}{\sqrt{1+4 x^2}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{15} x}{2 \sqrt{1+4 x^2}}\right )}{2 \sqrt{15}}\\ \end{align*}

Mathematica [A]  time = 0.0108849, size = 31, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{15} x}{2 \sqrt{4 x^2+1}}\right )}{2 \sqrt{15}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[15]*x)/(2*Sqrt[1 + 4*x^2])]/(2*Sqrt[15])

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Maple [A]  time = 0.008, size = 22, normalized size = 0.7 \begin{align*}{\frac{\sqrt{15}}{30}{\it Artanh} \left ({\frac{x\sqrt{15}}{2}{\frac{1}{\sqrt{4\,{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*arctanh(1/2*x*15^(1/2)/(4*x^2+1)^(1/2))*15^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.97988, size = 157, normalized size = 5.06 \begin{align*} \frac{1}{60} \, \sqrt{15} \log \left (\frac{961 \, x^{2} + 8 \, \sqrt{15}{\left (31 \, x^{2} + 4\right )} + 4 \, \sqrt{4 \, x^{2} + 1}{\left (31 \, \sqrt{15} x + 120 \, x\right )} + 124}{x^{2} + 4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*sqrt(15)*log((961*x^2 + 8*sqrt(15)*(31*x^2 + 4) + 4*sqrt(4*x^2 + 1)*(31*sqrt(15)*x + 120*x) + 124)/(x^2 +
4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.06585, size = 77, normalized size = 2.48 \begin{align*} -\frac{1}{60} \, \sqrt{15} \log \left (\frac{{\left (2 \, x - \sqrt{4 \, x^{2} + 1}\right )}^{2} - 8 \, \sqrt{15} + 31}{{\left (2 \, x - \sqrt{4 \, x^{2} + 1}\right )}^{2} + 8 \, \sqrt{15} + 31}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*sqrt(15)*log(((2*x - sqrt(4*x^2 + 1))^2 - 8*sqrt(15) + 31)/((2*x - sqrt(4*x^2 + 1))^2 + 8*sqrt(15) + 31)
)