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

Optimal. Leaf size=65 \[ -\frac{1}{3} \sqrt{4 x^2-1}+\frac{\tanh ^{-1}\left (\sqrt{3} \sqrt{4 x^2-1}\right )}{3 \sqrt{3}}+\frac{4 x}{3}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}} \]

[Out]

(4*x)/3 - Sqrt[-1 + 4*x^2]/3 - ArcTanh[Sqrt[3]*x]/(3*Sqrt[3]) + ArcTanh[Sqrt[3]*Sqrt[-1 + 4*x^2]]/(3*Sqrt[3])

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Rubi [A]  time = 0.13286, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6742, 444, 50, 63, 207, 388} \[ -\frac{1}{3} \sqrt{4 x^2-1}+\frac{\tanh ^{-1}\left (\sqrt{3} \sqrt{4 x^2-1}\right )}{3 \sqrt{3}}+\frac{4 x}{3}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/3 - Sqrt[-1 + 4*x^2]/3 - ArcTanh[Sqrt[3]*x]/(3*Sqrt[3]) + ArcTanh[Sqrt[3]*Sqrt[-1 + 4*x^2]]/(3*Sqrt[3])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 444

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 388

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

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+4 x^2}}{x+\sqrt{-1+4 x^2}} \, dx &=\int \left (-\frac{x \sqrt{-1+4 x^2}}{-1+3 x^2}+\frac{-1+4 x^2}{-1+3 x^2}\right ) \, dx\\ &=-\int \frac{x \sqrt{-1+4 x^2}}{-1+3 x^2} \, dx+\int \frac{-1+4 x^2}{-1+3 x^2} \, dx\\ &=\frac{4 x}{3}+\frac{1}{3} \int \frac{1}{-1+3 x^2} \, dx-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{-1+4 x}}{-1+3 x} \, dx,x,x^2\right )\\ &=\frac{4 x}{3}-\frac{1}{3} \sqrt{-1+4 x^2}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{(-1+3 x) \sqrt{-1+4 x}} \, dx,x,x^2\right )\\ &=\frac{4 x}{3}-\frac{1}{3} \sqrt{-1+4 x^2}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}-\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{4}+\frac{3 x^2}{4}} \, dx,x,\sqrt{-1+4 x^2}\right )\\ &=\frac{4 x}{3}-\frac{1}{3} \sqrt{-1+4 x^2}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}+\frac{\tanh ^{-1}\left (\sqrt{3} \sqrt{-1+4 x^2}\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0473054, size = 54, normalized size = 0.83 \[ \frac{1}{9} \left (-3 \sqrt{4 x^2-1}+\sqrt{3} \tanh ^{-1}\left (\sqrt{12 x^2-3}\right )+12 x-\sqrt{3} \tanh ^{-1}\left (\sqrt{3} x\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*x - 3*Sqrt[-1 + 4*x^2] - Sqrt[3]*ArcTanh[Sqrt[3]*x] + Sqrt[3]*ArcTanh[Sqrt[-3 + 12*x^2]])/9

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Maple [B]  time = 0.032, size = 262, normalized size = 4. \begin{align*}{\frac{4\,x}{3}}-{\frac{{\it Artanh} \left ( x\sqrt{3} \right ) \sqrt{3}}{9}}-{\frac{1}{18}\sqrt{36\, \left ( x-1/3\,\sqrt{3} \right ) ^{2}+24\,\sqrt{3} \left ( x-1/3\,\sqrt{3} \right ) +3}}-{\frac{\sqrt{3}\sqrt{4}}{18}\ln \left ( x\sqrt{4}+\sqrt{4\, \left ( x-1/3\,\sqrt{3} \right ) ^{2}+{\frac{8\,\sqrt{3}}{3} \left ( x-{\frac{\sqrt{3}}{3}} \right ) }+{\frac{1}{3}}} \right ) }+{\frac{\sqrt{3}}{18}{\it Artanh} \left ({\frac{3\,\sqrt{3}}{2} \left ({\frac{2}{3}}+{\frac{8\,\sqrt{3}}{3} \left ( x-{\frac{\sqrt{3}}{3}} \right ) } \right ){\frac{1}{\sqrt{36\, \left ( x-1/3\,\sqrt{3} \right ) ^{2}+24\,\sqrt{3} \left ( x-1/3\,\sqrt{3} \right ) +3}}}} \right ) }-{\frac{1}{18}\sqrt{36\, \left ( x+1/3\,\sqrt{3} \right ) ^{2}-24\,\sqrt{3} \left ( x+1/3\,\sqrt{3} \right ) +3}}+{\frac{\sqrt{3}\sqrt{4}}{18}\ln \left ( x\sqrt{4}+\sqrt{4\, \left ( x+1/3\,\sqrt{3} \right ) ^{2}-{\frac{8\,\sqrt{3}}{3} \left ( x+{\frac{\sqrt{3}}{3}} \right ) }+{\frac{1}{3}}} \right ) }+{\frac{\sqrt{3}}{18}{\it Artanh} \left ({\frac{3\,\sqrt{3}}{2} \left ({\frac{2}{3}}-{\frac{8\,\sqrt{3}}{3} \left ( x+{\frac{\sqrt{3}}{3}} \right ) } \right ){\frac{1}{\sqrt{36\, \left ( x+1/3\,\sqrt{3} \right ) ^{2}-24\,\sqrt{3} \left ( x+1/3\,\sqrt{3} \right ) +3}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*x-1/9*arctanh(x*3^(1/2))*3^(1/2)-1/18*(36*(x-1/3*3^(1/2))^2+24*3^(1/2)*(x-1/3*3^(1/2))+3)^(1/2)-1/18*3^(1/
2)*ln(x*4^(1/2)+(4*(x-1/3*3^(1/2))^2+8/3*3^(1/2)*(x-1/3*3^(1/2))+1/3)^(1/2))*4^(1/2)+1/18*3^(1/2)*arctanh(3/2*
(2/3+8/3*3^(1/2)*(x-1/3*3^(1/2)))*3^(1/2)/(36*(x-1/3*3^(1/2))^2+24*3^(1/2)*(x-1/3*3^(1/2))+3)^(1/2))-1/18*(36*
(x+1/3*3^(1/2))^2-24*3^(1/2)*(x+1/3*3^(1/2))+3)^(1/2)+1/18*3^(1/2)*ln(x*4^(1/2)+(4*(x+1/3*3^(1/2))^2-8/3*3^(1/
2)*(x+1/3*3^(1/2))+1/3)^(1/2))*4^(1/2)+1/18*3^(1/2)*arctanh(3/2*(2/3-8/3*3^(1/2)*(x+1/3*3^(1/2)))*3^(1/2)/(36*
(x+1/3*3^(1/2))^2-24*3^(1/2)*(x+1/3*3^(1/2))+3)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.654, size = 212, normalized size = 3.26 \begin{align*} \frac{1}{18} \, \sqrt{3} \log \left (\frac{6 \, x^{2} + \sqrt{3} \sqrt{4 \, x^{2} - 1} - 1}{3 \, x^{2} - 1}\right ) + \frac{1}{18} \, \sqrt{3} \log \left (\frac{3 \, x^{2} - 2 \, \sqrt{3} x + 1}{3 \, x^{2} - 1}\right ) + \frac{4}{3} \, x - \frac{1}{3} \, \sqrt{4 \, x^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*sqrt(3)*log((6*x^2 + sqrt(3)*sqrt(4*x^2 - 1) - 1)/(3*x^2 - 1)) + 1/18*sqrt(3)*log((3*x^2 - 2*sqrt(3)*x +
1)/(3*x^2 - 1)) + 4/3*x - 1/3*sqrt(4*x^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15428, size = 180, normalized size = 2.77 \begin{align*} \frac{1}{18} \, \sqrt{3} \log \left (\frac{{\left | 6 \, x - 2 \, \sqrt{3} \right |}}{{\left | 6 \, x + 2 \, \sqrt{3} \right |}}\right ) - \frac{1}{18} \, \sqrt{3} \log \left (-\frac{{\left | -12 \, x - 4 \, \sqrt{3} + 6 \, \sqrt{4 \, x^{2} - 1} + \frac{6}{2 \, x - \sqrt{4 \, x^{2} - 1}} \right |}}{2 \,{\left (6 \, x - 2 \, \sqrt{3} - 3 \, \sqrt{4 \, x^{2} - 1} - \frac{3}{2 \, x - \sqrt{4 \, x^{2} - 1}}\right )}}\right ) + \frac{4}{3} \, x - \frac{1}{3} \, \sqrt{4 \, x^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*sqrt(3)*log(abs(6*x - 2*sqrt(3))/abs(6*x + 2*sqrt(3))) - 1/18*sqrt(3)*log(-1/2*abs(-12*x - 4*sqrt(3) + 6*
sqrt(4*x^2 - 1) + 6/(2*x - sqrt(4*x^2 - 1)))/(6*x - 2*sqrt(3) - 3*sqrt(4*x^2 - 1) - 3/(2*x - sqrt(4*x^2 - 1)))
) + 4/3*x - 1/3*sqrt(4*x^2 - 1)