\(\int \frac {\sqrt {4-3 x^2}}{x} \, dx\) [346]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {4-3 x^2}-2 \text {arctanh}\left (\frac {1}{2} \sqrt {4-3 x^2}\right ) \]

[Out]

-2*arctanh(1/2*(-3*x^2+4)^(1/2))+(-3*x^2+4)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 52, 65, 212} \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {4-3 x^2}-2 \text {arctanh}\left (\frac {1}{2} \sqrt {4-3 x^2}\right ) \]

[In]

Int[Sqrt[4 - 3*x^2]/x,x]

[Out]

Sqrt[4 - 3*x^2] - 2*ArcTanh[Sqrt[4 - 3*x^2]/2]

Rule 52

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 65

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 212

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {4-3 x}}{x} \, dx,x,x^2\right ) \\ & = \sqrt {4-3 x^2}+2 \text {Subst}\left (\int \frac {1}{\sqrt {4-3 x} x} \, dx,x,x^2\right ) \\ & = \sqrt {4-3 x^2}-\frac {4}{3} \text {Subst}\left (\int \frac {1}{\frac {4}{3}-\frac {x^2}{3}} \, dx,x,\sqrt {4-3 x^2}\right ) \\ & = \sqrt {4-3 x^2}-2 \text {arctanh}\left (\frac {1}{2} \sqrt {4-3 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {4-3 x^2}-2 \text {arctanh}\left (\frac {1}{2} \sqrt {4-3 x^2}\right ) \]

[In]

Integrate[Sqrt[4 - 3*x^2]/x,x]

[Out]

Sqrt[4 - 3*x^2] - 2*ArcTanh[Sqrt[4 - 3*x^2]/2]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(\sqrt {-3 x^{2}+4}-2 \,\operatorname {arctanh}\left (\frac {2}{\sqrt {-3 x^{2}+4}}\right )\)	\(25\)
trager	\(\sqrt {-3 x^{2}+4}-2 \ln \left (\frac {\sqrt {-3 x^{2}+4}+2}{x}\right )\)	\(29\)
pseudoelliptic	\(\sqrt {-3 x^{2}+4}+\ln \left (\sqrt {-3 x^{2}+4}-2\right )-\ln \left (\sqrt {-3 x^{2}+4}+2\right )\)	\(37\)
meijerg	\(-\frac {-2 \left (2-4 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (3\right )+i \pi \right ) \sqrt {\pi }+4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {1-\frac {3 x^{2}}{4}}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {3 x^{2}}{4}}}{2}\right )}{2 \sqrt {\pi }}\)	\(66\)

[In]

int((-3*x^2+4)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

(-3*x^2+4)^(1/2)-2*arctanh(2/(-3*x^2+4)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {-3 \, x^{2} + 4} + 2 \, \log \left (\frac {\sqrt {-3 \, x^{2} + 4} - 2}{x}\right ) \]

[In]

integrate((-3*x^2+4)^(1/2)/x,x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\begin {cases} i \sqrt {3 x^{2} - 4} - 2 \log {\left (x \right )} + \log {\left (x^{2} \right )} + 2 i \operatorname {asin}{\left (\frac {2 \sqrt {3}}{3 x} \right )} & \text {for}\: \left |{x^{2}}\right | > \frac {4}{3} \\\sqrt {4 - 3 x^{2}} + \log {\left (x^{2} \right )} - 2 \log {\left (\sqrt {1 - \frac {3 x^{2}}{4}} + 1 \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((-3*x**2+4)**(1/2)/x,x)

[Out]

Piecewise((I*sqrt(3*x**2 - 4) - 2*log(x) + log(x**2) + 2*I*asin(2*sqrt(3)/(3*x)), Abs(x**2) > 4/3), (sqrt(4 -
3*x**2) + log(x**2) - 2*log(sqrt(1 - 3*x**2/4) + 1), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {-3 \, x^{2} + 4} - 2 \, \log \left (\frac {4 \, \sqrt {-3 \, x^{2} + 4}}{{\left | x \right |}} + \frac {8}{{\left | x \right |}}\right ) \]

[In]

integrate((-3*x^2+4)^(1/2)/x,x, algorithm="maxima")

[Out]

sqrt(-3*x^2 + 4) - 2*log(4*sqrt(-3*x^2 + 4)/abs(x) + 8/abs(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=\sqrt {-3 \, x^{2} + 4} - \log \left (\sqrt {-3 \, x^{2} + 4} + 2\right ) + \log \left (-\sqrt {-3 \, x^{2} + 4} + 2\right ) \]

[In]

integrate((-3*x^2+4)^(1/2)/x,x, algorithm="giac")

[Out]

sqrt(-3*x^2 + 4) - log(sqrt(-3*x^2 + 4) + 2) + log(-sqrt(-3*x^2 + 4) + 2)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {4-3 x^2}}{x} \, dx=2\,\ln \left (\sqrt {\frac {4}{3\,x^2}-1}-\frac {2\,\sqrt {3}\,\sqrt {\frac {1}{x^2}}}{3}\right )+\sqrt {3}\,\sqrt {\frac {4}{3}-x^2} \]

[In]

int((4 - 3*x^2)^(1/2)/x,x)

[Out]

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