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

Optimal. Leaf size=34 \[ -\frac {3}{32} \sin ^{-1}\left (1-\frac {8 x^2}{3}\right )-\frac {1}{8} \sqrt {3 x^2-4 x^4} \]

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Rubi [A]  time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2018, 640, 619, 216} \begin {gather*} -\frac {1}{8} \sqrt {3 x^2-4 x^4}-\frac {3}{32} \sin ^{-1}\left (1-\frac {8 x^2}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[3*x^2 - 4*x^4],x]

[Out]

-Sqrt[3*x^2 - 4*x^4]/8 - (3*ArcSin[1 - (8*x^2)/3])/32

Rule 216

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {3 x^2-4 x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\sqrt {3 x-4 x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{8} \sqrt {3 x^2-4 x^4}+\frac {3}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3 x-4 x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{8} \sqrt {3 x^2-4 x^4}-\frac {1}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{9}}} \, dx,x,3-8 x^2\right )\\ &=-\frac {1}{8} \sqrt {3 x^2-4 x^4}-\frac {3}{32} \sin ^{-1}\left (1-\frac {8 x^2}{3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 1.68 \begin {gather*} \frac {x \left (8 x^3+3 \sqrt {4 x^2-3} \tanh ^{-1}\left (\frac {2 x}{\sqrt {4 x^2-3}}\right )-6 x\right )}{16 \sqrt {3 x^2-4 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[3*x^2 - 4*x^4],x]

[Out]

(x*(-6*x + 8*x^3 + 3*Sqrt[-3 + 4*x^2]*ArcTanh[(2*x)/Sqrt[-3 + 4*x^2]]))/(16*Sqrt[3*x^2 - 4*x^4])

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IntegrateAlgebraic [C]  time = 0.12, size = 55, normalized size = 1.62 \begin {gather*} -\frac {1}{8} \sqrt {3 x^2-4 x^4}+\frac {3}{32} i \log \left (-8 i x^2+4 \sqrt {3 x^2-4 x^4}+3 i\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3/Sqrt[3*x^2 - 4*x^4],x]

[Out]

-1/8*Sqrt[3*x^2 - 4*x^4] + ((3*I)/32)*Log[3*I - (8*I)*x^2 + 4*Sqrt[3*x^2 - 4*x^4]]

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fricas [A]  time = 0.78, size = 37, normalized size = 1.09 \begin {gather*} -\frac {1}{8} \, \sqrt {-4 \, x^{4} + 3 \, x^{2}} - \frac {3}{16} \, \arctan \left (\frac {\sqrt {-4 \, x^{4} + 3 \, x^{2}}}{2 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-4*x^4 + 3*x^2) - 3/16*arctan(1/2*sqrt(-4*x^4 + 3*x^2)/x^2)

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giac [A]  time = 0.18, size = 26, normalized size = 0.76 \begin {gather*} -\frac {1}{8} \, \sqrt {-4 \, x^{4} + 3 \, x^{2}} + \frac {3}{32} \, \arcsin \left (\frac {8}{3} \, x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-4*x^4 + 3*x^2) + 3/32*arcsin(8/3*x^2 - 1)

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maple [A]  time = 0.01, size = 48, normalized size = 1.41 \begin {gather*} \frac {\sqrt {-4 x^{2}+3}\, \left (-2 \sqrt {-4 x^{2}+3}\, x +3 \arcsin \left (\frac {2 \sqrt {3}\, x}{3}\right )\right ) x}{16 \sqrt {-4 x^{4}+3 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x*(-4*x^2+3)^(1/2)*(-2*x*(-4*x^2+3)^(1/2)+3*arcsin(2/3*3^(1/2)*x))/(-4*x^4+3*x^2)^(1/2)

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maxima [A]  time = 3.03, size = 26, normalized size = 0.76 \begin {gather*} -\frac {1}{8} \, \sqrt {-4 \, x^{4} + 3 \, x^{2}} - \frac {3}{32} \, \arcsin \left (-\frac {8}{3} \, x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-4*x^4 + 3*x^2) - 3/32*arcsin(-8/3*x^2 + 1)

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mupad [B]  time = 4.33, size = 42, normalized size = 1.24 \begin {gather*} -\frac {\sqrt {3\,x^2-4\,x^4}}{8}-\frac {\ln \left (x^2-\frac {3}{8}-\frac {\sqrt {3-4\,x^2}\,\sqrt {x^2}\,1{}\mathrm {i}}{2}\right )\,3{}\mathrm {i}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(3*x^2 - 4*x^4)^(1/2),x)

[Out]

- (log(x^2 - ((3 - 4*x^2)^(1/2)*(x^2)^(1/2)*1i)/2 - 3/8)*3i)/32 - (3*x^2 - 4*x^4)^(1/2)/8

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {- x^{2} \left (4 x^{2} - 3\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt(-x**2*(4*x**2 - 3)), x)

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