3.8.86 \(\int \frac {x \sqrt {-1+3 x^2}}{\sqrt {2-3 x^2}} \, dx\)

Optimal. Leaf size=39 \[ -\frac {1}{6} \sqrt {2-3 x^2} \sqrt {3 x^2-1}-\frac {1}{12} \sin ^{-1}\left (3-6 x^2\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {444, 50, 53, 619, 216} \begin {gather*} -\frac {1}{6} \sqrt {2-3 x^2} \sqrt {3 x^2-1}-\frac {1}{12} \sin ^{-1}\left (3-6 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[-1 + 3*x^2])/Sqrt[2 - 3*x^2],x]

[Out]

-(Sqrt[2 - 3*x^2]*Sqrt[-1 + 3*x^2])/6 - ArcSin[3 - 6*x^2]/12

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 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

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 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 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]

Rubi steps

\begin {align*} \int \frac {x \sqrt {-1+3 x^2}}{\sqrt {2-3 x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {-1+3 x}}{\sqrt {2-3 x}} \, dx,x,x^2\right )\\ &=-\frac {1}{6} \sqrt {2-3 x^2} \sqrt {-1+3 x^2}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-3 x} \sqrt {-1+3 x}} \, dx,x,x^2\right )\\ &=-\frac {1}{6} \sqrt {2-3 x^2} \sqrt {-1+3 x^2}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-2+9 x-9 x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{6} \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {1}{36} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{9}}} \, dx,x,9 \left (1-2 x^2\right )\right )\\ &=-\frac {1}{6} \sqrt {2-3 x^2} \sqrt {-1+3 x^2}-\frac {1}{12} \sin ^{-1}\left (3-6 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.95 \begin {gather*} \frac {1}{6} \left (-\sin ^{-1}\left (\sqrt {2-3 x^2}\right )-\sqrt {-9 x^4+9 x^2-2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*Sqrt[-1 + 3*x^2])/Sqrt[2 - 3*x^2],x]

[Out]

(-Sqrt[-2 + 9*x^2 - 9*x^4] - ArcSin[Sqrt[2 - 3*x^2]])/6

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IntegrateAlgebraic [A]  time = 0.16, size = 59, normalized size = 1.51 \begin {gather*} \frac {1}{3} \tan ^{-1}\left (\frac {\sqrt {3 x^2-1}}{\sqrt {2-3 x^2}-1}\right )-\frac {1}{6} \sqrt {2-3 x^2} \sqrt {3 x^2-1} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*Sqrt[-1 + 3*x^2])/Sqrt[2 - 3*x^2],x]

[Out]

-1/6*(Sqrt[2 - 3*x^2]*Sqrt[-1 + 3*x^2]) + ArcTan[Sqrt[-1 + 3*x^2]/(-1 + Sqrt[2 - 3*x^2])]/3

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fricas [B]  time = 1.25, size = 65, normalized size = 1.67 \begin {gather*} -\frac {1}{6} \, \sqrt {3 \, x^{2} - 1} \sqrt {-3 \, x^{2} + 2} - \frac {1}{12} \, \arctan \left (\frac {3 \, \sqrt {3 \, x^{2} - 1} {\left (2 \, x^{2} - 1\right )} \sqrt {-3 \, x^{2} + 2}}{2 \, {\left (9 \, x^{4} - 9 \, x^{2} + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3*x^2 - 1)*sqrt(-3*x^2 + 2) - 1/12*arctan(3/2*sqrt(3*x^2 - 1)*(2*x^2 - 1)*sqrt(-3*x^2 + 2)/(9*x^4 -
9*x^2 + 2))

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giac [A]  time = 0.45, size = 33, normalized size = 0.85 \begin {gather*} -\frac {1}{6} \, \sqrt {3 \, x^{2} - 1} \sqrt {-3 \, x^{2} + 2} + \frac {1}{6} \, \arcsin \left (\sqrt {3 \, x^{2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3*x^2 - 1)*sqrt(-3*x^2 + 2) + 1/6*arcsin(sqrt(3*x^2 - 1))

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maple [A]  time = 0.01, size = 60, normalized size = 1.54 \begin {gather*} \frac {\sqrt {3 x^{2}-1}\, \sqrt {-3 x^{2}+2}\, \left (\arcsin \left (6 x^{2}-3\right )-2 \sqrt {-9 x^{4}+9 x^{2}-2}\right )}{12 \sqrt {-9 x^{4}+9 x^{2}-2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*x^2-1)^(1/2)*(-3*x^2+2)^(1/2)*(arcsin(6*x^2-3)-2*(-9*x^4+9*x^2-2)^(1/2))/(-9*x^4+9*x^2-2)^(1/2)

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maxima [A]  time = 2.00, size = 27, normalized size = 0.69 \begin {gather*} -\frac {1}{6} \, \sqrt {-9 \, x^{4} + 9 \, x^{2} - 2} + \frac {1}{12} \, \arcsin \left (6 \, x^{2} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(-9*x^4 + 9*x^2 - 2) + 1/12*arcsin(6*x^2 - 3)

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mupad [B]  time = 2.74, size = 206, normalized size = 5.28 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {\sqrt {3\,x^2-1}-\mathrm {i}}{\sqrt {2}-\sqrt {2-3\,x^2}}\right )}{3}-\frac {-\frac {\sqrt {3\,x^2-1}-\mathrm {i}}{\sqrt {2}-\sqrt {2-3\,x^2}}+\frac {{\left (\sqrt {3\,x^2-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {2}-\sqrt {2-3\,x^2}\right )}^3}+\frac {\sqrt {2}\,{\left (\sqrt {3\,x^2-1}-\mathrm {i}\right )}^2\,4{}\mathrm {i}}{3\,{\left (\sqrt {2}-\sqrt {2-3\,x^2}\right )}^2}}{\frac {2\,{\left (\sqrt {3\,x^2-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {2}-\sqrt {2-3\,x^2}\right )}^2}+\frac {{\left (\sqrt {3\,x^2-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {2}-\sqrt {2-3\,x^2}\right )}^4}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.11, size = 66, normalized size = 1.69 \begin {gather*} \frac {\begin {cases} - \frac {\sqrt {2 - 3 x^{2}} \sqrt {3 x^{2} - 1}}{2} + \frac {\operatorname {asin}{\left (\sqrt {3 x^{2} - 1} \right )}}{2} & \text {for}\: \left (x \geq \frac {\sqrt {3}}{3} \wedge x < \frac {\sqrt {6}}{3}\right ) \vee \left (x \leq - \frac {\sqrt {3}}{3} \wedge x > - \frac {\sqrt {6}}{3}\right ) \end {cases}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-sqrt(2 - 3*x**2)*sqrt(3*x**2 - 1)/2 + asin(sqrt(3*x**2 - 1))/2, ((x >= sqrt(3)/3) & (x < sqrt(6)/3
)) | ((x <= -sqrt(3)/3) & (x > -sqrt(6)/3))))/3

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