3.257 \(\int \frac {x^2}{(-6+x^2) \sqrt {-2+x^2}} \, dx\)

Optimal. Leaf size=41 \[ \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-2}}\right )-\sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{3}} x}{\sqrt {x^2-2}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {483, 217, 206, 377, 207} \[ \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-2}}\right )-\sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{3}} x}{\sqrt {x^2-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/((-6 + x^2)*Sqrt[-2 + x^2]),x]

[Out]

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

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])

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 217

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

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 483

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[e^n/b, Int[
(e*x)^(m - n)*(c + d*x^n)^q, x], x] - Dist[(a*e^n)/b, Int[((e*x)^(m - n)*(c + d*x^n)^q)/(a + b*x^n), x], x] /;
FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b
, c, d, e, m, n, -1, q, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 41, normalized size = 1.00 \[ \log \left (\sqrt {x^2-2}+x\right )-\sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{3}} x}{\sqrt {x^2-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/((-6 + x^2)*Sqrt[-2 + x^2]),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.12, size = 66, normalized size = 1.61 \[ -\log \left (\sqrt {x^2-2}-x\right )-\sqrt {\frac {3}{2}} \tanh ^{-1}\left (-\frac {x^2}{2 \sqrt {6}}+\frac {\sqrt {x^2-2} x}{2 \sqrt {6}}+\sqrt {\frac {3}{2}}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/((-6 + x^2)*Sqrt[-2 + x^2]),x]

[Out]

-(Sqrt[3/2]*ArcTanh[Sqrt[3/2] - x^2/(2*Sqrt[6]) + (x*Sqrt[-2 + x^2])/(2*Sqrt[6])]) - Log[-x + Sqrt[-2 + x^2]]

________________________________________________________________________________________

fricas [B]  time = 0.67, size = 77, normalized size = 1.88 \[ \frac {1}{4} \, \sqrt {3} \sqrt {2} \log \left (-\frac {2 \, \sqrt {3} \sqrt {2} {\left (5 \, x^{2} - 6\right )} - 25 \, x^{2} + 2 \, {\left (5 \, \sqrt {3} \sqrt {2} x - 12 \, x\right )} \sqrt {x^{2} - 2} + 30}{x^{2} - 6}\right ) - \log \left (-x + \sqrt {x^{2} - 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^2-6)/(x^2-2)^(1/2),x, algorithm="fricas")

[Out]

1/4*sqrt(3)*sqrt(2)*log(-(2*sqrt(3)*sqrt(2)*(5*x^2 - 6) - 25*x^2 + 2*(5*sqrt(3)*sqrt(2)*x - 12*x)*sqrt(x^2 - 2
) + 30)/(x^2 - 6)) - log(-x + sqrt(x^2 - 2))

________________________________________________________________________________________

giac [B]  time = 0.66, size = 72, normalized size = 1.76 \[ -\frac {1}{4} \, \sqrt {6} \log \left (\frac {{\left | 2 \, {\left (x - \sqrt {x^{2} - 2}\right )}^{2} - 8 \, \sqrt {6} - 20 \right |}}{{\left | 2 \, {\left (x - \sqrt {x^{2} - 2}\right )}^{2} + 8 \, \sqrt {6} - 20 \right |}}\right ) - \frac {1}{2} \, \log \left ({\left (x - \sqrt {x^{2} - 2}\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^2-6)/(x^2-2)^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(6)*log(abs(2*(x - sqrt(x^2 - 2))^2 - 8*sqrt(6) - 20)/abs(2*(x - sqrt(x^2 - 2))^2 + 8*sqrt(6) - 20))
- 1/2*log((x - sqrt(x^2 - 2))^2)

________________________________________________________________________________________

maple [C]  time = 0.31, size = 64, normalized size = 1.56

method	result	size
trager	\(-\ln \left (x -\sqrt {x^{2}-2}\right )-\frac {\RootOf \left (\textit {\_Z}^{2}-6\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-6\right ) x^{2}+12 \sqrt {x^{2}-2}\, x -6 \RootOf \left (\textit {\_Z}^{2}-6\right )}{x^{2}-6}\right )}{4}\)	\(64\)
default	\(\ln \left (x +\sqrt {x^{2}-2}\right )-\frac {\sqrt {6}\, \arctanh \left (\frac {8+2 \sqrt {6}\, \left (x -\sqrt {6}\right )}{4 \sqrt {\left (x -\sqrt {6}\right )^{2}+2 \sqrt {6}\, \left (x -\sqrt {6}\right )+4}}\right )}{4}+\frac {\sqrt {6}\, \arctanh \left (\frac {8-2 \sqrt {6}\, \left (x +\sqrt {6}\right )}{4 \sqrt {\left (x +\sqrt {6}\right )^{2}-2 \sqrt {6}\, \left (x +\sqrt {6}\right )+4}}\right )}{4}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^2-6)/(x^2-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-ln(x-(x^2-2)^(1/2))-1/4*RootOf(_Z^2-6)*ln(-(5*RootOf(_Z^2-6)*x^2+12*(x^2-2)^(1/2)*x-6*RootOf(_Z^2-6))/(x^2-6)
)

________________________________________________________________________________________

maxima [B]  time = 1.46, size = 107, normalized size = 2.61 \[ \frac {1}{12} \, \sqrt {6} {\left (2 \, \sqrt {6} \log \left (x + \sqrt {x^{2} - 2}\right ) - 3 \, \log \left (\sqrt {6} + \frac {4 \, \sqrt {x^{2} - 2}}{{\left | 2 \, x - 2 \, \sqrt {6} \right |}} + \frac {8}{{\left | 2 \, x - 2 \, \sqrt {6} \right |}}\right ) + 3 \, \log \left (-\sqrt {6} + \frac {4 \, \sqrt {x^{2} - 2}}{{\left | 2 \, x + 2 \, \sqrt {6} \right |}} + \frac {8}{{\left | 2 \, x + 2 \, \sqrt {6} \right |}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^2-6)/(x^2-2)^(1/2),x, algorithm="maxima")

[Out]

1/12*sqrt(6)*(2*sqrt(6)*log(x + sqrt(x^2 - 2)) - 3*log(sqrt(6) + 4*sqrt(x^2 - 2)/abs(2*x - 2*sqrt(6)) + 8/abs(
2*x - 2*sqrt(6))) + 3*log(-sqrt(6) + 4*sqrt(x^2 - 2)/abs(2*x + 2*sqrt(6)) + 8/abs(2*x + 2*sqrt(6))))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\sqrt {x^2-2}\,\left (x^2-6\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((x^2 - 2)^(1/2)*(x^2 - 6)),x)

[Out]

int(x^2/((x^2 - 2)^(1/2)*(x^2 - 6)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (x^{2} - 6\right ) \sqrt {x^{2} - 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**2-6)/(x**2-2)**(1/2),x)

[Out]

Integral(x**2/((x**2 - 6)*sqrt(x**2 - 2)), x)

________________________________________________________________________________________