3.11.22 \(\int (-1+x^2) (-1+x \sqrt {-1+3 x^2-x^4}) \, dx\)

Optimal. Leaf size=77 \[ -\frac {x^3}{3}+\frac {1}{48} \sqrt {-x^4+3 x^2-1} \left (8 x^4-18 x^2-1\right )+\frac {5}{32} i \log \left (-2 i x^2+2 \sqrt {-x^4+3 x^2-1}+3 i\right )+x \]

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Rubi [A]  time = 0.19, antiderivative size = 74, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6742, 1107, 612, 619, 216, 14, 1114, 640} \begin {gather*} -\frac {x^3}{3}-\frac {5}{32} \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )-\frac {1}{6} \left (-x^4+3 x^2-1\right )^{3/2}-\frac {1}{16} \left (3-2 x^2\right ) \sqrt {-x^4+3 x^2-1}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^2)*(-1 + x*Sqrt[-1 + 3*x^2 - x^4]),x]

[Out]

x - x^3/3 - ((3 - 2*x^2)*Sqrt[-1 + 3*x^2 - x^4])/16 - (-1 + 3*x^2 - x^4)^(3/2)/6 - (5*ArcSin[(3 - 2*x^2)/Sqrt[
5]])/32

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 612

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

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 1107

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

Rule 1114

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

Rule 6742

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

Rubi steps

\begin {align*} \int \left (-1+x^2\right ) \left (-1+x \sqrt {-1+3 x^2-x^4}\right ) \, dx &=\int \left (1-x \sqrt {-1+3 x^2-x^4}+x^2 \left (-1+x \sqrt {-1+3 x^2-x^4}\right )\right ) \, dx\\ &=x-\int x \sqrt {-1+3 x^2-x^4} \, dx+\int x^2 \left (-1+x \sqrt {-1+3 x^2-x^4}\right ) \, dx\\ &=x-\frac {1}{2} \operatorname {Subst}\left (\int \sqrt {-1+3 x-x^2} \, dx,x,x^2\right )+\int \left (-x^2+x^3 \sqrt {-1+3 x^2-x^4}\right ) \, dx\\ &=x-\frac {x^3}{3}+\frac {1}{8} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}-\frac {5}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+3 x-x^2}} \, dx,x,x^2\right )+\int x^3 \sqrt {-1+3 x^2-x^4} \, dx\\ &=x-\frac {x^3}{3}+\frac {1}{8} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}+\frac {1}{2} \operatorname {Subst}\left (\int x \sqrt {-1+3 x-x^2} \, dx,x,x^2\right )+\frac {1}{16} \sqrt {5} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{5}}} \, dx,x,3-2 x^2\right )\\ &=x-\frac {x^3}{3}+\frac {1}{8} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}-\frac {1}{6} \left (-1+3 x^2-x^4\right )^{3/2}+\frac {5}{16} \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \sqrt {-1+3 x-x^2} \, dx,x,x^2\right )\\ &=x-\frac {x^3}{3}-\frac {1}{16} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}-\frac {1}{6} \left (-1+3 x^2-x^4\right )^{3/2}+\frac {5}{16} \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )+\frac {15}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+3 x-x^2}} \, dx,x,x^2\right )\\ &=x-\frac {x^3}{3}-\frac {1}{16} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}-\frac {1}{6} \left (-1+3 x^2-x^4\right )^{3/2}+\frac {5}{16} \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )-\frac {1}{32} \left (3 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{5}}} \, dx,x,3-2 x^2\right )\\ &=x-\frac {x^3}{3}-\frac {1}{16} \left (3-2 x^2\right ) \sqrt {-1+3 x^2-x^4}-\frac {1}{6} \left (-1+3 x^2-x^4\right )^{3/2}-\frac {5}{32} \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 1.16 \begin {gather*} \frac {1}{96} \left (-32 x^3-15 \sin ^{-1}\left (\frac {3-2 x^2}{\sqrt {5}}\right )+16 \sqrt {-x^4+3 x^2-1} x^4-36 \sqrt {-x^4+3 x^2-1} x^2-2 \sqrt {-x^4+3 x^2-1}+96 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^2)*(-1 + x*Sqrt[-1 + 3*x^2 - x^4]),x]

[Out]

(96*x - 32*x^3 - 2*Sqrt[-1 + 3*x^2 - x^4] - 36*x^2*Sqrt[-1 + 3*x^2 - x^4] + 16*x^4*Sqrt[-1 + 3*x^2 - x^4] - 15
*ArcSin[(3 - 2*x^2)/Sqrt[5]])/96

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IntegrateAlgebraic [A]  time = 0.41, size = 77, normalized size = 1.00 \begin {gather*} x-\frac {x^3}{3}+\frac {1}{48} \sqrt {-1+3 x^2-x^4} \left (-1-18 x^2+8 x^4\right )+\frac {5}{32} i \log \left (3 i-2 i x^2+2 \sqrt {-1+3 x^2-x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^2)*(-1 + x*Sqrt[-1 + 3*x^2 - x^4]),x]

[Out]

x - x^3/3 + (Sqrt[-1 + 3*x^2 - x^4]*(-1 - 18*x^2 + 8*x^4))/48 + ((5*I)/32)*Log[3*I - (2*I)*x^2 + 2*Sqrt[-1 + 3
*x^2 - x^4]]

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fricas [A]  time = 0.56, size = 73, normalized size = 0.95 \begin {gather*} -\frac {1}{3} \, x^{3} + \frac {1}{48} \, {\left (8 \, x^{4} - 18 \, x^{2} - 1\right )} \sqrt {-x^{4} + 3 \, x^{2} - 1} + x - \frac {5}{32} \, \arctan \left (\frac {\sqrt {-x^{4} + 3 \, x^{2} - 1} {\left (2 \, x^{2} - 3\right )}}{2 \, {\left (x^{4} - 3 \, x^{2} + 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3 + 1/48*(8*x^4 - 18*x^2 - 1)*sqrt(-x^4 + 3*x^2 - 1) + x - 5/32*arctan(1/2*sqrt(-x^4 + 3*x^2 - 1)*(2*x^
2 - 3)/(x^4 - 3*x^2 + 1))

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giac [A]  time = 0.45, size = 52, normalized size = 0.68 \begin {gather*} -\frac {1}{3} \, x^{3} + \frac {1}{48} \, \sqrt {-x^{4} + 3 \, x^{2} - 1} {\left (2 \, {\left (4 \, x^{2} - 9\right )} x^{2} - 1\right )} + x + \frac {5}{32} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} {\left (2 \, x^{2} - 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3 + 1/48*sqrt(-x^4 + 3*x^2 - 1)*(2*(4*x^2 - 9)*x^2 - 1) + x + 5/32*arcsin(1/5*sqrt(5)*(2*x^2 - 3))

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maple [A]  time = 0.25, size = 75, normalized size = 0.97

method	result	size
elliptic	\(-\frac {x^{3}}{3}+x +\frac {x^{4} \sqrt {-x^{4}+3 x^{2}-1}}{6}-\frac {3 x^{2} \sqrt {-x^{4}+3 x^{2}-1}}{8}-\frac {\sqrt {-x^{4}+3 x^{2}-1}}{48}+\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x^{2}-\frac {3}{2}\right )}{5}\right )}{32}\)	\(75\)
trager	\(-\frac {x \left (x^{2}-3\right )}{3}+\left (\frac {1}{6} x^{4}-\frac {3}{8} x^{2}-\frac {1}{48}\right ) \sqrt {-x^{4}+3 x^{2}-1}+\frac {5 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {-x^{4}+3 x^{2}-1}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{32}\)	\(82\)
default	\(\frac {x^{4} \sqrt {-x^{4}+3 x^{2}-1}}{6}-\frac {x^{2} \sqrt {-x^{4}+3 x^{2}-1}}{8}-\frac {19 \sqrt {-x^{4}+3 x^{2}-1}}{48}+\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, \left (x^{2}-\frac {3}{2}\right )}{5}\right )}{32}+\frac {\left (-2 x^{2}+3\right ) \sqrt {-x^{4}+3 x^{2}-1}}{8}-\frac {x^{3}}{3}+x\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3+x+1/6*x^4*(-x^4+3*x^2-1)^(1/2)-3/8*x^2*(-x^4+3*x^2-1)^(1/2)-1/48*(-x^4+3*x^2-1)^(1/2)+5/32*arcsin(2/5
*5^(1/2)*(x^2-3/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{3} \, x^{3} + x + \int {\left (x^{3} - x\right )} \sqrt {x^{2} + x - 1} \sqrt {-x^{2} + x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.40, size = 111, normalized size = 1.44 \begin {gather*} x-\frac {\left (\frac {x^2}{2}-\frac {3}{4}\right )\,\sqrt {-x^4+3\,x^2-1}}{2}-\frac {\sqrt {-x^4+3\,x^2-1}\,\left (-8\,x^4+6\,x^2+19\right )}{48}-\frac {x^3}{3}-\frac {\ln \left (x^2-\frac {3}{2}-\sqrt {-x^4+3\,x^2-1}\,1{}\mathrm {i}\right )\,15{}\mathrm {i}}{32}+\frac {\ln \left (\sqrt {-x^4+3\,x^2-1}+x^2\,1{}\mathrm {i}-\frac {3}{2}{}\mathrm {i}\right )\,5{}\mathrm {i}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - (log(x^2 - (3*x^2 - x^4 - 1)^(1/2)*1i - 3/2)*15i)/32 + (log((3*x^2 - x^4 - 1)^(1/2) + x^2*1i - 3i/2)*5i)/1
6 - ((x^2/2 - 3/4)*(3*x^2 - x^4 - 1)^(1/2))/2 - ((3*x^2 - x^4 - 1)^(1/2)*(6*x^2 - 8*x^4 + 19))/48 - x^3/3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (x - 1\right ) \left (x + 1\right ) \left (x \sqrt {- x^{4} + 3 x^{2} - 1} - 1\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x*sqrt(-x**4 + 3*x**2 - 1) - 1), x)

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