3.328 \(\int (2+x^2-x^4)^{3/2} \, dx\)

Optimal. Leaf size=74 \[ \frac {1}{7} x \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{35} x \left (3 x^2+19\right ) \sqrt {-x^4+x^2+2}+\frac {48}{35} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {34}{35} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

1/7*x*(-x^4+x^2+2)^(3/2)+34/35*EllipticE(1/2*x*2^(1/2),I*2^(1/2))+48/35*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+1/3
5*x*(3*x^2+19)*(-x^4+x^2+2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1091, 1176, 1180, 524, 424, 419} \[ \frac {1}{7} x \left (-x^4+x^2+2\right )^{3/2}+\frac {1}{35} x \left (3 x^2+19\right ) \sqrt {-x^4+x^2+2}+\frac {48}{35} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {34}{35} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + x^2 - x^4)^(3/2),x]

[Out]

(x*(19 + 3*x^2)*Sqrt[2 + x^2 - x^4])/35 + (x*(2 + x^2 - x^4)^(3/2))/7 + (34*EllipticE[ArcSin[x/Sqrt[2]], -2])/
35 + (48*EllipticF[ArcSin[x/Sqrt[2]], -2])/35

Rule 419

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

Rule 424

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

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 1091

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[2*Sqrt[-c], Int[(d + e*x^2)/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c,
d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]

Rubi steps

\begin {align*} \int \left (2+x^2-x^4\right )^{3/2} \, dx &=\frac {1}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {3}{7} \int \left (4+x^2\right ) \sqrt {2+x^2-x^4} \, dx\\ &=\frac {1}{35} x \left (19+3 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{7} x \left (2+x^2-x^4\right )^{3/2}-\frac {1}{35} \int \frac {-82-34 x^2}{\sqrt {2+x^2-x^4}} \, dx\\ &=\frac {1}{35} x \left (19+3 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{7} x \left (2+x^2-x^4\right )^{3/2}-\frac {2}{35} \int \frac {-82-34 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {1}{35} x \left (19+3 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {34}{35} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx+\frac {96}{35} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx\\ &=\frac {1}{35} x \left (19+3 x^2\right ) \sqrt {2+x^2-x^4}+\frac {1}{7} x \left (2+x^2-x^4\right )^{3/2}+\frac {34}{35} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {48}{35} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 102, normalized size = 1.38 \[ \frac {5 x^9-13 x^7-31 x^5+45 x^3-75 i \sqrt {-2 x^4+2 x^2+4} F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+34 i \sqrt {-2 x^4+2 x^2+4} E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+58 x}{35 \sqrt {-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^2 - x^4)^(3/2),x]

[Out]

(58*x + 45*x^3 - 31*x^5 - 13*x^7 + 5*x^9 + (34*I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticE[I*ArcSinh[x], -1/2] - (75*
I)*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticF[I*ArcSinh[x], -1/2])/(35*Sqrt[2 + x^2 - x^4])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-x^4 + x^2 + 2)^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-x^4 + x^2 + 2)^(3/2), x)

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maple [B]  time = 0.00, size = 159, normalized size = 2.15 \[ -\frac {\sqrt {-x^{4}+x^{2}+2}\, x^{5}}{7}+\frac {8 \sqrt {-x^{4}+x^{2}+2}\, x^{3}}{35}+\frac {29 \sqrt {-x^{4}+x^{2}+2}\, x}{35}+\frac {41 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{35 \sqrt {-x^{4}+x^{2}+2}}-\frac {17 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{35 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(-x^4+x^2+2)^(1/2)*x^5+8/35*(-x^4+x^2+2)^(1/2)*x^3+29/35*(-x^4+x^2+2)^(1/2)*x+41/35*2^(1/2)*(-2*x^2+4)^(1
/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*2^(1/2)*x,I*2^(1/2))-17/35*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)
^(1/2)/(-x^4+x^2+2)^(1/2)*(EllipticF(1/2*2^(1/2)*x,I*2^(1/2))-EllipticE(1/2*2^(1/2)*x,I*2^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-x^4 + x^2 + 2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (-x^4+x^2+2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - x^4 + 2)^(3/2),x)

[Out]

int((x^2 - x^4 + 2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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