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

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

[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

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Rubi [A]  time = 0.0516522, antiderivative size = 74, normalized size of antiderivative = 1., 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 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]

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 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 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)])

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*}

Mathematica [C]  time = 0.0533497, size = 102, normalized size = 1.38 \[ \frac{-75 i \sqrt{-2 x^4+2 x^2+4} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )+5 x^9-13 x^7-31 x^5+45 x^3+34 i \sqrt{-2 x^4+2 x^2+4} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )+58 x}{35 \sqrt{-x^4+x^2+2}} \]

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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Maple [B]  time = 0.003, size = 159, normalized size = 2.2 \begin{align*} -{\frac{{x}^{5}}{7}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{8\,{x}^{3}}{35}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{29\,x}{35}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{41\,\sqrt{2}}{35}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}}-{\frac{17\,\sqrt{2}}{35}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1} \left ({\it EllipticF} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- x^{4} + x^{2} + 2\right )^{\frac{3}{2}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}\,{d x} \end{align*}

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)