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

Optimal. Leaf size=172 \[ \frac{31 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{35 \sqrt{x^4+3 x^2+2}}+\frac{1}{7} x \left (x^4+3 x^2+2\right )^{3/2}+\frac{1}{35} x \left (9 x^2+29\right ) \sqrt{x^4+3 x^2+2}+\frac{6 x \left (x^2+2\right )}{5 \sqrt{x^4+3 x^2+2}}-\frac{6 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+3 x^2+2}} \]

[Out]

(6*x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) + (x*(29 + 9*x^2)*Sqrt[2 + 3*x^2 + x^4])/35 + (x*(2 + 3*x^2 + x^4)^(
3/2))/7 - (6*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(5*Sqrt[2 + 3*x^2 + x^4])
+ (31*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(35*Sqrt[2 + 3*x^2 + x^4])

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Rubi [A]  time = 0.058111, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {1091, 1176, 1189, 1099, 1135} \[ \frac{1}{7} x \left (x^4+3 x^2+2\right )^{3/2}+\frac{1}{35} x \left (9 x^2+29\right ) \sqrt{x^4+3 x^2+2}+\frac{6 x \left (x^2+2\right )}{5 \sqrt{x^4+3 x^2+2}}+\frac{31 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{35 \sqrt{x^4+3 x^2+2}}-\frac{6 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(6*x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) + (x*(29 + 9*x^2)*Sqrt[2 + 3*x^2 + x^4])/35 + (x*(2 + 3*x^2 + x^4)^(
3/2))/7 - (6*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(5*Sqrt[2 + 3*x^2 + x^4])
+ (31*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(35*Sqrt[2 + 3*x^2 + x^4])

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 1189

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[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \left (2+3 x^2+x^4\right )^{3/2} \, dx &=\frac{1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac{3}{7} \int \left (4+3 x^2\right ) \sqrt{2+3 x^2+x^4} \, dx\\ &=\frac{1}{35} x \left (29+9 x^2\right ) \sqrt{2+3 x^2+x^4}+\frac{1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac{1}{35} \int \frac{62+42 x^2}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{1}{35} x \left (29+9 x^2\right ) \sqrt{2+3 x^2+x^4}+\frac{1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac{6}{5} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{62}{35} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{6 x \left (2+x^2\right )}{5 \sqrt{2+3 x^2+x^4}}+\frac{1}{35} x \left (29+9 x^2\right ) \sqrt{2+3 x^2+x^4}+\frac{1}{7} x \left (2+3 x^2+x^4\right )^{3/2}-\frac{6 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{2+3 x^2+x^4}}+\frac{31 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{35 \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0421048, size = 114, normalized size = 0.66 \[ \frac{-20 i \sqrt{x^2+1} \sqrt{x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )+5 x^9+39 x^7+121 x^5+165 x^3-42 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+78 x}{35 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(78*x + 165*x^3 + 121*x^5 + 39*x^7 + 5*x^9 - (42*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]]
, 2] - (20*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/(35*Sqrt[2 + 3*x^2 + x^4])

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Maple [C]  time = 0.003, size = 155, normalized size = 0.9 \begin{align*}{\frac{{x}^{5}}{7}\sqrt{{x}^{4}+3\,{x}^{2}+2}}+{\frac{24\,{x}^{3}}{35}\sqrt{{x}^{4}+3\,{x}^{2}+2}}+{\frac{39\,x}{35}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{{\frac{31\,i}{35}}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}+{{\frac{3\,i}{5}}\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*x^5*(x^4+3*x^2+2)^(1/2)+24/35*x^3*(x^4+3*x^2+2)^(1/2)+39/35*x*(x^4+3*x^2+2)^(1/2)-31/35*I*2^(1/2)*(2*x^2+4
)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2^(1/2))+3/5*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^
2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*x*2^(1/2),2^(1/2))-EllipticE(1/2*I*x*2^(1/2),2^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (x^{4} + 3 \, 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+3*x^2+2)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((x^4 + 3*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} + 3 x^{2} + 2\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**4 + 3*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} + 3 \, 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+3*x^2+2)^(3/2),x, algorithm="giac")

[Out]

integrate((x^4 + 3*x^2 + 2)^(3/2), x)