3.352 \(\int \sqrt{4+3 x^2+x^4} \, dx\)

Optimal. Leaf size=169 \[ \frac{7 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{3 \sqrt{2} \sqrt{x^4+3 x^2+4}}+\frac{\sqrt{x^4+3 x^2+4} x}{x^2+2}+\frac{1}{3} \sqrt{x^4+3 x^2+4} x-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{x^4+3 x^2+4}} \]

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/3 + (x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) - (Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/
(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] + (7*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/
(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(3*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

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Rubi [A]  time = 0.0508969, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1091, 1197, 1103, 1195} \[ \frac{\sqrt{x^4+3 x^2+4} x}{x^2+2}+\frac{1}{3} \sqrt{x^4+3 x^2+4} x+\frac{7 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3 \sqrt{2} \sqrt{x^4+3 x^2+4}}-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[4 + 3*x^2 + x^4],x]

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/3 + (x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) - (Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/
(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] + (7*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/
(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(3*Sqrt[2]*Sqrt[4 + 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 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1103

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

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \sqrt{4+3 x^2+x^4} \, dx &=\frac{1}{3} x \sqrt{4+3 x^2+x^4}+\frac{1}{3} \int \frac{8+3 x^2}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{1}{3} x \sqrt{4+3 x^2+x^4}-2 \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{14}{3} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{1}{3} x \sqrt{4+3 x^2+x^4}+\frac{x \sqrt{4+3 x^2+x^4}}{2+x^2}-\frac{\sqrt{2} \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{4+3 x^2+x^4}}+\frac{7 \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{3 \sqrt{2} \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.353942, size = 331, normalized size = 1.96 \[ \frac{\sqrt{2} \left (3 \sqrt{7}-7 i\right ) \sqrt{\frac{-2 i x^2+\sqrt{7}-3 i}{\sqrt{7}-3 i}} \sqrt{\frac{2 i x^2+\sqrt{7}+3 i}{\sqrt{7}+3 i}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{\sqrt{7}-3 i}} x\right ),\frac{-\sqrt{7}+3 i}{\sqrt{7}+3 i}\right )+4 \sqrt{-\frac{i}{\sqrt{7}-3 i}} x \left (x^4+3 x^2+4\right )-3 \sqrt{2} \left (\sqrt{7}+3 i\right ) \sqrt{\frac{-2 i x^2+\sqrt{7}-3 i}{\sqrt{7}-3 i}} \sqrt{\frac{2 i x^2+\sqrt{7}+3 i}{\sqrt{7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )}{12 \sqrt{-\frac{i}{\sqrt{7}-3 i}} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[4 + 3*x^2 + x^4],x]

[Out]

(4*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(4 + 3*x^2 + x^4) - 3*Sqrt[2]*(3*I + Sqrt[7])*Sqrt[(-3*I + Sqrt[7] - (2*I)*x^
2)/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])]*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I +
Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + Sqrt[2]*(-7*I + 3*Sqrt[7])*Sqrt[(-3*I + Sqrt[7] - (2*I)*x^2)
/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])]*EllipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I + S
qrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(12*Sqrt[(-I)/(-3*I + Sqrt[7])]*Sqrt[4 + 3*x^2 + x^4])

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Maple [C]  time = 0.003, size = 224, normalized size = 1.3 \begin{align*}{\frac{x}{3}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{32}{3\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1- \left ( -{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7} \right ){x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}-32\,{\frac{\sqrt{1- \left ( -3/8+i/8\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -3/8-i/8\sqrt{7} \right ){x}^{2}} \left ({\it EllipticF} \left ( 1/4\,x\sqrt{-6+2\,i\sqrt{7}},1/4\,\sqrt{2+6\,i\sqrt{7}} \right ) -{\it EllipticE} \left ( 1/4\,x\sqrt{-6+2\,i\sqrt{7}},1/4\,\sqrt{2+6\,i\sqrt{7}} \right ) \right ) }{\sqrt{-6+2\,i\sqrt{7}}\sqrt{{x}^{4}+3\,{x}^{2}+4} \left ( i\sqrt{7}+3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(x^4+3*x^2+4)^(1/2)+32/3/(-6+2*I*7^(1/2))^(1/2)*(1-(-3/8+1/8*I*7^(1/2))*x^2)^(1/2)*(1-(-3/8-1/8*I*7^(1/2
))*x^2)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))-32/(-6+2*I
*7^(1/2))^(1/2)*(1-(-3/8+1/8*I*7^(1/2))*x^2)^(1/2)*(1-(-3/8-1/8*I*7^(1/2))*x^2)^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7
^(1/2)+3)*(EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))-EllipticE(1/4*x*(-6+2*I*7^(1/2))^
(1/2),1/4*(2+6*I*7^(1/2))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + 3 \, x^{2} + 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 3*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + 3 x^{2} + 4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**4 + 3*x**2 + 4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + 3 \, x^{2} + 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 3*x^2 + 4), x)