3.227 \(\int (1+x^2) \sqrt{1+x^2+x^4} \, dx\)

Optimal. Leaf size=145 \[ \frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}}+\frac{1}{5} \left (x^2+2\right ) \sqrt{x^4+x^2+1} x+\frac{3 \sqrt{x^4+x^2+1} x}{5 \left (x^2+1\right )}-\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}} \]

[Out]

(3*x*Sqrt[1 + x^2 + x^4])/(5*(1 + x^2)) + (x*(2 + x^2)*Sqrt[1 + x^2 + x^4])/5 - (3*(1 + x^2)*Sqrt[(1 + x^2 + x
^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(5*Sqrt[1 + x^2 + x^4]) + (3*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 +
x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(5*Sqrt[1 + x^2 + x^4])

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Rubi [A]  time = 0.0425715, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1176, 1197, 1103, 1195} \[ \frac{1}{5} \left (x^2+2\right ) \sqrt{x^4+x^2+1} x+\frac{3 \sqrt{x^4+x^2+1} x}{5 \left (x^2+1\right )}+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}}-\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)*Sqrt[1 + x^2 + x^4],x]

[Out]

(3*x*Sqrt[1 + x^2 + x^4])/(5*(1 + x^2)) + (x*(2 + x^2)*Sqrt[1 + x^2 + x^4])/5 - (3*(1 + x^2)*Sqrt[(1 + x^2 + x
^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(5*Sqrt[1 + x^2 + x^4]) + (3*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 +
x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/(5*Sqrt[1 + x^2 + x^4])

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 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 \left (1+x^2\right ) \sqrt{1+x^2+x^4} \, dx &=\frac{1}{5} x \left (2+x^2\right ) \sqrt{1+x^2+x^4}+\frac{1}{15} \int \frac{9+9 x^2}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{1}{5} x \left (2+x^2\right ) \sqrt{1+x^2+x^4}-\frac{3}{5} \int \frac{1-x^2}{\sqrt{1+x^2+x^4}} \, dx+\frac{6}{5} \int \frac{1}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{3 x \sqrt{1+x^2+x^4}}{5 \left (1+x^2\right )}+\frac{1}{5} x \left (2+x^2\right ) \sqrt{1+x^2+x^4}-\frac{3 \left (1+x^2\right ) \sqrt{\frac{1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{1+x^2+x^4}}+\frac{3 \left (1+x^2\right ) \sqrt{\frac{1+x^2+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{1+x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.183043, size = 168, normalized size = 1.16 \[ \frac{\frac{3}{2} \sqrt{2+\left (1-i \sqrt{3}\right ) x^2} \sqrt{2+\left (1+i \sqrt{3}\right ) x^2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{2} \left (x+i \sqrt{3} x\right )\right ),\frac{1}{2} i \left (\sqrt{3}+i\right )\right )+x^7+3 x^5+3 x^3+3 \sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 x}{5 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x + 3*x^3 + 3*x^5 + x^7 + 3*(-1)^(1/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*EllipticE[I*ArcSin
h[(-1)^(5/6)*x], (-1)^(2/3)] + (3*Sqrt[2 + (1 - I*Sqrt[3])*x^2]*Sqrt[2 + (1 + I*Sqrt[3])*x^2]*EllipticF[ArcSin
[(x + I*Sqrt[3]*x)/2], (I/2)*(I + Sqrt[3])])/2)/(5*Sqrt[1 + x^2 + x^4])

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Maple [C]  time = 0.006, size = 233, normalized size = 1.6 \begin{align*}{\frac{{x}^{3}}{5}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{2\,x}{5}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{6}{5\,\sqrt{-2+2\,i\sqrt{3}}}\sqrt{1- \left ( -{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ){x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}}-{\frac{12}{5\,\sqrt{-2+2\,i\sqrt{3}} \left ( i\sqrt{3}+1 \right ) }\sqrt{1- \left ( -{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ){x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{-2+2\,i\sqrt{3}}}{2}},{\frac{\sqrt{-2+2\,i\sqrt{3}}}{2}} \right ) \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^3*(x^4+x^2+1)^(1/2)+2/5*x*(x^4+x^2+1)^(1/2)+6/5/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2
)*(1-(-1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1/2)*EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1
/2))^(1/2))-12/5/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2*I*3^(1/2))*x^2)^(1/2)/
(x^4+x^2+1)^(1/2)/(I*3^(1/2)+1)*(EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))-EllipticE(
1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + x^{2} + 1}{\left (x^{2} + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + x^2 + 1)*(x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{x^{4} + x^{2} + 1}{\left (x^{2} + 1\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + x^2 + 1)*(x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 1\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{4} + x^{2} + 1}{\left (x^{2} + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + x^2 + 1)*(x^2 + 1), x)