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

Optimal. Leaf size=137 \[ \frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}}+\frac{\sqrt{x^4+x^2+1} x}{x^2+1}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )-\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{\sqrt{x^4+x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.0861591, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1208, 1139, 1103, 1195, 1210, 1698, 203} \[ \frac{\sqrt{x^4+x^2+1} x}{x^2+1}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}}-\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{\sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^2 + x^4]/(1 + x^2),x]

[Out]

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

Rule 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 0]

Rule 1139

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

Rule 1210

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

Rule 1698

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{1+x^2+x^4}}{1+x^2} \, dx &=\int \frac{x^2}{\sqrt{1+x^2+x^4}} \, dx+\int \frac{1}{\left (1+x^2\right ) \sqrt{1+x^2+x^4}} \, dx\\ &=\frac{1}{2} \int \frac{1}{\sqrt{1+x^2+x^4}} \, dx+\frac{1}{2} \int \frac{1-x^2}{\left (1+x^2\right ) \sqrt{1+x^2+x^4}} \, dx+\int \frac{1}{\sqrt{1+x^2+x^4}} \, dx-\int \frac{1-x^2}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{x \sqrt{1+x^2+x^4}}{1+x^2}-\frac{\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 )}{\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 )}{4 \sqrt{1+x^2+x^4}}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{x}{\sqrt{1+x^2+x^4}}\right )\\ &=\frac{x \sqrt{1+x^2+x^4}}{1+x^2}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{1+x^2+x^4}}\right )-\frac{\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 )}{\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 )}{4 \sqrt{1+x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0951932, size = 118, normalized size = 0.86 \[ -\frac{\sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \left (\text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+\sqrt [3]{-1} \Pi \left (\sqrt [3]{-1};-i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )\right )}{\sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.052, size = 293, normalized size = 2.1 \begin{align*} -4\,{\frac{\sqrt{1+1/2\,{x}^{2}-i/2{x}^{2}\sqrt{3}}\sqrt{1+1/2\,{x}^{2}+i/2{x}^{2}\sqrt{3}}{\it EllipticF} \left ( 1/2\,x\sqrt{-2+2\,i\sqrt{3}},1/2\,\sqrt{-2+2\,i\sqrt{3}} \right ) }{\sqrt{-2+2\,i\sqrt{3}}\sqrt{{x}^{4}+{x}^{2}+1} \left ( i\sqrt{3}+1 \right ) }}+4\,{\frac{\sqrt{1+1/2\,{x}^{2}-i/2{x}^{2}\sqrt{3}}\sqrt{1+1/2\,{x}^{2}+i/2{x}^{2}\sqrt{3}}{\it EllipticE} \left ( 1/2\,x\sqrt{-2+2\,i\sqrt{3}},1/2\,\sqrt{-2+2\,i\sqrt{3}} \right ) }{\sqrt{-2+2\,i\sqrt{3}}\sqrt{{x}^{4}+{x}^{2}+1} \left ( i\sqrt{3}+1 \right ) }}+{\frac{1}{\sqrt{-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3}}}\sqrt{1+{\frac{{x}^{2}}{2}}-{\frac{i}{2}}{x}^{2}\sqrt{3}}\sqrt{1+{\frac{{x}^{2}}{2}}+{\frac{i}{2}}{x}^{2}\sqrt{3}}{\it EllipticPi} \left ( \sqrt{-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3}}x,- \left ( -{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) ^{-1},{\frac{\sqrt{-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3}}}{\sqrt{-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{{x}^{4}+{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(-2+2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(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))+4/(-2+2*I*3^(1/2))^(1/2
)*(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)/(I*3^(1/2)+1)*Elli
pticE(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))+1/(-1/2+1/2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x
^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)*EllipticPi((-1/2+1/2*I*3^(1/2))^(1/2)*
x,-1/(-1/2+1/2*I*3^(1/2)),(-1/2-1/2*I*3^(1/2))^(1/2)/(-1/2+1/2*I*3^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x^2+1)^(1/2)/(x^2+1),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 (\frac{\sqrt{x^{4} + x^{2} + 1}}{x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x^2+1)^(1/2)/(x^2+1),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 \frac{\sqrt{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}{x^{2} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+x**2+1)**(1/2)/(x**2+1),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 \frac{\sqrt{x^{4} + x^{2} + 1}}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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