3.238 \(\int \frac{(1+x^2)^3}{(1+x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{\sqrt{x^4+x^2+1}}+\frac{2 \sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}-\frac{\left (1-x^2\right ) x}{3 \sqrt{x^4+x^2+1}}-\frac{2 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.0445113, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1205, 1197, 1103, 1195} \[ \frac{2 \sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}-\frac{\left (1-x^2\right ) x}{3 \sqrt{x^4+x^2+1}}+\frac{\left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{\sqrt{x^4+x^2+1}}-\frac{2 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)^3/(1 + x^2 + x^4)^(3/2),x]

[Out]

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

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], 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[q, 1] && LtQ[p, -1]

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 \frac{\left (1+x^2\right )^3}{\left (1+x^2+x^4\right )^{3/2}} \, dx &=-\frac{x \left (1-x^2\right )}{3 \sqrt{1+x^2+x^4}}+\frac{1}{3} \int \frac{4+2 x^2}{\sqrt{1+x^2+x^4}} \, dx\\ &=-\frac{x \left (1-x^2\right )}{3 \sqrt{1+x^2+x^4}}-\frac{2}{3} \int \frac{1-x^2}{\sqrt{1+x^2+x^4}} \, dx+2 \int \frac{1}{\sqrt{1+x^2+x^4}} \, dx\\ &=-\frac{x \left (1-x^2\right )}{3 \sqrt{1+x^2+x^4}}+\frac{2 x \sqrt{1+x^2+x^4}}{3 \left (1+x^2\right )}-\frac{2 \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 )}{3 \sqrt{1+x^2+x^4}}+\frac{\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 )}{\sqrt{1+x^2+x^4}}\\ \end{align*}

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(1 + x^2)^3/(1 + x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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Maple [C]  time = 0.033, size = 268, normalized size = 1.9 \begin{align*} -4\,{\frac{-x/6+1/6\,{x}^{3}}{\sqrt{{x}^{4}+{x}^{2}+1}}}+{\frac{8}{3\,\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{8}{3\,\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}}}}-6\,{\frac{1/6\,{x}^{3}+x/3}{\sqrt{{x}^{4}+{x}^{2}+1}}}-6\,{\frac{-1/3\,{x}^{3}-x/6}{\sqrt{{x}^{4}+{x}^{2}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(-1/6*x+1/6*x^3)/(x^4+x^2+1)^(1/2)+8/3/(-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))-
8/3/(-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)))-6*(1/6*x^3+1/3*x)/(x^4+x^2+1)^(1/2)-6*(-1/3*x^3-1/6*x)/(x^4+x^2+1
)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{2} + 1\right )}^{3}}{{\left (x^{4} + x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (x^{6} + 3 \, x^{4} + 3 \, x^{2} + 1\right )} \sqrt{x^{4} + x^{2} + 1}}{x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^6 + 3*x^4 + 3*x^2 + 1)*sqrt(x^4 + x^2 + 1)/(x^8 + 2*x^6 + 3*x^4 + 2*x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} + 1\right )^{3}}{\left (\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2 + 1)**3/((x**2 - x + 1)*(x**2 + x + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{2} + 1\right )}^{3}}{{\left (x^{4} + x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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