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

Optimal. Leaf size=159 \[ \frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}}+\frac{1}{5} \sqrt{x^4+x^2+1} x^3+\frac{14 \sqrt{x^4+x^2+1} x}{15 \left (x^2+1\right )}+\frac{11}{15} \sqrt{x^4+x^2+1} x-\frac{14 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{15 \sqrt{x^4+x^2+1}} \]

[Out]

(11*x*Sqrt[1 + x^2 + x^4])/15 + (x^3*Sqrt[1 + x^2 + x^4])/5 + (14*x*Sqrt[1 + x^2 + x^4])/(15*(1 + x^2)) - (14*
(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(15*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 = 0.0714685, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1206, 1679, 1197, 1103, 1195} \[ \frac{1}{5} \sqrt{x^4+x^2+1} x^3+\frac{14 \sqrt{x^4+x^2+1} x}{15 \left (x^2+1\right )}+\frac{11}{15} \sqrt{x^4+x^2+1} x+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{5 \sqrt{x^4+x^2+1}}-\frac{14 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{15 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*x*Sqrt[1 + x^2 + x^4])/15 + (x^3*Sqrt[1 + x^2 + x^4])/5 + (14*x*Sqrt[1 + x^2 + x^4])/(15*(1 + x^2)) - (14*
(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(15*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 1206

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(
a + b*x^2 + c*x^4)^(p + 1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !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}{\sqrt{1+x^2+x^4}} \, dx &=\frac{1}{5} x^3 \sqrt{1+x^2+x^4}+\frac{1}{5} \int \frac{5+12 x^2+11 x^4}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{11}{15} x \sqrt{1+x^2+x^4}+\frac{1}{5} x^3 \sqrt{1+x^2+x^4}+\frac{1}{15} \int \frac{4+14 x^2}{\sqrt{1+x^2+x^4}} \, dx\\ &=\frac{11}{15} x \sqrt{1+x^2+x^4}+\frac{1}{5} x^3 \sqrt{1+x^2+x^4}-\frac{14}{15} \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{11}{15} x \sqrt{1+x^2+x^4}+\frac{1}{5} x^3 \sqrt{1+x^2+x^4}+\frac{14 x \sqrt{1+x^2+x^4}}{15 \left (1+x^2\right )}-\frac{14 \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 )}{15 \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 = 0.173071, size = 157, normalized size = 0.99 \[ \frac{2 \sqrt [3]{-1} \left (2 \sqrt [3]{-1}-7\right ) \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )+x \left (3 x^6+14 x^4+14 x^2+11\right )+14 \sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )}{15 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(11 + 14*x^2 + 14*x^4 + 3*x^6) + 14*(-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)] + 2*(-1)^(1/3)*(-7 + 2*(-1)^(1/3))*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)
^(2/3)*x^2]*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)])/(15*Sqrt[1 + x^2 + x^4])

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Maple [C]  time = 0.027, size = 233, normalized size = 1.5 \begin{align*}{\frac{{x}^{3}}{5}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{11\,x}{15}\sqrt{{x}^{4}+{x}^{2}+1}}+{\frac{8}{15\,\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{56}{15\,\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)^3/(x^4+x^2+1)^(1/2),x)

[Out]

1/5*x^3*(x^4+x^2+1)^(1/2)+11/15*x*(x^4+x^2+1)^(1/2)+8/15/(-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))-56/15/(-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))-Ellipt
icE(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 \frac{{\left (x^{2} + 1\right )}^{3}}{\sqrt{x^{4} + x^{2} + 1}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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

[Out]

integral((x^6 + 3*x^4 + 3*x^2 + 1)/sqrt(x^4 + 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}}{\sqrt{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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