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

Optimal. Leaf size=166 \[ \frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}(x),\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}}+\frac{2 \sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}-\frac{\left (2 x^2+1\right ) x}{3 \sqrt{x^4+x^2+1}}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )-\frac{2 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.103757, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1221, 1119, 1197, 1103, 1195, 1210, 1698, 203} \[ \frac{2 \sqrt{x^4+x^2+1} x}{3 \left (x^2+1\right )}-\frac{\left (2 x^2+1\right ) x}{3 \sqrt{x^4+x^2+1}}+\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^4+x^2+1}}\right )+\frac{3 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{4 \sqrt{x^4+x^2+1}}-\frac{2 \left (x^2+1\right ) \sqrt{\frac{x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac{1}{4}\right )}{3 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1221

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

Rule 1119

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(d*x)^(m - 1)*(b + 2*c*
x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*(p + 1)*(b^2 - 4*a*c)), x] - Dist[d^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(d*x
)^(m - 2)*(b*(m - 1) + 2*c*(m + 4*p + 5)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

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]

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{1}{\left (1+x^2\right ) \left (1+x^2+x^4\right )^{3/2}} \, dx &=-\int \frac{x^2}{\left (1+x^2+x^4\right )^{3/2}} \, dx+\int \frac{1}{\left (1+x^2\right ) \sqrt{1+x^2+x^4}} \, dx\\ &=-\frac{x \left (1+2 x^2\right )}{3 \sqrt{1+x^2+x^4}}+\frac{1}{3} \int \frac{1+2 x^2}{\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\\ &=-\frac{x \left (1+2 x^2\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 )}{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{2}{3} \int \frac{1-x^2}{\sqrt{1+x^2+x^4}} \, dx+\int \frac{1}{\sqrt{1+x^2+x^4}} \, dx\\ &=-\frac{x \left (1+2 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{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{1+x^2+x^4}}\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{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 = 0.213119, size = 204, normalized size = 1.23 \[ \frac{\sqrt [3]{-1} \left (\sqrt [3]{-1}-2\right ) \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \text{EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-2 x^3+2 \sqrt [3]{-1} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-3 (-1)^{2/3} \sqrt{\sqrt [3]{-1} x^2+1} \sqrt{1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};-i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-x}{3 \sqrt{x^4+x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.017, size = 398, normalized size = 2.4 \begin{align*} -2\,{\frac{1/3\,{x}^{3}+x/6}{\sqrt{{x}^{4}+{x}^{2}+1}}}+{\frac{2}{3\,\sqrt{-2+2\,i\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 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+{\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 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+{\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 EllipticE} \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{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(1/(x^2+1)/(x^4+x^2+1)^(3/2),x)

[Out]

-2*(1/3*x^3+1/6*x)/(x^4+x^2+1)^(1/2)+2/3/(-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)*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*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))+8/3/(-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)*
EllipticE(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{1}{{\left (x^{4} + x^{2} + 1\right )}^{\frac{3}{2}}{\left (x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^4 + x^2 + 1)^(3/2)*(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^{10} + 3 \, x^{8} + 5 \, x^{6} + 5 \, x^{4} + 3 \, x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + x^2 + 1)/(x^10 + 3*x^8 + 5*x^6 + 5*x^4 + 3*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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