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

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

[Out]

1/2*arctan(x/(x^4+x^2+1)^(1/2))-1/3*x*(2*x^2+1)/(x^4+x^2+1)^(1/2)+2/3*x*(x^4+x^2+1)^(1/2)/(x^2+1)-2/3*(x^2+1)*
(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4
+x^2+1)^(1/2)+3/4*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2)*((x^4+x^
2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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])

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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.21, size = 204, normalized size = 1.23 \[ \frac {-2 x^3+\sqrt [3]{-1} \left (\sqrt [3]{-1}-2\right ) \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 \sqrt [3]{-1} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} E\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+3 (-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \Pi \left (\sqrt [3]{-1};i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-x}{3 \sqrt {x^4+x^2+1}} \]

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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x^{10} + 3 \, x^{8} + 5 \, x^{6} + 5 \, x^{4} + 3 \, x^{2} + 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + x^{2} + 1\right )}^{\frac {3}{2}} {\left (x^{2} + 1\right )}}\,{d x} \]

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)

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maple [C]  time = 0.02, size = 398, normalized size = 2.40 \[ \frac {8 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticE \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {2 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {8 \sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticF \left (\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {\frac {x^{2}}{2}-\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \sqrt {\frac {x^{2}}{2}+\frac {i \sqrt {3}\, x^{2}}{2}+1}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {2 \left (\frac {1}{3} x^{3}+\frac {1}{6} x \right )}{\sqrt {x^{4}+x^{2}+1}} \]

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)*(1/2*x^2-1/2*I*3^(1/2)*x^2+1)^(1/2)*(1/2*x^2+1
/2*I*3^(1/2)*x^2+1)^(1/2)/(x^4+x^2+1)^(1/2)*EllipticF(1/2*(-2+2*I*3^(1/2))^(1/2)*x,1/2*(-2+2*I*3^(1/2))^(1/2))
-8/3/(-2+2*I*3^(1/2))^(1/2)*(1/2*x^2-1/2*I*3^(1/2)*x^2+1)^(1/2)*(1/2*x^2+1/2*I*3^(1/2)*x^2+1)^(1/2)/(x^4+x^2+1
)^(1/2)/(1+I*3^(1/2))*EllipticF(1/2*(-2+2*I*3^(1/2))^(1/2)*x,1/2*(-2+2*I*3^(1/2))^(1/2))+8/3/(-2+2*I*3^(1/2))^
(1/2)*(1/2*x^2-1/2*I*3^(1/2)*x^2+1)^(1/2)*(1/2*x^2+1/2*I*3^(1/2)*x^2+1)^(1/2)/(x^4+x^2+1)^(1/2)/(1+I*3^(1/2))*
EllipticE(1/2*(-2+2*I*3^(1/2))^(1/2)*x,1/2*(-2+2*I*3^(1/2))^(1/2))+1/(-1/2+1/2*I*3^(1/2))^(1/2)*(1/2*x^2-1/2*I
*3^(1/2)*x^2+1)^(1/2)*(1/2*x^2+1/2*I*3^(1/2)*x^2+1)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + x^{2} + 1\right )}^{\frac {3}{2}} {\left (x^{2} + 1\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (x^2+1\right )\,{\left (x^4+x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {3}{2}} \left (x^{2} + 1\right )}\, dx \]

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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