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

Optimal. Leaf size=468 \[ \frac {2 \sqrt {2} \sqrt {2 x^4+2 x^2+1} x}{15 \left (\sqrt {2} x^2+1\right )}+\frac {2 \left (3 x^2+1\right ) x}{15 \sqrt {2 x^4+2 x^2+1}}-\frac {x}{3 \sqrt {2 x^4+2 x^2+1}}-\frac {\sqrt {2 x^4+2 x^2+1}}{3 x}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{15 \sqrt {15}}+\frac {\left (3 \sqrt {2}-7\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{3\ 2^{3/4} \left (3 \sqrt {2}-2\right ) \sqrt {2 x^4+2 x^2+1}}-\frac {2 \sqrt [4]{2} \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{15 \sqrt {2 x^4+2 x^2+1}}-\frac {\sqrt [4]{2} \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{45 \left (2-3 \sqrt {2}\right ) \sqrt {2 x^4+2 x^2+1}} \]

[Out]

-2/225*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/3*x/(2*x^4+2*x^2+1)^(1/2)+2/15*x*(3*x^2+1)/(2*x
^4+2*x^2+1)^(1/2)-1/3*(2*x^4+2*x^2+1)^(1/2)/x+2/15*x*(2*x^4+2*x^2+1)^(1/2)*2^(1/2)/(1+x^2*2^(1/2))-2/15*(cos(2
*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticE(sin(2*arctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2)
)*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(2*x^4+2*x^2+1)^(1/2)-1/45*(cos(2*arctan(2
^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticPi(sin(2*arctan(2^(1/4)*x)),1/2-11/24*2^(1/2),1/2*(2-2^(1
/2))^(1/2))*(3+2^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(2-3*2^(1/2))/(2*x^4
+2*x^2+1)^(1/2)+1/6*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticF(sin(2*arctan(2^(1/4)
*x)),1/2*(2-2^(1/2))^(1/2))*(-7+3*2^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(
-2+3*2^(1/2))/(2*x^4+2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.45, antiderivative size = 644, normalized size of antiderivative = 1.38, number of steps used = 15, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {1335, 1121, 1281, 1197, 1103, 1195, 1221, 1178, 1216, 1706} \[ \frac {2 \sqrt {2} \sqrt {2 x^4+2 x^2+1} x}{15 \left (\sqrt {2} x^2+1\right )}+\frac {2 \left (3 x^2+1\right ) x}{15 \sqrt {2 x^4+2 x^2+1}}-\frac {x}{3 \sqrt {2 x^4+2 x^2+1}}-\frac {\sqrt {2 x^4+2 x^2+1}}{3 x}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{15 \sqrt {15}}-\frac {\left (3+2 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{15\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {2^{3/4} \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{105 \sqrt {2 x^4+2 x^2+1}}-\frac {\left (1-\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{6 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {2 \sqrt [4]{2} \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{15 \sqrt {2 x^4+2 x^2+1}}+\frac {\left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{315 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(3*Sqrt[1 + 2*x^2 + 2*x^4]) + (2*x*(1 + 3*x^2))/(15*Sqrt[1 + 2*x^2 + 2*x^4]) - Sqrt[1 + 2*x^2 + 2*x^4]/(3*x
) + (2*Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(15*(1 + Sqrt[2]*x^2)) - (2*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*
x^4]])/(15*Sqrt[15]) - (2*2^(1/4)*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticE[2*
ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(15*Sqrt[1 + 2*x^2 + 2*x^4]) - ((1 - Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 +
2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(6*2^(1/4)*Sqrt[1 + 2*x^
2 + 2*x^4]) - (2^(3/4)*(3 + Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF
[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(105*Sqrt[1 + 2*x^2 + 2*x^4]) - ((3 + 2*Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqr
t[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(15*2^(3/4)*Sqrt[1
+ 2*x^2 + 2*x^4]) + ((3 + Sqrt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*Elliptic
Pi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(315*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

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 1121

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[((d*x)^(m + 1)*(b^2 - 2*a
*c + b*c*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*d*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*
c)), Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*
x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (Integer
Q[p] || IntegerQ[m])

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

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 1216

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Di
st[(c*d + a*e*q)/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2)
, Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), 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] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]

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 1281

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

Rule 1335

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*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] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (3+2 x^2\right ) \left (1+2 x^2+2 x^4\right )^{3/2}} \, dx &=\int \left (\frac {1}{3 x^2 \left (1+2 x^2+2 x^4\right )^{3/2}}-\frac {2}{3 \left (3+2 x^2\right ) \left (1+2 x^2+2 x^4\right )^{3/2}}\right ) \, dx\\ &=\frac {1}{3} \int \frac {1}{x^2 \left (1+2 x^2+2 x^4\right )^{3/2}} \, dx-\frac {2}{3} \int \frac {1}{\left (3+2 x^2\right ) \left (1+2 x^2+2 x^4\right )^{3/2}} \, dx\\ &=-\frac {x}{3 \sqrt {1+2 x^2+2 x^4}}-\frac {1}{15} \int \frac {2-4 x^2}{\left (1+2 x^2+2 x^4\right )^{3/2}} \, dx+\frac {1}{12} \int \frac {4-4 x^2}{x^2 \sqrt {1+2 x^2+2 x^4}} \, dx-\frac {4}{15} \int \frac {1}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {x}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {2 x \left (1+3 x^2\right )}{15 \sqrt {1+2 x^2+2 x^4}}-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}-\frac {1}{60} \int \frac {16+24 x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{12} \int \frac {4-8 x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{105} \left (4 \left (3+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx+\frac {1}{105} \left (4 \left (2+3 \sqrt {2}\right )\right ) \int \frac {1+\sqrt {2} x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {x}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {2 x \left (1+3 x^2\right )}{15 \sqrt {1+2 x^2+2 x^4}}-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{15 \sqrt {15}}-\frac {2^{3/4} \left (3+\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{105 \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{315 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}+\frac {1}{5} \sqrt {2} \int \frac {1-\sqrt {2} x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{3} \sqrt {2} \int \frac {1-\sqrt {2} x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{3} \left (1-\sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{15} \left (4+3 \sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {x}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {2 x \left (1+3 x^2\right )}{15 \sqrt {1+2 x^2+2 x^4}}-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {2 \sqrt {2} x \sqrt {1+2 x^2+2 x^4}}{15 \left (1+\sqrt {2} x^2\right )}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{15 \sqrt {15}}-\frac {2 \sqrt [4]{2} \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{15 \sqrt {1+2 x^2+2 x^4}}-\frac {\left (1-\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{6 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}-\frac {2^{3/4} \left (3+\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{105 \sqrt {1+2 x^2+2 x^4}}-\frac {\left (3+2 \sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{15\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{315 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 211, normalized size = 0.45 \[ \frac {-(27-39 i) \sqrt {1-i} x \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-12 i \sqrt {1-i} x \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} E\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-2 \left (12 x^4+39 x^2+2 (1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} x \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+15\right )}{90 x \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*I)*Sqrt[1 - I]*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticE[I*ArcSinh[Sqrt[1 - I]*x], I] - (2
7 - 39*I)*Sqrt[1 - I]*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] - 2
*(15 + 39*x^2 + 12*x^4 + 2*(1 - I)^(3/2)*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticPi[1/3 + I/3, I
*ArcSinh[Sqrt[1 - I]*x], I]))/(90*x*Sqrt[1 + 2*x^2 + 2*x^4])

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fricas [F]  time = 1.31, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1}}{8 \, x^{12} + 28 \, x^{10} + 40 \, x^{8} + 32 \, x^{6} + 14 \, x^{4} + 3 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2*x^4 + 2*x^2 + 1)/(8*x^12 + 28*x^10 + 40*x^8 + 32*x^6 + 14*x^4 + 3*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 553, normalized size = 1.18 \[ -\frac {x}{3 \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{5 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{5 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{15 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{5 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {4 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (\sqrt {-1+i}\, x , \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{45 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {2 x^{4}+2 x^{2}+1}}{3 x}+\frac {\left (-\frac {1}{3}+\frac {i}{3}\right ) \sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )+\EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\frac {2}{5} x^{3}+\frac {2}{15} x}{\sqrt {2 x^{4}+2 x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*x^4+2*x^2+1)^(1/2)/x-1/3/(2*x^4+2*x^2+1)^(1/2)*x-1/3/(-1+I)^(1/2)*((1-I)*x^2+1)^(1/2)*((1+I)*x^2+1)^(1
/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))+(-1/3+1/3*I)/(-1+I)^(1/2)*((1-I)
*x^2+1)^(1/2)*((1+I)*x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*(EllipticF((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))-E
llipticE((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2)))+8/3*(3/20*x^3+1/20*x)/(2*x^4+2*x^2+1)^(1/2)-1/15/(-1+I)^(1
/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*
2^(1/2))-1/5*I/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF((-1+I)^(1
/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))-1/5/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/
2)*EllipticE((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))+1/5*I/(-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(
1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticE((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))-4/45/(-1+I)^(1/2)*(-I*x^2+x^2+1
)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi((-1+I)^(1/2)*x,1/3+1/3*I,(-1-I)^(1/2)/(-1+I)^(1/2
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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