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

Optimal. Leaf size=399 \[ \frac{\left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{21 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}+\frac{\sqrt{2} \sqrt{2 x^4+2 x^2+1} x}{3 \left (\sqrt{2} x^2+1\right )}-\frac{\sqrt{2 x^4+2 x^2+1}}{3 x}-\frac{1}{6} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )-\frac{\sqrt [4]{2} \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{2 x^4+2 x^2+1}}+\frac{5 \left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{252 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

[Out]

-Sqrt[1 + 2*x^2 + 2*x^4]/(3*x) + (Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(3*(1 + Sqrt[2]*x^2)) - (Sqrt[5/3]*ArcTan
[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/6 - (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])/(3*Sqrt[1 + 2*x^2 + 2*x^4]) + ((3 + Sqrt[2])*(1 + Sqr
t[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])/(21*2
^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + (5*(3 + Sqrt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x
^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(252*2^(1/4)*Sqrt[1 + 2*x^2 + 2
*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.242102, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1311, 1281, 1197, 1103, 1195, 1216, 1706} \[ \frac{\sqrt{2} \sqrt{2 x^4+2 x^2+1} x}{3 \left (\sqrt{2} x^2+1\right )}-\frac{\sqrt{2 x^4+2 x^2+1}}{3 x}-\frac{1}{6} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )+\frac{\left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{21 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}}-\frac{\sqrt [4]{2} \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{3 \sqrt{2 x^4+2 x^2+1}}+\frac{5 \left (3+\sqrt{2}\right )^2 \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{252 \sqrt [4]{2} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + 2*x^2 + 2*x^4]/(3*x) + (Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(3*(1 + Sqrt[2]*x^2)) - (Sqrt[5/3]*ArcTan
[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/6 - (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])/(3*Sqrt[1 + 2*x^2 + 2*x^4]) + ((3 + Sqrt[2])*(1 + Sqr
t[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])/(21*2
^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + (5*(3 + Sqrt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x
^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(252*2^(1/4)*Sqrt[1 + 2*x^2 + 2
*x^4])

Rule 1311

Int[(((f_.)*(x_))^(m_)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[
1/(d*e), Int[(f*x)^m*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 - b*d*e + a*e^2)/(d*e*f
^2), Int[((f*x)^(m + 2)*(a + b*x^2 + c*x^4)^(p - 1))/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne
Q[b^2 - 4*a*c, 0] && GtQ[p, 0] && LtQ[m, 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 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 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 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{\sqrt{1+2 x^2+2 x^4}}{x^2 \left (3+2 x^2\right )} \, dx &=\frac{1}{6} \int \frac{2+6 x^2}{x^2 \sqrt{1+2 x^2+2 x^4}} \, dx-\frac{5}{3} \int \frac{1}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{1}{6} \int \frac{-6-4 x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx-\frac{1}{21} \left (5 \left (3+\sqrt{2}\right )\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx+\frac{1}{21} \left (5 \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{\sqrt{1+2 x^2+2 x^4}}{3 x}-\frac{1}{6} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )-\frac{5 \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 )}{42 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}+\frac{5 \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 )}{252 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}-\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 (-3-\sqrt{2}\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx\\ &=-\frac{\sqrt{1+2 x^2+2 x^4}}{3 x}+\frac{\sqrt{2} x \sqrt{1+2 x^2+2 x^4}}{3 \left (1+\sqrt{2} x^2\right )}-\frac{1}{6} \sqrt{\frac{5}{3}} \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )-\frac{\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 )}{3 \sqrt{1+2 x^2+2 x^4}}+\frac{\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 )}{21 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}+\frac{5 \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 )}{252 \sqrt [4]{2} \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.200288, size = 208, normalized size = 0.52 \[ \frac{(9-3 i) \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{1-i} x\right ),i\right )-12 x^4-12 x^2-6 i \sqrt{1-i} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x E\left (\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )-5 (1-i)^{3/2} \sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} x \Pi \left (\frac{1}{3}+\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )-6}{18 x \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6 - 12*x^2 - 12*x^4 - (6*I)*Sqrt[1 - I]*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticE[I*ArcSinh[Sq
rt[1 - I]*x], I] + (9 - 3*I)*Sqrt[1 - I]*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticF[I*ArcSinh[Sqr
t[1 - I]*x], I] - 5*(1 - I)^(3/2)*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticPi[1/3 + I/3, I*ArcSin
h[Sqrt[1 - I]*x], I])/(18*x*Sqrt[1 + 2*x^2 + 2*x^4])

________________________________________________________________________________________

Maple [C]  time = 0.013, size = 511, normalized size = 1.3 \begin{align*}{\frac{2\,{\it EllipticF} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{3\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{{\frac{i}{3}}{\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{3\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{{\frac{i}{3}}{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) }{\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{5}{9\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\it EllipticPi} \left ( x\sqrt{-1+i},{\frac{1}{3}}+{\frac{i}{3}},{\frac{\sqrt{-1-i}}{\sqrt{-1+i}}} \right ){\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{1}{3\,x}\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}+{\frac{2\,{\it EllipticF} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{3\,\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{ \left ({\frac{2}{3}}-{\frac{2\,i}{3}} \right ) \left ({\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) -{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) \right ) }{\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/(-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(x*(-1+I)^(1/2),1/2*2
^(1/2)+1/2*I*2^(1/2))-1/3*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)*Ellipt
icF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-1/3/(-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(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+1/3*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(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-5/9/(-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(x*(-1+I)^(1/2),1/3+1/3*I,(-1-I)^(1/2)
/(-1+I)^(1/2))-1/3*(2*x^4+2*x^2+1)^(1/2)/x+2/3/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x
^2+1)^(1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+(-2/3+2/3*I)/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*
(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*(EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-EllipticE(x*(-1
+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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