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

Optimal. Leaf size=360 \[ \frac {1}{9} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )+\frac {5 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{63 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{9 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {5 \left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{378 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {\sqrt {2 x^4+2 x^2+1}}{9 x^3} \]

[Out]

1/27*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/9*(2*x^4+2*x^2+1)^(1/2)/x^3-1/18*(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))*(1+x^2*2
^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2+1)^(1/2)+5/126*(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))*(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^(3/4)/(2*x^4+2*x^2+1)^(1/2)-5/756*(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))^2*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2+1)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1309, 1281, 12, 1103, 1216, 1706} \[ -\frac {\sqrt {2 x^4+2 x^2+1}}{9 x^3}+\frac {1}{9} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )+\frac {5 \left (3+\sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{63 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{9 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {5 \left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{378 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + 2*x^2 + 2*x^4]/(9*x^3) + (Sqrt[5/3]*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/9 - ((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])/(9*2^(1/4
)*Sqrt[1 + 2*x^2 + 2*x^4]) + (5*(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])/(63*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])/(378*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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 1309

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

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^4 \left (3+2 x^2\right )} \, dx &=\frac {1}{9} \int \frac {3+4 x^2}{x^4 \sqrt {1+2 x^2+2 x^4}} \, dx+\frac {10}{9} \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}}{9 x^3}-\frac {1}{27} \int \frac {6}{\sqrt {1+2 x^2+2 x^4}} \, dx+\frac {1}{63} \left (10 \left (3+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{63} \left (10 \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}}{9 x^3}+\frac {1}{9} \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 )}{63 \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 )}{378 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}-\frac {2}{9} \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {\sqrt {1+2 x^2+2 x^4}}{9 x^3}+\frac {1}{9} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )-\frac {\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 )}{9 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}+\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 )}{63 \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 )}{378 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 154, normalized size = 0.43 \[ -\frac {6 x^4+6 x^2+3 (1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} x^3 F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-5 (1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} x^3 \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )+3}{27 x^3 \sqrt {2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/27*(3 + 6*x^2 + 6*x^4 + 3*(1 - I)^(3/2)*x^3*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticF[I*ArcSinh
[Sqrt[1 - I]*x], I] - 5*(1 - I)^(3/2)*x^3*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*EllipticPi[1/3 + I/3, I*
ArcSinh[Sqrt[1 - I]*x], I])/(x^3*Sqrt[1 + 2*x^2 + 2*x^4])

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, x^{4} + 2 \, x^{2} + 1}}{2 \, x^{6} + 3 \, x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 448, normalized size = 1.24 \[ \frac {2 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {2 i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {4 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {2 i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {10 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (\sqrt {-1+i}\, x , \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{27 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {2 x^{4}+2 x^{2}+1}}{9 x^{3}}+\frac {\left (\frac {2}{9}-\frac {2 i}{9}\right ) \sqrt {\left (1-i\right ) x^{2}+1}\, \sqrt {\left (1+i\right ) x^{2}+1}\, \left (-\EllipticE \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )+\EllipticF \left (\sqrt {-1+i}\, x , \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/9-2/9*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))-EllipticE((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2)))-4/9/(-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)*EllipticF((-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))+2/9*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))+2/9/(-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)*EllipticE((
-1+I)^(1/2)*x,1/2*2^(1/2)+1/2*I*2^(1/2))-2/9*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))+10/27/(-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))-1/9*(2*x^4
+2*x^2+1)^(1/2)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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